01 — Linear Algebra

Comprehensive study guide for Bocconi Math Module 2 (cod. 30063), General Exam. Source materials: Alice Sicconi's Proofs - Linear Algebra.pdf, Linear Algebra and Differential Calculus.pdf, lect{1,2,3,5}_lin_alg.pdf, TA41_lin_alg.pdf, General_24524_ENG_SOL.pdf, Final exercises solution.pdf.

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§1. Overview & Exam Relevance

Linear Algebra is the first block of the first partial and accounts for approximately 27% of the general exam (typically 1–2 MCQs worth 5 pts each plus one open-ended question worth up to 20 pts — about 30–40 pts of the 150-pt exam).

Topic scope. The exam tests:

• matrix arithmetic: products, determinants, minors, transposition, the trace;
• eigenvalues, eigenvectors, eigenspaces, characteristic polynomial;
• diagonalization of symmetric matrices via the Spectral Theorem;
• quadratic forms $f(\mathbf{x}) = \mathbf{x}^T A \mathbf{x}$ and the bijection with symmetric matrices;
• classification (sign) of quadratic forms using (i) eigenvalue signs or (ii) Sylvester–Jacobi leading principal minors.

Typical MCQ patterns (from May 2024 General exam).

• MCQ1 (Mode A) / MCQ6 (Mode B): "Which of the following is an eigenvector of $A$?" — plug each candidate into $A\mathbf{v}$ and test $A\mathbf{v} = \lambda \mathbf{v}$.
• Similar MCQs: "The eigenvalues of $A$ are …" with multiplicity traps — shortcut via $\sum \lambda_i = \operatorname{tr} A$ and $\prod \lambda_i = \det A$.

Typical open-ended pattern.

• Q7 (Mode A) / Q12 (Mode B), May 2024: state the correspondence theorem (Prop 1201), write the quadratic form attached to a symmetric matrix, prove it is positive semi-definite but not positive definite, find a basis of $\mathbb{R}^3$ made of eigenvectors.
• "Final exercises" Q1: quadratic form $f(x,y,z) = kx^2 + ky^2 - 4z^2 - 2xz$ — find $k$ so that the origin is a strong global maximum, determine eigenvalue signs, find the kernel for $k=0$.

Why this topic is high-leverage for the final partial.

• The Spectral Theorem, Sylvester–Jacobi, and quadratic form signing show up again in Differential Calculus (Hessian classification of stationary points — the Hessian is a symmetric matrix and its quadratic form is the second-order Taylor remainder). Mastering linear algebra pays double dividends.

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§2. Definitions

2.1 Vector and matrix

A vector $\mathbf{x} \in \mathbb{R}^n$ is a column of $n$ real numbers: $\mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix}.$

A matrix $A$ of order $m \times n$ is a rectangular array of real entries $a_{ij}$ with $1 \le i \le m$, $1 \le j \le n$. A square matrix of order $n$ has $m = n$.

Dot product of two vectors: $\mathbf{x} \cdot \mathbf{y} = \sum_{i=1}^n x_i y_i = \mathbf{x}^T \mathbf{y}$. Orthogonal vectors: $\mathbf{x} \cdot \mathbf{y} = 0$. Orthonormal vectors: orthogonal AND $\|\mathbf{x}\| = \|\mathbf{y}\| = 1$. Norm: $\|\mathbf{x}\| = \sqrt{\mathbf{x} \cdot \mathbf{x}}$.

2.2 Transpose, symmetric, orthogonal

Transpose $A^T$: swap rows and columns — $(A^T)_{ij} = a_{ji}$.

Symmetric matrix: $A^T = A$, equivalently $a_{ij} = a_{ji}$ for all $i,j$. Symmetric matrices are always square.

Orthogonal matrix $B$ (of order $n$): $B \cdot B^T = B^T \cdot B = I$. Equivalently $B^{-1} = B^T$. Key properties. An orthogonal matrix is invertible, its rows and columns each form an orthonormal set, and $\det B = \pm 1$. An orthogonal matrix is not necessarily symmetric.

2.3 Determinant and minors

The determinant $\det(A)$ is a scalar function on square matrices.

• $2 \times 2$: $\det \begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc$.
• $3 \times 3$: Laplace expansion along any row/column using the sign pattern $\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}$.
• Rules: $\det(A^T) = \det(A)$, $\det(A^{-1}) = 1/\det(A)$, $\det(AB) = \det(A) \det(B)$.
• Singular / non-singular: $A$ is singular iff $\det A = 0$ (rows/columns are linearly dependent); $A$ is non-singular iff $\det A \neq 0$.

Minor: determinant of a submatrix obtained by removing any number of rows and columns. Principal minor: minor obtained by removing the same rows as columns (e.g., remove row 2 and column 2). Leading (North-West) principal minor $\det A_k$: the $k \times k$ minor obtained by keeping the first $k$ rows AND first $k$ columns (i.e., removing the last $n-k$ rows and columns). For a $3 \times 3$ matrix there are three leading principal minors: $\det A_1 = a_{11}$, $\det A_2 = \det \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$, $\det A_3 = \det A$.

2.4 Rank and linear independence

A set of vectors $\{\mathbf{v}_1, \ldots, \mathbf{v}_k\}$ is linearly independent iff the only solution to $\alpha_1 \mathbf{v}_1 + \cdots + \alpha_k \mathbf{v}_k = \mathbf{0}$ is $\alpha_1 = \cdots = \alpha_k = 0$. Otherwise linearly dependent.

Rank of a matrix = number of linearly independent rows = number of linearly independent columns. For a square matrix of order $n$, $\operatorname{rank}(A) = n \iff \det A \neq 0$.

2.5 Basis, dimension, subspace

A vector subspace $V \subseteq \mathbb{R}^n$ is a non-empty subset closed under linear combinations: $\mathbf{x}, \mathbf{y} \in V$ and $\alpha, \beta \in \mathbb{R}$ imply $\alpha \mathbf{x} + \beta \mathbf{y} \in V$ (in particular $\mathbf{0} \in V$).

A spanning set of $V$ is a set of vectors whose linear combinations generate all of $V$. A basis is a spanning set that is also linearly independent. Dimension $\dim V$ is the number of vectors in any basis.

In $\mathbb{R}^2$ there are 3 kinds of subspaces: $\{\mathbf{0}\}$ (dim 0), lines through $\mathbf{0}$ (dim 1), and all of $\mathbb{R}^2$ (dim 2). In $\mathbb{R}^3$: $\{\mathbf{0}\}$, lines through $\mathbf{0}$, planes through $\mathbf{0}$, and all of $\mathbb{R}^3$.

2.6 Eigenvalue, eigenvector, eigenpair

Let $A$ be a symmetric matrix of order $n$. A scalar $\lambda \in \mathbb{R}$ is an eigenvalue of $A$ and a non-zero vector $\mathbf{v} \in \mathbb{R}^n$ is an associated eigenvector if $\boxed{A \mathbf{v} = \lambda \mathbf{v}, \qquad \mathbf{v} \neq \mathbf{0}.}$

The pair $(\lambda, \mathbf{v})$ is an eigenpair. The spectrum of $A$ is $\sigma(A) = \{\lambda \in \mathbb{R} : \det(A - \lambda I) = 0\}$.

Remark. The restriction to symmetric matrices is Sicconi's convention and matches the Bocconi syllabus — all eigenvalue theorems in §3 assume $A$ symmetric, which guarantees every root of the characteristic polynomial is real.

2.7 Eigenspace

The eigenspace associated with eigenvalue $\lambda$ is $W_\lambda = \{ \mathbf{x} \in \mathbb{R}^n : (A - \lambda I)\mathbf{x} = \mathbf{0} \}.$ It contains every eigenvector associated with $\lambda$, plus the zero vector. By Theorem 1181 (§3.2), $W_\lambda$ is a vector subspace of $\mathbb{R}^n$.

Dimension / multiplicity: $\dim W_\lambda$ = algebraic multiplicity of $\lambda$ as a root of $p_A(\lambda)$ = $n - \operatorname{rank}(A - \lambda I)$. For a symmetric matrix the algebraic and geometric multiplicities coincide.

2.8 Characteristic polynomial

The characteristic polynomial is $p_A(\lambda) = \det(A - \lambda I)$, a polynomial of degree $n$. Its roots are exactly the eigenvalues of $A$ (Theorem 1179).

Shortcut for $2\times 2$: $p_A(\lambda) = \lambda^2 - (\operatorname{tr} A)\lambda + \det A$, so $\lambda_1 + \lambda_2 = \operatorname{tr} A$ and $\lambda_1 \lambda_2 = \det A$. For general $n$: $\sum_i \lambda_i = \operatorname{tr} A$, $\prod_i \lambda_i = \det A$ (each counted with multiplicity).

2.9 Eigenbasis

An eigenbasis of $\mathbb{R}^n$ is a basis made of orthonormal eigenvectors of a symmetric matrix $A$. Any pair of eigenvectors from different eigenspaces is automatically orthogonal (Theorem 1184); eigenvectors from the same eigenspace may need Gram–Schmidt to be orthogonalised and then normalised.

2.10 Diagonalization

A matrix $A$ is orthogonally diagonalizable if there exists an orthogonal matrix $B$ (i.e., $B^T = B^{-1}$) such that $B^T A B = \Lambda,$ where $\Lambda = \operatorname{diag}(\lambda_1, \ldots, \lambda_n)$ is the diagonal matrix of eigenvalues. Equivalently, $A = B \Lambda B^T$ (the spectral decomposition). The Spectral Theorem (§3.7) says every symmetric real matrix is orthogonally diagonalizable.

2.11 Quadratic form

A form $f : \mathbb{R}^n \to \mathbb{R}$ is a sum of monomials of the same degree. A quadratic form is a form of degree 2: a sum of monomials each of degree 2, e.g., $f(x,y,z) = 3x^2 - 5xy + 7yz + z^2$.

Key identity (Prop 1201, §3.5): every quadratic form can be written as $f(\mathbf{x}) = \mathbf{x}^T A \mathbf{x}$ for a unique symmetric $A$. The entries are read off as:

• $a_{ii}$ = coefficient of $x_i^2$;
• $a_{ij} = a_{ji}$ = $\tfrac12 \cdot$ (coefficient of $x_i x_j$).

Note. $f(\mathbf{0}) = 0$ always.

2.12 Positive/negative (semi)definite — indefinite

Let $A$ be a symmetric matrix of order $n$ and $f(\mathbf{x}) = \mathbf{x}^T A \mathbf{x}$. Then $A$ (equivalently, $f$) is:

| Class | Condition on $f$ | Geometric picture at $\mathbf{0}$ | |-------|--------------------|--------------------------| | positive definite | $f(\mathbf{x}) > 0\;\forall \mathbf{x}\neq\mathbf{0}$ | strict (unique) global minimum | | positive semi-definite (not def.) | $f(\mathbf{x}) \ge 0\;\forall\mathbf{x}$, $\exists \mathbf{x}\neq\mathbf{0}$ with $f(\mathbf{x})=0$ | weak global minimum (infinitely many minimizers along a line) | | negative definite | $f(\mathbf{x}) < 0\;\forall \mathbf{x}\neq\mathbf{0}$ | strict global maximum | | negative semi-definite (not def.) | $f(\mathbf{x}) \le 0\;\forall\mathbf{x}$, $\exists \mathbf{x}\neq\mathbf{0}$ with $f(\mathbf{x})=0$ | weak global maximum | | indefinite | takes both signs | saddle at $\mathbf{0}$ |

Pictures (Sicconi's lecture notes). Bowl $f=x^2+y^2$ = positive definite. Half-pipe $f=x^2$ = positive semi-definite (not definite). Pringle $f=x^2-y^2$ = indefinite.

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§3. Theorems, Propositions & Proofs

3.1 Theorem 1179 — Characterization of eigenvalues as roots of the characteristic equation

Statement. Let $A$ be a symmetric matrix of order $n$. A real scalar $\lambda \in \mathbb{R}$ is an eigenvalue of $A$ if and only if $\det(A - \lambda I) = 0$.

Source: Proofs - Linear Algebra.pdf p.3 (handwritten theorem 1179).

Proof.

($\Rightarrow$) $\lambda$ is an eigenvalue $\Longrightarrow \det(A - \lambda I) = 0$.

By definition there exists $\mathbf{x} \neq \mathbf{0} \in \mathbb{R}^n$ such that $A \mathbf{x} = \lambda \mathbf{x}.$ This is the definition of eigenvalue/eigenvector. Subtracting $\lambda \mathbf{x}$ from both sides: $A \mathbf{x} - \lambda \mathbf{x} = \mathbf{0}, \qquad A \mathbf{x} - \lambda I \mathbf{x} = \mathbf{0}, \qquad (A - \lambda I)\mathbf{x} = \mathbf{0}.$ Set $B := A - \lambda I$. Then $B \mathbf{x} = \mathbf{0}$ is a homogeneous linear system. Since $\mathbf{x} = \mathbf{0}$ is a trivial solution AND the eigenvector $\mathbf{x}\neq\mathbf{0}$ is also a solution, the homogeneous system has infinitely many solutions. This forces $\det B = 0$, i.e., $\det(A - \lambda I) = 0.$

($\Leftarrow$) $\det(A - \lambda I) = 0 \Longrightarrow \lambda$ is an eigenvalue.

The homogeneous system $(A - \lambda I)\mathbf{x} = \mathbf{0}$ with $\det(A - \lambda I) = 0$ has infinitely many solutions. Pick one non-trivial solution $\mathbf{x} \neq \mathbf{0} \in \mathbb{R}^n$ with $(A - \lambda I)\mathbf{x} = \mathbf{0}$. Then $A \mathbf{x} - \lambda I \mathbf{x} = \mathbf{0} \;\Longrightarrow\; A \mathbf{x} - \lambda \mathbf{x} = \mathbf{0} \;\Longrightarrow\; A \mathbf{x} = \lambda \mathbf{x}.$ This is the definition of an eigenpair. Hence $\lambda$ is an eigenvalue of $A$. $\blacksquare$

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3.2 Theorem 1181 — Eigenspace is a subspace of $\mathbb{R}^n$

Statement. Let $A$ be a symmetric matrix of order $n$ and $\lambda \in \mathbb{R}$ an eigenvalue of $A$. Then the eigenspace $W_\lambda = \{\mathbf{x} \in \mathbb{R}^n : (A - \lambda I)\mathbf{x} = \mathbf{0}\}$ is a vector subspace of $\mathbb{R}^n$.

Source: Proofs - Linear Algebra.pdf p.4 (handwritten theorem 1181).

Proof.

To show $W_\lambda$ is a subspace, we verify it is closed under linear combinations. Take $\mathbf{x}_1, \mathbf{x}_2 \in W_\lambda$ and scalars $\alpha, \beta \in \mathbb{R}$. We compute: $(A - \lambda I)(\alpha \mathbf{x}_1 + \beta \mathbf{x}_2) \;=\; (A - \lambda I)\alpha \mathbf{x}_1 + (A - \lambda I)\beta \mathbf{x}_2 \;=\; \alpha\underbrace{(A - \lambda I)\mathbf{x}_1}_{=\mathbf{0}} + \beta\underbrace{(A - \lambda I)\mathbf{x}_2}_{=\mathbf{0}} \;=\; \alpha \mathbf{0} + \beta \mathbf{0} \;=\; \mathbf{0}.$ Linear combination; then apply the definition of eigenpair $(A - \lambda I)\mathbf{x}_i = \mathbf{0}$ to each term. Therefore $\alpha \mathbf{x}_1 + \beta \mathbf{x}_2 \in W_\lambda$. In particular choosing $\alpha = \beta = 0$ shows $\mathbf{0} \in W_\lambda$. Hence $W_\lambda$ is a vector subspace of $\mathbb{R}^n$. $\blacksquare$

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3.3 Theorem 1184 — Eigenvectors of distinct eigenvalues are orthogonal

Statement. Let $A$ be a symmetric matrix of order $n$ and $(\lambda_1, \mathbf{x}_1)$, $(\lambda_2, \mathbf{x}_2)$ be eigenpairs with $\lambda_1 \neq \lambda_2$. Then $\mathbf{x}_1 \perp \mathbf{x}_2$, i.e., $\mathbf{x}_1 \cdot \mathbf{x}_2 = 0$.

Source: Proofs - Linear Algebra.pdf p.5 (handwritten theorem 1184).

Proof.

Start from the two eigen-equations $A \mathbf{x}_1 = \lambda_1 \mathbf{x}_1$ and $A \mathbf{x}_2 = \lambda_2 \mathbf{x}_2$. Taking the scalar $\mathbf{x}_1^T (A \mathbf{x}_2)$: \mathbf{x}_1^T (A \mathbf{x}_2) = \mathbf{x}_1^T (\lambda_2 \mathbf{x}_2) = \lambda_2 (\mathbf{x}_1 \cdot \mathbf{x}_2). \tag{I} And similarly: \mathbf{x}_2^T (A \mathbf{x}_1) = \mathbf{x}_2^T (\lambda_1 \mathbf{x}_1) = \lambda_1 (\mathbf{x}_1 \cdot \mathbf{x}_2). \tag{II}

Now the key manipulation. Since $\mathbf{x}_1^T (A \mathbf{x}_2)$ is a $1 \times 1$ scalar $\alpha$, we have $\alpha = \alpha^T$. Transposing: $\mathbf{x}_1^T (A \mathbf{x}_2) = \bigl(\mathbf{x}_1^T (A \mathbf{x}_2)\bigr)^T = (A \mathbf{x}_2)^T \mathbf{x}_1 = \mathbf{x}_2^T A^T \mathbf{x}_1.$ When transposing a product, the order switches: $(PQ)^T = Q^T P^T$. Because $A$ is symmetric, $A^T = A$, and so \mathbf{x}_1^T (A \mathbf{x}_2) = \mathbf{x}_2^T A \mathbf{x}_1. \tag{III}

Combining (I), (II), (III): $\lambda_2 (\mathbf{x}_1 \cdot \mathbf{x}_2) \;=\; \mathbf{x}_1^T A \mathbf{x}_2 \;=\; \mathbf{x}_2^T A \mathbf{x}_1 \;=\; \lambda_1 (\mathbf{x}_1 \cdot \mathbf{x}_2).$ Hence $(\lambda_2 - \lambda_1)(\mathbf{x}_1 \cdot \mathbf{x}_2) = 0$. Since $\lambda_1 \neq \lambda_2$, we conclude $\mathbf{x}_1 \cdot \mathbf{x}_2 = 0$, i.e., $\mathbf{x}_1 \perp \mathbf{x}_2$. $\blacksquare$

Why this matters. Theorem 1184 is the engine that lets us build orthonormal eigenbases: eigenvectors from distinct eigenspaces are free orthogonality. Only inside a single eigenspace do we sometimes have to run Gram–Schmidt.

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3.4 Theorem 1191 — Invertible symmetric matrices and reciprocal eigenvalues

Statement. Let $A$ be a symmetric matrix of order $n$. (1) $A$ is invertible if and only if all of its eigenvalues are non-zero. (2) If $A$ is invertible, then $\lambda \in \sigma(A) \iff \tfrac{1}{\lambda} \in \sigma(A^{-1})$.

Source: Proofs - Linear Algebra.pdf p.6 (handwritten theorem 1191).

Proof of (1). The product rule for eigenvalues gives $\lambda_1 \cdot \lambda_2 \cdots \lambda_n = \det A.$ Therefore $\lambda_1,\ldots,\lambda_n \neq 0 \iff \det A \neq 0 \iff A \text{ is invertible}. \quad\checkmark$

Proof of (2), direction $\lambda \in \sigma(A) \Rightarrow 1/\lambda \in \sigma(A^{-1})$.

$A$ is invertible so $\det A \neq 0$, hence $\lambda \neq 0$ by (1). Let $\mathbf{x} \neq \mathbf{0}$ be an eigenvector for $\lambda$: $A \mathbf{x} = \lambda \mathbf{x}$. Divide by $\lambda$: $\mathbf{x} = \tfrac{1}{\lambda} A \mathbf{x}.$ Multiply both sides on the left by $A^{-1}$: $A^{-1} \mathbf{x} = A^{-1}\bigl(\tfrac{1}{\lambda} A \mathbf{x}\bigr) = \tfrac{1}{\lambda}(A^{-1} A)\mathbf{x} = \tfrac{1}{\lambda} I \mathbf{x} = \tfrac{1}{\lambda} \mathbf{x}.$ Thus $A^{-1}\mathbf{x} = \tfrac{1}{\lambda}\mathbf{x}$, which says $\tfrac{1}{\lambda}$ is an eigenvalue of $A^{-1}$ with the same eigenvector $\mathbf{x}$.

Proof of (2), converse. If $\tfrac{1}{\lambda}$ is an eigenvalue of $A^{-1}$, apply the result just proved to the matrix $A^{-1}$ (which is symmetric when $A$ is symmetric): $\tfrac{1}{1/\lambda} = \lambda$ is an eigenvalue of $(A^{-1})^{-1} = A$. $\blacksquare$

Remark (lecture note). The eigenvectors stay the same when we pass from $A$ to $A^{-1}$; only the eigenvalues are reciprocated.

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3.5 Proposition 1201 — Bijective correspondence between quadratic forms and symmetric matrices

Statement. There is a bijective correspondence between quadratic forms $f : \mathbb{R}^n \to \mathbb{R}$ and symmetric matrices of order $n$, given by $f(\mathbf{x}) = \mathbf{x}^T A \mathbf{x} = \sum_{i=1}^n \sum_{j=1}^n a_{ij} x_i x_j \qquad \forall\,\mathbf{x}\in\mathbb{R}^n.$

Source: Proofs - Linear Algebra.pdf p.7 (handwritten prop 1201). Also: lect3 p.6 proposition 1295 in the textbook numbering.

Proof.

($\Rightarrow$) Symmetric $A$ yields a unique quadratic form $f$.

Expand $\mathbf{x}^T A \mathbf{x}$: $\mathbf{x}^T A \mathbf{x} = \sum_{i=1}^n a_{ii} x_i^2 + 2\sum_{1 \le i < j \le n} a_{ij} x_i x_j.$ The doubling $2 \cdot a_{ij} x_i x_j$ comes from combining the $(i,j)$ and $(j,i)$ entries, which are equal because $A$ is symmetric. This is a sum of monomials of degree 2, i.e., a quadratic form. Setting $f(\mathbf{x}) := \mathbf{x}^T A \mathbf{x}$ gives a well-defined quadratic form.

($\Leftarrow$) Quadratic form $f$ yields a unique symmetric $A$.

Given a quadratic form $f$, we want to write $f(\mathbf{x}) = \mathbf{x}^T A \mathbf{x}$ for some symmetric $A$. Reading coefficients:

• The coefficient of $x_i^2$ must equal $a_{ii}$, fixing the diagonal.
• The coefficient of the monomial $x_i x_j$ (for $i \neq j$) must equal $a_{ij} + a_{ji}$.

Requiring $A$ to be symmetric ($a_{ij} = a_{ji}$) forces $a_{ij} = a_{ji} = \tfrac{1}{2}\bigl(\text{coefficient of } x_i x_j\bigr)$. So the symmetric matrix $A$ is uniquely determined by $f$. $\blacksquare$

Practical rule. To build $A$ from $f$: put coefficients of squared terms on the diagonal; split each mixed term evenly across the two symmetric off-diagonal positions.

Example (Sicconi lect3 p.7). $A = \begin{pmatrix} 3 & 1 \\ 1 & -2 \end{pmatrix} \leadsto f(x_1, x_2) = 3x_1^2 - 2x_2^2 + 2 x_1 x_2$ (the off-diagonal $1 + 1$ becomes a $+2 x_1 x_2$ term).

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3.6 Theorem 1207 — Positive definite matrices are invertible (and inverse is PD)

Statement. If $A$ is positive definite, then (1) $A$ is invertible; (2) $A^{-1}$ is also positive definite.

Source: Proofs - Linear Algebra.pdf p.8 (handwritten theorem 1207).

Proof of (1). Suppose by contradiction $A$ is not invertible, so $\det A = 0$. Then the homogeneous system $A \mathbf{x} = \mathbf{0}$ admits a non-trivial solution $\mathbf{x} \neq \mathbf{0}$. But then $\mathbf{x}^T A \mathbf{x} = \mathbf{x}^T \mathbf{0} = 0$ with $\mathbf{x} \neq \mathbf{0}$, contradicting the definition of positive definite (which demands $\mathbf{x}^T A \mathbf{x} > 0$ for all non-zero $\mathbf{x}$). Absurd. Hence $\det A \neq 0$ and $A$ is invertible.

Proof of (2). Take any $\mathbf{x} \neq \mathbf{0}$ and set $\mathbf{y} = A^{-1}\mathbf{x}$ (equivalently $\mathbf{x} = A \mathbf{y}$). Since $\det A \neq 0$, $A^{-1}$ is also invertible, so $\mathbf{y} \neq \mathbf{0}$. We now compute $\mathbf{x}^T A^{-1} \mathbf{x}$: $\mathbf{x}^T A^{-1} \mathbf{x} = \mathbf{x} \cdot \mathbf{y} = \sum_{i=1}^n x_i y_i = \sum_{i=1}^n \Bigl(\sum_{j=1}^n a_{ij} y_j\Bigr) y_i = \sum_{i=1}^n \sum_{j=1}^n a_{ij} y_i y_j = \mathbf{y}^T A \mathbf{y}.$ The substitution $x_i = \sum_j a_{ij} y_j$ uses $\mathbf{x} = A\mathbf{y}$. Since $A$ is positive definite and $\mathbf{y} \neq \mathbf{0}$: $\mathbf{x}^T A^{-1} \mathbf{x} = \mathbf{y}^T A \mathbf{y} > 0.$ This holds for every $\mathbf{x} \neq \mathbf{0}$, so $A^{-1}$ is positive definite. $\blacksquare$

Semi-definite companion (Proof p.8, bottom). If $A$ is positive semi-definite, then $A$ is invertible iff $A$ is positive definite. (Proof via the equivalence $\mathbf{x}^T A \mathbf{x} = 0 \iff A \mathbf{x} = \mathbf{0}$ which holds for semi-definite matrices.)

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3.7 Spectral Theorem (Theorem 1291) — Orthogonal diagonalization of symmetric matrices

Statement. A symmetric matrix $A$ of order $n$ is orthogonally diagonalizable. That is, there exists an orthogonal matrix $B$ (so $B^T = B^{-1}$) such that $B^T A B = \Lambda,$ where $\Lambda = \operatorname{diag}(\lambda_1, \ldots, \lambda_n)$ is the diagonal matrix of eigenvalues, each repeated according to its multiplicity. Equivalently, $A = B \Lambda B^T$ (spectral decomposition).

Source: lect3_linalg.pdf p.1 — Sicconi reproduces textbook Theorem 1291. (Proof is course-level, constructed by assembling an orthonormal eigenbasis as the columns of $B$ — see algorithm M-LA-6 in §5.)

Idea of construction.

1. Find all eigenvalues $\lambda_1, \ldots, \lambda_n$ of $A$ (Theorem 1179).
2. For each distinct eigenvalue, find a basis of its eigenspace (Theorem 1181); eigenvectors from different eigenspaces are automatically orthogonal (Theorem 1184).
3. Inside each eigenspace, apply Gram–Schmidt if necessary to turn the basis orthogonal, then normalise every vector to length 1.
4. Place the resulting $n$ orthonormal eigenvectors as the columns of $B$. Then $B$ is orthogonal and $B^T A B = \Lambda$ with the $\lambda_i$ on the diagonal (ordered to match the columns).

Useful consequence. $\det A = \lambda_1 \cdots \lambda_n$ (since $|A| = |B|^2 |\Lambda| = |\Lambda|$, noting $|B| = \pm 1$ for orthogonal $B$).

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3.8 Sign of eigenvalues ↔ definiteness (Theorem 1210)

Statement. Let $A$ be a symmetric matrix of order $n$. (i) $A$ is positive definite $\iff$ all eigenvalues are $> 0$. (ii) $A$ is positive semi-definite $\iff$ all eigenvalues are $\ge 0$. (iii) $A$ is negative definite $\iff$ all eigenvalues are $< 0$. (iv) $A$ is negative semi-definite $\iff$ all eigenvalues are $\le 0$. (v) $A$ is indefinite $\iff$ $A$ has at least one positive and one negative eigenvalue.

Source: Proofs - Linear Algebra.pdf pp.9–10 (handwritten theorem 1210).

Proof of (i), $\Rightarrow$: positive definite implies eigenvalues $> 0$.

Let $\lambda$ be an eigenvalue with a normalised eigenvector $\mathbf{x}$ (so $\mathbf{x} \cdot \mathbf{x} = 1$ and $\mathbf{x}\neq\mathbf{0}$). Then $\lambda = \lambda \cdot (\mathbf{x} \cdot \mathbf{x}) = \mathbf{x}^T (\lambda \mathbf{x}) = \mathbf{x}^T A \mathbf{x}.$ Because $A$ is positive definite and $\mathbf{x} \neq \mathbf{0}$, $\mathbf{x}^T A \mathbf{x} > 0$, hence $\lambda > 0$.

Proof of (i), $\Leftarrow$: eigenvalues $> 0$ implies $A$ positive definite.

Assume $\lambda_1, \ldots, \lambda_n > 0$. By the Spectral Theorem there exists an orthogonal $B$ with $B^T A B = \Lambda$. Take any $\mathbf{x} \neq \mathbf{0}$ and set $\mathbf{y} = B^T \mathbf{x}$. Since $B$ (and hence $B^T$) is orthogonal, $B^T$ is invertible and $\det B^T \neq 0$, so $\mathbf{y} \neq \mathbf{0}$. Note that $B \mathbf{y} = B B^T \mathbf{x} = I \mathbf{x} = \mathbf{x}$. Substituting: $\mathbf{x}^T A \mathbf{x} = (B\mathbf{y})^T A (B\mathbf{y}) = \mathbf{y}^T (B^T A B) \mathbf{y} = \mathbf{y}^T \Lambda \mathbf{y} = \sum_{i=1}^n \lambda_i y_i^2.$ At least one $y_i \neq 0$ (since $\mathbf{y}\neq\mathbf{0}$), and every $\lambda_i > 0$, so $\sum \lambda_i y_i^2 > 0$. Hence $\mathbf{x}^T A \mathbf{x} > 0$ for every $\mathbf{x}\neq\mathbf{0}$, which says $A$ is positive definite.

For semi-definiteness, replace $>$ by $\ge$ throughout (the proof only uses the sign of the $\lambda_i$). Negative (semi)definite is obtained by applying the result to $-A$: $A$ negative definite $\iff$ $-A$ positive definite $\iff$ all eigenvalues of $-A$ are $>0$ $\iff$ all eigenvalues of $A$ are $<0$. $\blacksquare$

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3.9 Sylvester–Jacobi's criterion (Theorem 1217)

Statement. Let $A$ be a symmetric matrix of order $n$ and denote its leading (North-West) principal minors $\det A_1, \det A_2, \ldots, \det A_n$. Then: (1) $A$ is positive definite $\iff$ all leading principal minors are strictly positive: $\det A_1 > 0,\ \det A_2 > 0,\ \ldots,\ \det A_n > 0.$ (2) $A$ is negative definite $\iff$ the leading principal minors are all non-zero and alternate in sign starting with a strictly negative sign: $\det A_1 < 0,\ \det A_2 > 0,\ \det A_3 < 0, \ldots \qquad \det A_n \begin{cases} > 0 & n\text{ even}\\ < 0 & n\text{ odd}\end{cases}$ (3) If the leading principal minors are all non-zero but the sign sequence is neither (1) nor (2), then $A$ is indefinite.

Source: Proofs - Linear Algebra.pdf pp.13–14 (handwritten theorem 1217, with its preliminary result 1216 on general principal minors).

Proof of (1), $\Rightarrow$: positive definite implies all leading principal minors $> 0$.

By Theorem 1216 (preliminary), if $A$ is positive definite then every principal minor of $A$ is strictly positive: for any principal submatrix $A_{ii}$ and any $\mathbf{x} \in \mathbb{R}^{n-1}$ extended to $\tilde{\mathbf{x}} \in \mathbb{R}^n$ by inserting a $0$ in the $i$-th position, $\mathbf{x}^T A_{ii} \mathbf{x} = \tilde{\mathbf{x}}^T A \tilde{\mathbf{x}} > 0 \quad \forall \mathbf{x}\neq\mathbf{0},$ so $A_{ii}$ is positive definite and (by Theorem 1207) $\det A_{ii} > 0$. In particular every leading principal minor is strictly positive.

Proof of (1), $\Leftarrow$: all leading principal minors $> 0$ implies positive definite.

Take any $\mathbf{x} \neq \mathbf{0} \in \mathbb{R}^n$. Let $C$ be the upper-triangular matrix produced by Briosch's theorem (a course result we import), which is invertible since $\det C \neq 0$. Set $\mathbf{z} = C \mathbf{x} \neq \mathbf{0}$. Then $\mathbf{x}^T A \mathbf{x} = \sum_{k=1}^n \frac{\det A_k}{\det A_{k-1}}\, z_k^2,$ with the convention $\det A_0 = 1$. By hypothesis every ratio $\det A_k / \det A_{k-1}$ is strictly positive, and at least one $z_k^2 > 0$, so $\mathbf{x}^T A \mathbf{x} > 0$. Hence $A$ is positive definite.

Proof of (2), $\Rightarrow$. Suppose $A$ is negative definite and set $B := -A$. Then $B$ is positive definite, so $\det B_k > 0$ for every $k$. But $\det A_k = \det(-B_k) = (-1)^k \det B_k,$ which gives $\det A_k < 0$ for $k$ odd and $\det A_k > 0$ for $k$ even — exactly the alternating pattern (starting negative).

Proof of (2), $\Leftarrow$. Reverse the argument: given $A$ with the alternating sign pattern, set $B := -A$; check that every $\det B_k = (-1)^k \det A_k > 0$; conclude $B$ is positive definite by part (1); hence $A = -B$ is negative definite.

(3) Residual case. If the signs match neither (1) nor (2), then $A$ cannot be positive/negative (semi)definite (those imply the specific patterns), so $A$ must be indefinite. $\blacksquare$

Caveat (boxed by Sicconi). If some $\det A_k = 0$, the leading-minors test is inconclusive — $A$ may be positive semi-definite, negative semi-definite, or indefinite. Then you must check all principal minors (not just the leading ones) or compute eigenvalues directly.

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3.10 Auxiliary results worth remembering

Result on signs of diagonal entries (Theorem 1213). If $A$ is positive (semi)definite then $a_{ii} > 0$ (resp. $\ge 0$) for every $i$. Proof: evaluate $f$ on the coordinate vector $\mathbf{e}_i$: $f(\mathbf{e}_i) = \mathbf{e}_i^T A \mathbf{e}_i = a_{ii}$.

Product rule $\det A = \prod \lambda_i$. Follows from the Spectral Theorem: $\det A = \det(B\Lambda B^T) = |B|^2 \cdot \det \Lambda = 1 \cdot \lambda_1 \cdots \lambda_n$.

Trace rule $\operatorname{tr} A = \sum \lambda_i$. Trace is invariant under $A \mapsto B^T A B$ for orthogonal $B$, and $\operatorname{tr} \Lambda = \sum \lambda_i$.

Invertibility of symmetric $A$. If $A$ is symmetric and invertible, then $A^{-1}$ is symmetric too: $(A^{-1})^T = (A^T)^{-1} = A^{-1}$.

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§4. Worked Examples

Example 4.1 — Eigenvalues of a $2\times2$ matrix via the shortcut

Let $A = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}$. (Source: Linear Algebra and Differential Calculus.pdf p.10.)

Using trace/determinant.

• $\operatorname{tr} A = 1 + 4 = 5 \Rightarrow \lambda_1 + \lambda_2 = 5$.
• $\det A = 1 \cdot 4 - 2\cdot 2 = 0 \Rightarrow \lambda_1 \lambda_2 = 0$.

So one eigenvalue is $0$ and the other is $5$: $\sigma(A) = \{0, 5\}$.

Cross-check with characteristic polynomial. $p_A(\lambda) = (1-\lambda)(4-\lambda) - 4 = \lambda^2 - 5\lambda = \lambda(\lambda - 5)$, confirming $\lambda \in \{0, 5\}$.

Example 4.2 — Eigenspaces of the same matrix

$\lambda = 0$. Solve $(A - 0\cdot I)\mathbf{x} = \mathbf{0}$: $\begin{cases} x + 2y = 0 \\ 2x + 4y = 0 \end{cases} \Longrightarrow x = -2y.$ The second equation is dependent. Eigenspace $W_0 = \Bigl\{\mathbf{x} \in \mathbb{R}^2 : \mathbf{x} = \alpha \begin{pmatrix} -2 \\ 1 \end{pmatrix},\ \alpha \in \mathbb{R}\Bigr\}.$

$\lambda = 5$. Solve $(A - 5I)\mathbf{x} = \mathbf{0}$: $-4x + 2y = 0 \Rightarrow y = 2x$. Eigenspace $W_5 = \Bigl\{\mathbf{x} \in \mathbb{R}^2 : \mathbf{x} = \alpha \begin{pmatrix} 1 \\ 2 \end{pmatrix},\ \alpha \in \mathbb{R}\Bigr\}.$

Eigenbasis. Check orthogonality: $(-2,1)\cdot(1,2) = -2+2 = 0$ — orthogonal (as guaranteed by Theorem 1184 since $0\neq 5$). Normalise (each vector has norm $\sqrt{5}$): $\mathbf{u}_1 = \begin{pmatrix} -2/\sqrt{5} \\ 1/\sqrt{5} \end{pmatrix}, \qquad \mathbf{u}_2 = \begin{pmatrix} 1/\sqrt{5} \\ 2/\sqrt{5} \end{pmatrix}.$ These form an eigenbasis of $\mathbb{R}^2$. The orthogonal matrix $B = [\mathbf{u}_1\ \mathbf{u}_2]$ gives the spectral decomposition $B^T A B = \Lambda = \begin{pmatrix} 0 & 0 \\ 0 & 5 \end{pmatrix}.$

Example 4.3 — Eigenvalues of a 3×3 (textbook TA question)

$A = \begin{pmatrix} -3 & 0 & -4 \\ 0 & -5 & 0 \\ -4 & 0 & 3 \end{pmatrix}$. (Source: Linear Algebra and Differential Calculus.pdf p.12 — Question 4.)

Shortcut via trace + answer-choice elimination. $\operatorname{tr} A = -3 - 5 + 3 = -5$, so $\lambda_1 + \lambda_2 + \lambda_3 = -5$. Among the MCQ options, only "$-5$ with multiplicity 2 and $5$ with multiplicity 1" satisfies the sum: $-5 + (-5) + 5 = -5$.

Cross-check with determinant. Expanding along row 2 (only the central term survives), $\det A = -5 \cdot \det\begin{pmatrix}-3&-4\\-4&3\end{pmatrix} = -5(-9 - 16) = -5(-25) = 125$. And indeed $(-5)\cdot(-5)\cdot 5 = 125$. $\checkmark$

Example 4.4 — Writing a quadratic form from a matrix (and vice versa)

From matrix. Given $A = \begin{pmatrix} 3 & 1 \\ 1 & -2 \end{pmatrix}$. (Source: lect3 p.7.) Reading off: $a_{11} = 3$ (coefficient of $x_1^2$), $a_{22} = -2$ (coefficient of $x_2^2$), $a_{12}+a_{21} = 2$ (coefficient of $x_1 x_2$). Hence $f(x_1, x_2) = 3x_1^2 - 2x_2^2 + 2 x_1 x_2.$

From form. Given $f(x_1, x_2, x_3) = -x_1^2 + 2 x_1 x_3 + x_2^2 - 3 x_2 x_3$. (Source: lect3 p.8.)

• Diagonal: $a_{11} = -1$, $a_{22} = 1$, $a_{33} = 0$.
• Off-diagonal: coefficient of $x_1 x_2$ is $0 \Rightarrow a_{12} = 0$. Coefficient of $x_1 x_3$ is $2 \Rightarrow a_{13} = 1$. Coefficient of $x_2 x_3$ is $-3 \Rightarrow a_{23} = -3/2$. $A = \begin{pmatrix} -1 & 0 & 1 \\ 0 & 1 & -3/2 \\ 1 & -3/2 & 0 \end{pmatrix}.$

Example 4.5 — Classifying a quadratic form two ways

Given $f(x,y) = 3x^2 + 2xy + 5y^2$, i.e., $A = \begin{pmatrix} 3 & 1 \\ 1 & 5 \end{pmatrix}$. (Source: lect2 p.8.)

Via Sylvester–Jacobi.

• $\det A_1 = 3 > 0$
• $\det A_2 = \det A = 15 - 1 = 14 > 0$

Both leading principal minors positive $\Rightarrow$ positive definite $\Rightarrow$ $\mathbf{0}$ is a strict (unique) global minimum of $f$.

Via eigenvalue signs. $\operatorname{tr} A = 8$, $\det A = 14$. Both $>0$, so both eigenvalues are positive, giving the same conclusion.

Example 4.6 — The three $3\times 3$ classification cases (lecture demo)

(Source: Linear Algebra and Differential Calculus.pdf pp.18–19.)

(a) Indefinite. $f(x,y) = 2x^2 - 6xy + y^2$, $A = \begin{pmatrix}2 & -3 \\ -3 & 1\end{pmatrix}$. $\det A_1 = 2 > 0$, $\det A_2 = 2 - 9 = -7 < 0$. Sign pattern $(+, -)$ matches neither positive nor negative definite, so $A$ is indefinite and $\mathbf{0}$ is a saddle point.

(b) Indefinite with a zero leading minor. $f(x,y,z) = x^2 + 3xz - yz$, $A = \begin{pmatrix}1 & 0 & 3/2 \\ 0 & 0 & -1/2 \\ 3/2 & -1/2 & 0\end{pmatrix}$. $\det A_1 = 1$, $\det A_2 = 0$, $\det A_3 = -1/4$. Since $\det A_3 \neq 0$ but the Sylvester pattern fails (a zero appears in the leading sequence yet the final determinant is non-zero and negative), $A$ is indefinite — not definite or semi-definite.

(c) Positive semi-definite (not definite). $f(x,y,z) = x^2 + 2y^2 + 3z^2 + 2xz + 4yz$, $A = \begin{pmatrix}1 & 0 & 1 \\ 0 & 2 & 2 \\ 1 & 2 & 3\end{pmatrix}$. Leading: $\det A_1 = 1$, $\det A_2 = 2$, $\det A_3 = 0$. The zero means Sylvester–Jacobi is inconclusive. Check all principal minors: order-1 minors are $1, 2, 3$ (all $\ge 0$); order-2 principal minors are $\det\begin{pmatrix}2&2\\2&3\end{pmatrix}=2$, $\det\begin{pmatrix}1&1\\1&3\end{pmatrix}=2$, $\det\begin{pmatrix}1&0\\0&2\end{pmatrix}=2$ (all $\ge 0$); order-3 is $\det A = 0$. All principal minors $\ge 0 \Rightarrow$ positive semi-definite $\Rightarrow$ $\mathbf{0}$ is a non-strict (weak) minimizer.

Example 4.7 — Full diagonalization of a $3\times 3$ (lecture example)

$A = \begin{pmatrix}1 & -1 & 2 \\ -1 & 1 & -2 \\ 2 & -2 & 4\end{pmatrix}$. Find an eigenbasis of $\mathbb{R}^3$. (Source: Linear Algebra and Differential Calculus.pdf pp.16–17.)

Step 1. Characteristic polynomial. Expanding and simplifying (full calculation in the source): $p_A(\lambda) = -\lambda^3 + 6\lambda^2 = \lambda^2 (-\lambda + 6)$. Roots: $\lambda = 0$ (multiplicity 2), $\lambda = 6$ (multiplicity 1).

Step 2. Eigenspaces.

• $\lambda = 0$: solve $A \mathbf{x} = \mathbf{0}$. Reduces to $x - y + 2z = 0$, so $x = y - 2z$. Two-parameter family: $\begin{pmatrix}x\\y\\z\end{pmatrix} = y \begin{pmatrix}1\\1\\0\end{pmatrix} + z\begin{pmatrix}-2\\0\\1\end{pmatrix}.$ Basis of $W_0$: $\mathbf{x}_1 = (1,1,0)^T$, $\mathbf{x}_2 = (-2,0,1)^T$.
• $\lambda = 6$: solve $(A - 6I)\mathbf{x} = \mathbf{0}$. Gaussian elimination gives $x = -y$, $z = -2y$. Basis of $W_6$: $\mathbf{x}_3 = (-1, 1, -2)^T$.

Step 3. Eigenbasis. $\mathbf{x}_1$ and $\mathbf{x}_3$ sit in different eigenspaces so are orthogonal (and so are $\mathbf{x}_2, \mathbf{x}_3$). But $\mathbf{x}_1, \mathbf{x}_2$ are in the same eigenspace $W_0$ and happen NOT to be orthogonal: $\mathbf{x}_1 \cdot \mathbf{x}_2 = -2$. Apply Gram–Schmidt:

• Normalise $\mathbf{x}_1$: $\|\mathbf{x}_1\| = \sqrt{2}$, so $\mathbf{u}_1 = (1/\sqrt{2}, 1/\sqrt{2}, 0)^T$.
• Orthogonal auxiliary: $\mathbf{z}_2 = \mathbf{x}_2 - (\mathbf{x}_2 \cdot \mathbf{u}_1)\mathbf{u}_1 = (-2, 0, 1)^T - (-2/\sqrt{2})(1/\sqrt{2}, 1/\sqrt{2}, 0)^T = (-1, 1, 1)^T$. Then $\|\mathbf{z}_2\| = \sqrt{3}$, giving $\mathbf{u}_2 = (-1/\sqrt{3}, 1/\sqrt{3}, 1/\sqrt{3})^T$.
• Normalise $\mathbf{x}_3$: $\|\mathbf{x}_3\| = \sqrt{6}$, $\mathbf{u}_3 = (-1/\sqrt{6}, 1/\sqrt{6}, -2/\sqrt{6})^T$.

$\{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3\}$ is the eigenbasis of $\mathbb{R}^3$ for $A$, and $B = [\mathbf{u}_1\ \mathbf{u}_2\ \mathbf{u}_3]$ satisfies $B^T A B = \operatorname{diag}(0, 0, 6)$.

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§5. Solution Methods

Each method is a named algorithm with input → steps → output → common pitfalls. Cross-references in the right column point to the May 2024 exam problems where the method is applied.

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M-LA-1 — Verify a candidate vector is an eigenvector

Used on: May 2024 MCQ1 (Mode A), MCQ6 (Mode B).

Input. A symmetric matrix $A$ of order $n$ and a candidate vector $\mathbf{v} \in \mathbb{R}^n$.

Steps.

1. Compute the product $A \mathbf{v}$.
2. Check whether the resulting vector is a scalar multiple of $\mathbf{v}$: look for a single $\lambda \in \mathbb{R}$ so that $A \mathbf{v} = \lambda \mathbf{v}$.
3. If yes, $\mathbf{v}$ is an eigenvector with eigenvalue $\lambda$. If no, $\mathbf{v}$ is not an eigenvector.

Output. "Yes, eigenvector with eigenvalue $\lambda$" or "No".

Pitfalls.

• Forgetting that the zero vector is never an eigenvector (even though $A \mathbf{0} = \mathbf{0}$ trivially), by definition.
• On MCQs, always test the shortest vector first (fewer arithmetic operations). The May 2024 MCQ1 correct answer is $(0,1,0)^T$ because $A(0,1,0)^T = (0,-2,0)^T = -2 \cdot (0,1,0)^T$.

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M-LA-2 — Compute the eigenvalues of a $3\times 3$ (or larger) matrix

Used on: Final exercises Q1, May 2024 Q7/Q12 (eigenvalue determination for $A = \mathbf{1}\mathbf{1}^T$).

Input. Symmetric matrix $A$ of order $n$.

Steps.

1. Write down $A - \lambda I$ (subtract $\lambda$ from every diagonal entry).
2. Compute $p_A(\lambda) = \det(A - \lambda I)$ by Laplace expansion along the simplest row/column (prefer rows/columns with zeros).
3. Factor $p_A(\lambda)$ over $\mathbb{R}$. Useful tricks:
   • If a row/column of $A - \lambda I$ becomes identically zero at $\lambda = \lambda^\star$, then $\lambda^\star$ is a root and $p_A$ is divisible by $(\lambda - \lambda^\star)$.
   • Rational-root test: integer roots of $p_A$ divide the constant term $p_A(0) = \det A$ (up to sign).
   • Sum/product shortcut: $\sum \lambda_i = \operatorname{tr} A$ and $\prod \lambda_i = \det A$; use as a sanity check after solving.
4. Solve $p_A(\lambda) = 0$ and list the eigenvalues with multiplicities. The total of multiplicities must equal $n$.

Output. Spectrum $\sigma(A) = \{\lambda_1, \ldots, \lambda_k\}$ with algebraic multiplicities.

Pitfalls.

• On May 2024 Q7, $A = \begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}$ has $\operatorname{tr} A = 3$ and $\det A = 0$, and since $A$ is rank-1, $\lambda = 0$ has multiplicity $n-1 = 2$ and the remaining eigenvalue is $\operatorname{tr} A = 3$: $\sigma(A) = \{0\text{ (mult.\ 2)}, 3\text{ (mult.\ 1)}\}$. Always check your calculation against these totals.
• When the characteristic polynomial is a cubic with an obvious factor of $\lambda$ (happens whenever $\det A = 0$), factor it out first and solve the remaining quadratic.

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M-LA-3 — Find a basis of an eigenspace

Used on: May 2024 Q7 (Mode A) / Q12 (Mode B).

Input. Symmetric matrix $A$, a specific eigenvalue $\lambda$.

Steps.

1. Write the system $(A - \lambda I)\mathbf{x} = \mathbf{0}$.
2. Gaussian eliminate. Expect $\operatorname{rank}(A - \lambda I) = n - \mu$ where $\mu$ is the algebraic multiplicity of $\lambda$; you'll get $\mu$ free parameters.
3. Express the solution set as a span of $\mu$ independent vectors — those are the basis of $W_\lambda$.

Output. A basis $\{\mathbf{v}_1, \ldots, \mathbf{v}_\mu\}$ of $W_\lambda$.

Pitfalls.

• $\dim W_\lambda$ must equal the algebraic multiplicity of $\lambda$ (this is the symmetric-matrix-always-diagonalizable fact). If you find fewer independent vectors, you've made an arithmetic mistake.
• Don't forget that eigenspaces contain $\mathbf{0}$ even though $\mathbf{0}$ is not an eigenvector.

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M-LA-4 — Build a basis of eigenvectors for $\mathbb{R}^n$

Used on: May 2024 Q7(c) / Q12(c) — explicit request "determine a basis of $\mathbb{R}^3$ constituted by eigenvectors of $A$".

Input. Symmetric matrix $A$ of order $n$.

Steps.

1. Find all eigenvalues (M-LA-2).
2. For each distinct eigenvalue, find a basis of its eigenspace (M-LA-3).
3. Union of all eigenspace bases is a basis of $\mathbb{R}^n$ (since bases from distinct eigenspaces are automatically linearly independent by Theorem 1184, and the total count equals $n$ by the symmetric spectral structure).
4. If the question asks "not necessarily orthogonal or orthonormal" (as in May 2024 Q7(c)), you can stop here.
5. If the question asks for an eigenbasis (orthonormal), continue with M-LA-6.

Output. A list of $n$ linearly independent eigenvectors of $A$ spanning $\mathbb{R}^n$.

Pitfalls.

• Inside a single eigenspace of multiplicity $\ge 2$, the basis vectors are not automatically orthogonal. Theorem 1184 only applies to eigenvectors from distinct eigenspaces.

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M-LA-5 — Classify a quadratic form (eigenvalue method vs Sylvester–Jacobi)

Used on: May 2024 Q7/Q12, Final exercises Q1.

Input. A symmetric matrix $A$ (or a quadratic form from which you extract $A$ via M-LA-rules in §2.11).

Decision tree — which test to use.

• Prefer Sylvester–Jacobi when the matrix is small ($2\times2$ or $3\times3$) and the leading principal minors are easy to compute.
• Prefer the eigenvalue-sign test when (i) the matrix has an obvious rank-deficiency (so some eigenvalue is automatically $0$), (ii) you're told the eigenvalues or can get them cheaply, (iii) a leading principal minor is $0$ (Sylvester is inconclusive in that case), or (iv) you're also asked the sign of eigenvalues.

Steps (Sylvester–Jacobi branch).

1. Compute $\det A_1, \det A_2, \ldots, \det A_n$.
2. Match against the pattern:
   • All $>0$ $\Rightarrow$ positive definite.
   • Alternating starting with $<0$ ($-, +, -, +, \ldots$) $\Rightarrow$ negative definite.
   • All non-zero but neither pattern $\Rightarrow$ indefinite.
   • Some leading minor is $0$ and $\det A \neq 0$ $\Rightarrow$ indefinite (since neither positive nor negative definite, and not semi-definite since $\det A\neq 0$ for semi-definite-not-definite is impossible).
   • Some leading minor is $0$ and $\det A = 0$ $\Rightarrow$ inconclusive — check ALL principal minors.
3. If inconclusive, evaluate every principal minor (not just leading): all $\ge 0$ $\Rightarrow$ positive semi-definite; all of the signed form $(\le 0$ for odd order, $\ge 0$ for even order) $\Rightarrow$ negative semi-definite; otherwise indefinite.

Steps (Eigenvalue-sign branch).

1. Compute the eigenvalues (M-LA-2).
2. Apply Theorem 1210: all $>0$ → positive definite, all $\ge 0$ → positive semi-definite, mixed → indefinite, etc.

Output. One of: positive definite, positive semi-definite (non-def), negative definite, negative semi-definite (non-def), indefinite.

Pitfalls (critical — these are exam traps!).

• Negative definite needs ALTERNATING signs, not all negative. $(-, +, -, \ldots)$ — starting negative, alternating.
• Sylvester fails with leading zeros. If $\det A_k = 0$ for any $k$, do NOT conclude indefinite from that alone; it may be semi-definite. Switch to the full principal-minor test (all minors, not just leading).
• Semi-definite ≠ definite. May 2024 Q7 is the canonical trap: $A = \mathbf{1}\mathbf{1}^T$ has eigenvalues $0, 0, 3$, all $\ge 0$, so $A$ is positive semi-definite — but not positive definite because $\lambda = 0$ occurs. The quadratic form $f(\mathbf{x}) = (x_1+x_2+x_3)^2$ vanishes on the hyperplane $x_1+x_2+x_3 = 0$, confirming it is not strictly positive.

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M-LA-6 — Diagonalize a symmetric matrix (full spectral decomposition)

Used on: conceptual backbone of quadratic-form sign analysis; also direct exam topics when a question asks for $B^T A B = \Lambda$.

Input. Symmetric matrix $A$ of order $n$.

Steps.

1. Compute eigenvalues (M-LA-2).
2. For each eigenvalue, find a basis of the eigenspace (M-LA-3).
3. Orthogonalise within each eigenspace (Gram–Schmidt): keep the first basis vector, and for each subsequent one $\mathbf{v}_k$ subtract its projections onto the already-processed vectors.
4. Normalise every vector to unit length. The resulting $n$ vectors form an eigenbasis.
5. Assemble $B$ by stacking the eigenbasis as columns; $B$ is automatically orthogonal ($B^T = B^{-1}$).
6. The diagonal matrix $\Lambda = B^T A B$ has the eigenvalues on the diagonal, ordered to match the column order in $B$.

Output. Orthogonal $B$ and diagonal $\Lambda$ with $B^T A B = \Lambda$ (equivalently $A = B \Lambda B^T$).

Pitfalls.

• Don't forget to normalise after Gram–Schmidt. Orthogonalising preserves orthogonality but doesn't scale to unit length.
• If an eigenspace already has an orthonormal basis (e.g., the spectral theorem delivered pre-orthogonal basis vectors), skip Gram–Schmidt — it's idempotent but wasted work.
• Order matters: the $\Lambda$ you produce has the eigenvalues in the order the columns of $B$ are arranged. Swap two columns of $B$ ↔ swap two diagonal entries of $\Lambda$.

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§6. Practice Problems with Solutions

Problem 6.1 — TA session 1, exercise 1: positive-definite quadratic form parametric

Consider $f : \mathbb{R}^2 \to \mathbb{R}$ with $f(\mathbf{x}) = (\alpha^2 - 1)x_1^2 + (2 - \alpha)x_2^2$, $\alpha \in \mathbb{R}$. The quadratic form is positive definite in $\mathbb{R}^2$ if and only if … ?

(A) $|\alpha| > 1$, (B) $-1 < \alpha < 2$, (C) $\alpha < 2$, (D) none of the preceding.

Solution. Associated matrix $A = \begin{pmatrix} \alpha^2 - 1 & 0 \\ 0 & 2 - \alpha \end{pmatrix}$. Apply Sylvester–Jacobi:

• $\det A_1 = \alpha^2 - 1 > 0 \iff \alpha < -1 \text{ or } \alpha > 1$.
• $\det A_2 = \det A = (\alpha^2 - 1)(2 - \alpha) > 0$.

Combining: $\alpha^2 - 1 > 0$ AND $2 - \alpha > 0$ $\iff$ ($\alpha<-1$ or $\alpha>1$) AND $\alpha < 2$ $\iff$ $\alpha<-1$ OR $1 < \alpha < 2$.

So $A$ is positive definite iff $\alpha < -1$ or $1 < \alpha < 2$. Answer: (D) none of the preceding.

Source: TA41_lin_alg.pdf pp.1–2.

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Problem 6.2 — May 2024 MCQ1 (Mode A) / MCQ6 (Mode B): which vector is an eigenvector?

Which of the following vectors is an eigenvector of $A = \begin{pmatrix}2&0&1\\0&-2&-1\\1&0&3\end{pmatrix}$?

(A) $(2, 0, -1)^T$, (B) $(3, 1, 2)^T$, (C) $(0, 1, 0)^T$, (D) none of the others.

Solution. Apply M-LA-1. Testing $\mathbf{v} = (0,1,0)^T$: $A \mathbf{v} = \begin{pmatrix} 2\cdot 0 + 0\cdot 1 + 1\cdot 0 \\ 0\cdot 0 + (-2)\cdot 1 + (-1)\cdot 0 \\ 1\cdot 0 + 0\cdot 1 + 3\cdot 0\end{pmatrix} = \begin{pmatrix}0\\-2\\0\end{pmatrix} = -2 \begin{pmatrix}0\\1\\0\end{pmatrix} = -2 \mathbf{v}.$ So $(0,1,0)^T$ is an eigenvector with eigenvalue $-2$. Answer: (C).

Source: General_24524_ENG_SOL.pdf p.1 MCQ1.

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Problem 6.3 — May 2024 Q7 (Mode A) / Q12 (Mode B): Sicconi's flagship exam problem

(a) State the theorem on the correspondence between symmetric matrices $A$ of order $n$ and quadratic forms $q : \mathbb{R}^n \to \mathbb{R}$. (b) Consider the matrix $A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}.$ Write the quadratic form associated to $A$ and prove that this form is positive semi-definite but not positive definite. (c) Determine a basis of $\mathbb{R}^3$ constituted by eigenvectors of $A$ (not necessarily orthogonal or orthonormal).

Solution.

(a) Proposition 1201: there is a bijective correspondence between quadratic forms $q : \mathbb{R}^n \to \mathbb{R}$ and symmetric matrices $A$ of order $n$, given by $q(\mathbf{x}) = \mathbf{x}^T A \mathbf{x} = \sum_{i,j} a_{ij} x_i x_j$.

(b) Read off the quadratic form: $q(x_1, x_2, x_3) = x_1^2 + x_2^2 + x_3^2 + 2 x_1 x_2 + 2 x_1 x_3 + 2 x_2 x_3 = (x_1 + x_2 + x_3)^2.$ Eigenvalues via characteristic polynomial. Note $A$ is rank 1 (every row equals $(1,1,1)$), so $\lambda = 0$ has multiplicity 2. The remaining eigenvalue equals $\operatorname{tr} A = 3$. Formally: $\det(A - \lambda I) = (1-\lambda)^3 + 2 - 3(1-\lambda) = -3\lambda + 3\lambda^2 - \lambda^3 + 3\lambda = -\lambda^2(\lambda - 3),$ giving $\lambda_1 = 0$ (multiplicity 2), $\lambda_2 = 3$ (multiplicity 1). Since all eigenvalues are $\ge 0$, by Theorem 1210(ii) $A$ is positive semi-definite. Since $\lambda = 0$ occurs (not all strictly positive), $A$ is not positive definite. (Concretely, $q(1, -1, 0) = 0$ with $(1,-1,0)^T \neq \mathbf{0}$.)

(c) Bases of the eigenspaces.

• $W_0$: solve $A\mathbf{x} = \mathbf{0}$, which reduces to $x_1 + x_2 + x_3 = 0$, equivalently $x_3 = -x_1 - x_2$. A basis is $\{(1, 0, -1)^T, (0, 1, -1)^T\}$.
• $W_3$: solve $(A - 3I)\mathbf{x} = \mathbf{0}$: $\begin{cases} -2x_1 + x_2 + x_3 = 0 \\ x_1 - 2x_2 + x_3 = 0 \\ x_1 + x_2 - 2x_3 = 0 \end{cases}$ From the first: $x_2 = 2x_1 - x_3$. Substitute into the second: $x_1 - 2(2x_1 - x_3) + x_3 = 0 \Rightarrow x_3 = x_1$. Then $x_2 = x_1$. Basis: $(1, 1, 1)^T$.

Eigenbasis of $\mathbb{R}^3$ (union by M-LA-4): $\{(1, 0, -1)^T,\ (0, 1, -1)^T,\ (1, 1, 1)^T\}$.

Source: General_24524_ENG_SOL.pdf p.5 Q7 (Mode A) / p.11 Q12 (Mode B).

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Problem 6.4 — Final exercises Q1: quadratic form with parameter $k$

Given $f(x, y, z) = k x^2 + k y^2 - 4 z^2 - 2 x z$, $k \in \mathbb{R}$. (a) Find conditions on $k$ such that the origin is a strong global maximum point. (b) Determine the sign of the eigenvalues for $k = -1$. (c) Set $k = 0$, verify that $\lambda = 0$ is an eigenvalue and find the kernel of the linear function associated to the matrix $A$.

Solution.

(a) The origin is always a stationary point of a quadratic form ($f(\mathbf{0}) = 0$; $\nabla f(\mathbf{0}) = \mathbf{0}$). For it to be a strong global maximum, we need $f(\mathbf{x}) = \mathbf{x}^T A \mathbf{x} < 0$ for every $\mathbf{x} \neq \mathbf{0}$, i.e., $A$ must be negative definite. The associated matrix (half the mixed coefficient on the off-diagonal): $A = \begin{pmatrix} k & 0 & -1 \\ 0 & k & 0 \\ -1 & 0 & -4 \end{pmatrix}.$ Apply Sylvester–Jacobi (negative-definite pattern: $\det A_1 < 0$, $\det A_2 > 0$, $\det A_3 < 0$):

• $\det A_1 = k < 0$.
• $\det A_2 = k \cdot k - 0 = k^2 > 0$ — requires $k \neq 0$.
• $\det A_3 = \det A$. Expand along row 2: $\det A = k \cdot \det\begin{pmatrix}k & -1 \\ -1 & -4\end{pmatrix} = k(-4k - 1) = -4k^2 - k$. Need $-4k^2 - k < 0$, i.e., $k(4k+1) > 0$ $\iff$ $k > 0$ or $k < -1/4$.

Combining all three: $k < 0$ AND $k \neq 0$ AND ($k > 0$ or $k < -1/4$) $\Longrightarrow k < -\tfrac{1}{4}$.

Answer. $A$ is negative definite $\iff k < -\tfrac{1}{4}$, in which case the origin is a strong global max.

(b) For $k = -1$, the three conditions above become $-1 < 0\ \checkmark$, $1 > 0\ \checkmark$, $-4 \cdot 1 - (-1) = -3 < 0\ \checkmark$. $A$ is negative definite. By Theorem 1210(iii), all three eigenvalues are $< 0$: $\lambda_1, \lambda_2, \lambda_3 < 0$.

(c) For $k = 0$, $A = \begin{pmatrix} 0 & 0 & -1 \\ 0 & 0 & 0 \\ -1 & 0 & -4 \end{pmatrix}$. Characteristic polynomial: expanding along row 2, $p_A(\lambda) = -\lambda \cdot \det\begin{pmatrix}-\lambda & -1 \\ -1 & -4-\lambda\end{pmatrix} = -\lambda\bigl(\lambda(4+\lambda) - 1\bigr) = -\lambda(\lambda^2 + 4\lambda - 1)$. Clearly $\lambda = 0$ is a root (factor of $\lambda$), so $\lambda = 0$ is an eigenvalue of $A$.

Kernel of $A$ = $W_0$: solve $A\mathbf{x} = \mathbf{0}$: $\begin{cases} -x_3 = 0 \\ 0 = 0 \\ -x_1 - 4 x_3 = 0 \end{cases} \Longrightarrow x_3 = 0,\ x_1 = 0,\ x_2 \text{ free}.$ Hence $W_0 = \Bigl\{\mathbf{x} \in \mathbb{R}^3 : \mathbf{x} = x_2 \begin{pmatrix}0\\1\\0\end{pmatrix},\ x_2 \in \mathbb{R}\Bigr\}.$

Source: Final exercises solution.pdf pp.1–3 Problem 1.

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Problem 6.5 — TA session 1 exercise 2: characteristic polynomial & eigenbasis of a $2\times 2$

Given $A = \begin{pmatrix} 1 & -2 \\ -2 & 3 \end{pmatrix}$, compute the characteristic polynomial, find the eigenvalues and eigenvectors, determine a basis of $\mathbb{R}^2$, and decide whether it is an eigenbasis.

Solution.

Characteristic polynomial. $p_A(\lambda) = \lambda^2 - (\operatorname{tr} A)\lambda + \det A = \lambda^2 - 4\lambda + (3 - 4) = \lambda^2 - 4\lambda - 1$.

Eigenvalues. Solving $\lambda^2 - 4\lambda - 1 = 0$ by the quadratic formula: $\lambda = \tfrac{4 \pm \sqrt{16+4}}{2} = 2 \pm \sqrt{5}$. So $\sigma(A) = \{2 - \sqrt{5},\ 2 + \sqrt{5}\}$.

Eigenvectors. For $\lambda_1 = 2 - \sqrt{5}$: $(A - \lambda_1 I)\mathbf{x} = \mathbf{0}$ gives $(-1 + \sqrt{5})x_1 - 2 x_2 = 0$, i.e., $x_1 = \tfrac{2}{\sqrt{5}-1} x_2$. Eigenvector $\mathbf{x} = (2/(\sqrt{5}-1),\ 1)^T$.

For $\lambda_2 = 2 + \sqrt{5}$: similarly $(-1 - \sqrt{5})x_1 - 2 x_2 = 0 \Rightarrow x_1 = -\tfrac{2}{1+\sqrt{5}} x_2$. Eigenvector $\mathbf{x} = (-2/(1+\sqrt{5}),\ 1)^T$.

Basis of $\mathbb{R}^2$. Since $\lambda_1 \neq \lambda_2$, the two eigenvectors are linearly independent (and in fact orthogonal by Theorem 1184). They therefore form a basis of $\mathbb{R}^2$.

Is it an eigenbasis? An eigenbasis requires the vectors to be orthonormal. Check norms — clearly neither has norm 1. So this basis is not an eigenbasis, but an orthonormal eigenbasis is obtained by dividing each vector by its norm.

Source: TA41_lin_alg.pdf pp.3–7 exercise 2.

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Problem 6.6 — TA session 1 exercise 7: write the quadratic form from a matrix

Given $A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 0 \\ 1 & 0 & 3 \end{pmatrix}$, provide the analytical expression of the quadratic form associated to $A$.

Solution. $A$ is symmetric. Read coefficients: diagonal gives $x_1^2 + 2 x_2^2 + 3 x_3^2$; each off-diagonal contributes $2 \cdot 1 \cdot x_i x_j$ because $a_{ij} + a_{ji} = 2$ for every mixed pair $(i,j) = (1,2), (1,3)$, and $2 \cdot 0$ for $(2,3)$. Therefore $f(x_1, x_2, x_3) = x_1^2 + 2 x_2^2 + 3 x_3^2 + 2 x_1 x_2 + 2 x_1 x_3 + 0\cdot x_2 x_3 = x_1^2 + 2 x_2^2 + 3 x_3^2 + 2 x_1 x_2 + 2 x_1 x_3.$

Source: TA41_lin_alg.pdf p.8 exercise 7.

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Problem 6.7 — TA session 1 exercise 8: classify four quadratic forms

Establish the sign of: (a) $f(\mathbf{x}) = x_1^2 + x_2^2 + \alpha x_3^2$, (b) $f(\mathbf{x}) = 2x_1^2 + 2x_2^2 + 2x_3^2 - 2x_1 x_2 - 2 x_1 x_3 - 2 x_2 x_3$, (c) $f(\mathbf{x}) = 2x_1^2 - 8 x_1 x_2 + 2 x_2^2 + 2 x_3^2$, (d) $f(\mathbf{x}) = 2x_1^2 + 5 x_2^2 + 5 x_3^2 + 4 x_1 x_2 - 4 x_1 x_3 - 8 x_2 x_3$.

Solution (part (a) — worked fully; others use the same method).

$A = \operatorname{diag}(1, 1, \alpha)$. Eigenvalues (trivially, diagonal matrix): $1, 1, \alpha$, so $\sigma(A) = \{1, 1, \alpha\}$ with algebraic multiplicity of $1$ equal to $2$.

• $\alpha > 0$: all eigenvalues $>0$ $\Rightarrow$ positive definite.
• $\alpha = 0$: eigenvalues $\{1, 1, 0\}$, all $\ge 0$, not all $>0$ $\Rightarrow$ positive semi-definite (but not positive definite). Confirm by exhibiting a non-zero vector on which $f$ vanishes: $f(0, 0, x_3) = 0$.
• $\alpha < 0$: eigenvalues $\{1, 1, \alpha\}$, mixed signs $\Rightarrow$ indefinite.

Parts (b), (c), (d) follow the same pattern: compute $A$ from $f$ (diagonal entries = $x_i^2$ coefficients; off-diagonal = half the $x_i x_j$ coefficient), then either compute eigenvalues or apply Sylvester–Jacobi.

Source: TA41_lin_alg.pdf pp.9–10 exercise 8.

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Problem 6.8 — Invertibility inferred from signs (Theorem 1207 applied)

Let $A$ be a symmetric $4 \times 4$ matrix that is positive definite. What can you say about $\det A^{-1}$ and the eigenvalues of $A^{-1}$?

Solution. By Theorem 1207 (§3.6), $A^{-1}$ exists and is also positive definite. By Theorem 1210, its four eigenvalues are all strictly positive. By Theorem 1191, they are exactly the reciprocals of the eigenvalues of $A$. Therefore:

• $\det A^{-1} = \tfrac{1}{\det A} > 0$.
• All four eigenvalues of $A^{-1}$ are strictly positive — in fact, they are $1/\lambda_i$ where $\lambda_i$ are the eigenvalues of $A$, each $>0$.

(This is the kind of short theoretical question that Sicconi's practice materials refer to as "theorem-chaining" problems; useful as warm-up.)

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§7. Common Pitfalls

Below are the errors Sicconi flags most often and the traps the Bocconi exam-writers exploit. Read this list the day before the exam.

1. Eigenvector vs eigenspace dimension.

   • An eigenvector is a single non-zero vector. An eigenspace is a subspace (possibly multi-dimensional) containing all eigenvectors for a given $\lambda$, plus the zero vector. Saying "the eigenspace is $(1, 0, -1)^T$" is wrong; it's the span of $(1,0,-1)^T$.
   • Eigenvalue multiplicity = dimension of the corresponding eigenspace (for symmetric matrices).

2. Sylvester–Jacobi for negative definite — alternating signs, not all negative.

   • Pattern is $(-, +, -, +, \ldots)$, not $(-, -, -, -, \ldots)$. A matrix with all leading principal minors negative is NOT negative definite in general (only in the trivial $1 \times 1$ case).
   • Mnemonic: negative definite $\iff$ $-A$ is positive definite; so $\det(-A)_k = (-1)^k \det A_k > 0$, forcing $\det A_k$ to alternate.

3. "Positive semi-definite" ≠ "positive definite". (May 2024 Q7 trap.)

   • Positive definite: $\mathbf{x}^T A \mathbf{x} > 0$ for every $\mathbf{x} \neq \mathbf{0}$. Requires all eigenvalues strictly $>0$, or all leading principal minors strictly $>0$.
   • Positive semi-definite: $\mathbf{x}^T A \mathbf{x} \ge 0$ for every $\mathbf{x}$. Allows some eigenvalues $=0$.
   • If $\det A = 0$, $A$ is NOT positive definite — even if every other leading minor is positive. At best it's semi-definite.

4. Sylvester is inconclusive when a leading minor is zero.

   • If $\det A_k = 0$ for some $k$, you cannot conclude definiteness or indefiniteness from leading minors alone.
   • Fallback: compute all principal minors (not just leading), or find the eigenvalues directly.

5. Gram–Schmidt is only needed inside a single eigenspace.

   • Eigenvectors from different eigenspaces of a symmetric matrix are automatically orthogonal (Theorem 1184). Trying to re-orthogonalize them wastes time and introduces arithmetic errors.
   • Inside one eigenspace of multiplicity $\ge 2$, arbitrary basis vectors may NOT be orthogonal — apply Gram–Schmidt there.

6. Normalise after, not before, Gram–Schmidt.

   • Order: orthogonalise $\to$ normalise. Reversing can leave vectors off the unit sphere.

7. Quadratic form coefficients: half, not whole.

   • When building $A$ from $f$, the off-diagonal entries are half the coefficient of $x_i x_j$, not the whole coefficient. $a_{ij} + a_{ji} =$ coefficient of $x_i x_j$; symmetry ($a_{ij} = a_{ji}$) then gives $a_{ij} = \tfrac{1}{2}\cdot$(coef).

8. Eigenvalues of $A$ are eigenvalues of the shifted matrix, with a shift.

   • If $B = A + cI$, then $\sigma(B) = \{\lambda + c : \lambda \in \sigma(A)\}$, with the SAME eigenvectors. Useful for quick checks.

9. Spectrum shortcut traps.

   • $\sum \lambda_i = \operatorname{tr} A$ and $\prod \lambda_i = \det A$ always hold (with multiplicity!). On MCQs with multiplicities, use the sum and product to eliminate wrong options fast (Sicconi's favourite shortcut on Linear Algebra and Differential Calculus.pdf p.12).
   • But: knowing only the sum does NOT determine the eigenvalues; you might need to compute $\det$ or the full characteristic polynomial.

10. Eigenbasis vs basis of eigenvectors.

    • May 2024 Q7(c) explicitly says "not necessarily orthonormal" — meaning a basis of eigenvectors suffices (no need for Gram–Schmidt + normalisation). Read the problem statement carefully.
    • "Eigenbasis" in Sicconi's notes means orthonormal basis of eigenvectors; "basis made of eigenvectors" is the weaker requirement.

11. Forgetting that "symmetric" matters.

    • All the eigenvalue theorems here (1179, 1181, 1184, 1191, 1210, 1217, Spectral, 1207) assume $A$ symmetric. Non-symmetric matrices may have complex eigenvalues, and their eigenvectors from different eigenspaces need not be orthogonal.

12. Confusing the kernel of a linear map with the image.

    • Kernel of $A$ = null space = $\{\mathbf{x} : A\mathbf{x} = \mathbf{0}\}$ = eigenspace $W_0$ (when $0$ is an eigenvalue).
    • For a symmetric matrix, $\dim \ker(A) + \operatorname{rank}(A) = n$ (rank–nullity).

13. Misreading the exam MCQ format.

    • Wrong answer: $-1$ pt. Blank: $0$ pt. Correct: $+5$ pt. On MCQs where you can eliminate $\ge 2$ options, guess between the remaining. On a pure 50-50, skip.

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Final checklist before the exam (linear algebra slice)

• [ ] I can state and prove Theorem 1179 (eigenvalue $\iff$ root of char. poly.).
• [ ] I can state and prove Theorem 1181 (eigenspace is a subspace).
• [ ] I can state and prove Theorem 1184 (orthogonality of eigenvectors from distinct eigenvalues).
• [ ] I can state and prove Theorem 1191 parts (1) and (2).
• [ ] I can state (proof not required for some questions) Proposition 1201 and derive the matrix-form bijection.
• [ ] I can state and prove Theorem 1207 (positive definite $\Rightarrow$ invertible and $A^{-1}$ PD).
• [ ] I can state the Spectral Theorem and sketch the construction of $B$.
• [ ] I know the Sylvester–Jacobi sign patterns: $(+, +, +, \ldots)$ for PD, $(-, +, -, \ldots)$ for ND, and when the test is inconclusive.
• [ ] I can classify any quadratic form both by eigenvalue signs and by Sylvester–Jacobi, and I know when to pick each method.
• [ ] I can run M-LA-1 through M-LA-6 on a $3 \times 3$ symmetric matrix in under 15 minutes.

End of Linear Algebra study guide.