Eigenvalues and Eigenvectors

Recall prompts

Attempt these from memory before reading.

Define an eigenvalue and an eigenvector.
How do you find them? Write the equation whose roots they are.
What does it mean for a matrix to be diagonalisable, and what is the condition?
State the Cayley-Hamilton theorem.
What does the spectral theorem guarantee for a real symmetric matrix, and why is that matrix class special?

Check your answers against the text. Prompt 3 is the one that separates computing eigenvalues from understanding the map: the condition is about eigenvectors, not eigenvalues.

Motivation: eigenvectors are the directions the map does not rotate

A linear map mixes coordinates. Almost every vector you feed in comes out pointing somewhere unrelated. The eigenvectors are the exceptions: they come out pointing along themselves, changed only in length.

That is the whole idea, and everything else follows from wanting to exploit it. If a basis of eigenvectors exists, then in that basis the map does nothing but scale each coordinate independently — which is what diagonalisation is. A hard coupled problem becomes $n$ easy uncoupled ones.

The characteristic polynomial is where the eigenvalues hide. An eigenvector satisfies $(A - \lambda I)v = 0$ with $v \neq 0$ , which is possible only when $A - \lambda I$ is singular, so

\det(A - \lambda I) = 0 .

That determinant is a polynomial of degree $n$ , so an $n \times n$ matrix has $n$ eigenvalues counted with multiplicity — over $\mathbb{C}$ , always. Over $\mathbb{R}$ it may have none at all, and a rotation is the standard example.

Not every matrix is diagonalisable, and Jordan form is what you get instead. Diagonalisability is a statement about eigenvectors: you need $n$ of them, independent. A repeated eigenvalue may come with fewer independent eigenvectors than its algebraic multiplicity, and the missing directions are supplied by generalised eigenvectors — which is what a Jordan block records. The distinction between algebraic and geometric multiplicity is the entire content of section 5.5.

Why symmetric matrices are the special case. A real symmetric matrix is always diagonalisable, by an orthogonal matrix, and always has real eigenvalues. That is the spectral theorem, and it is why symmetric matrices are the ones that appear as observables in quantum mechanics and as quadratic forms in geometry: their eigenbases are orthonormal, so the change of basis costs nothing geometrically.

5.1 Definitions

Let $A \in \mathcal{M}_{n \times n}(F)$ . A scalar $\lambda \in F$ is an eigenvalue of $A$ if there Exists a non-zero vector $\mathbf{v} \in F^n$ such that

A\mathbf{v} = \lambda \mathbf{v}

The vector $\mathbf{v}$ is called an eigenvector corresponding to $\lambda$ .

The eigenspace corresponding to $\lambda$ is $E_\lambda = \ker(A - \lambda I)$ . Its dimension is the geometric multiplicity of $\lambda$ .

5.2 Characteristic Polynomial

Theorem 5.1. $\lambda$ is an eigenvalue of $A$ if and only if $\det(A - \lambda I) = 0$ .

The polynomial $p(\lambda) = \det(A - \lambda I)$ is called the characteristic polynomial of $A$ . Its roots (in the algebraic closure of $F$ ) are the eigenvalues of $A$ .

If $p(\lambda) = (\\lambda - \\lambda_1)^{m_1}(\\lambda - \\lambda_2)^{m_2}\cdots(\\lambda - \\lambda_k)^{m_k}$ with $\\lambda_1, \ldots, \\lambda_k$ distinct, then $m_i$ is the algebraic multiplicity of $\\lambda_i$ .

Proposition 5.2. For each eigenvalue $\lambda$ , $1 \leq \mathrm{dim}(E_\lambda) \leq m_\lambda$ (geometric multiplicity does not exceed algebraic multiplicity).

5.3 Diagonalisation

Definition. $A$ is diagonalisable if there exists an invertible matrix $P$ and a diagonal Matrix $D$ such that

A = PDP^{-1}

Theorem 5.3. $A \in \mathcal{M}_{n \times n}(F)$ is diagonalisable (over $F$ ) if and only if $A$ has $n$ linearly independent Eigenvectors (over $F$ ). Equivalently, the sum of the geometric multiplicities equals $n$ .

Corollary 5.4. If $A$ has $n$ distinct eigenvalues, then $A$ is diagonalisable.

Proof. Eigenvectors corresponding to distinct eigenvalues are linearly independent. With $n$ distinct Eigenvalues, we obtain $n$ linearly independent eigenvectors, which form a basis of $F^n$ . $\blacksquare$

5.4 Cayley—Hamilton Theorem

Theorem 5.5 (Cayley—Hamilton). Every square matrix satisfies its own characteristic polynomial: If $p(\lambda) = \det(\lambda I - A)$ Then $p(A) = 0$ (the zero matrix).

Proof sketch. Let $p(\lambda) = \lambda^n + c_{n-1}\lambda^{n-1} + \cdots + c_1\lambda + c_0$ . By the adjugate formula (Theorem 3.5), $(\lambda I - A) \cdot \mathrm{adj}(\lambda I - A) = p(\lambda) \cdot I$ . Each entry of $\mathrm{adj}(\lambda I - A)$ is a polynomial in $\lambda$ of degree at most $n - 1$ So we can write $\mathrm{adj}(\lambda I - A) = B_{n-1}\lambda^{n-1} + \cdots + B_1\lambda + B_0$ for Matrices $B_i$ . Multiplying out and comparing coefficients of $\lambda^k$ :

B_{n-1} = I, \quad B_{n-2} - AB_{n-1} = c_{n-1}I, \quad \ldots, \quad -AB_0 = c_0 I

Multiplying the $k$ -th equation on the left by $A^k$ and summing over $k$ :

A^n B_{n-1} + A^{n-1}(B_{n-2} - AB_{n-1}) + \cdots + A^0(-AB_0) = A^n + c_{n-1}A^{n-1} + \cdots + c_0 I = p(A)

But the left side telescopes to zero, so $p(A) = 0$ . $\blacksquare$

5.5 Jordan Normal Form

When a matrix is not diagonalisable, the Jordan normal form provides the next-best canonical Representation.

Theorem 5.6. Let $A \in \mathcal{M}_{n \times n}(\mathbb{C})$ . Then $A$ is similar to a block-diagonal Matrix

J = \begin{pmatrix} J_1 & & \\ & \ddots & \\ & & J_k \end{pmatrix}

Where each Jordan block has the form

J_i = \begin{pmatrix} \lambda_i & 1 & & \\ & \lambda_i & \ddots & \\ & & \ddots & 1 \\ & & & \lambda_i \end{pmatrix}

The Jordan form is unique up to permutation of the blocks.

Intuition. Each Jordan block corresponds to one eigenvalue. The size of the block equals the Number of steps in the chain $\mathbf{v}, (A - \lambda I)\mathbf{v}, (A - \lambda I)^2\mathbf{v}, \ldots$ Of generalised eigenvectors. A diagonalisable matrix has all Jordan blocks of size $1 \times 1$ .

Problem. Find the Jordan normal form of

A = \begin{pmatrix} 3 & 1 \\ 0 & 3 \end{pmatrix}

Solution

The characteristic polynomial is $\det(A - \lambda I) = (3 - \lambda)^2$ So $\lambda = 3$ is the Only eigenvalue with algebraic multiplicity 2.

$A - 3I = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ Which has rank 1, so the geometric Multiplicity is $\dim(\ker(A - 3I)) = 2 - 1 = 1$ .

Since the geometric multiplicity (1) is less than the algebraic multiplicity (2), $A$ is not Diagonalisable. The Jordan form has one block of size 2:

J = \begin{pmatrix} 3 & 1 \\ 0 & 3 \end{pmatrix}

(In this case, $A$ is already in Jordan form.) $\blacksquare$

5.6 Spectral Theorem for Real Symmetric Matrices

Theorem 5.7 (Spectral Theorem). If $A \in \mathcal{M}_{n \times n}(\mathbb{R})$ is symmetric ( $A = A^T$ ), then:

All eigenvalues of $A$ are real.
$A$ has $n$ linearly independent orthonormal eigenvectors.
$A$ is orthogonally diagonalisable: $A = QDQ^T$ where $Q$ is orthogonal ( $Q^TQ = I$ ).

Proof. We prove (1) and then (2) and (3) by induction on $n$ .

(1) Let $\lambda \in \mathbb{C}$ be an eigenvalue with eigenvector $\mathbf{v} \in \mathbb{C}^n$ $\mathbf{v} \neq \mathbf{0}$ . Then

\overline{\mathbf{v}}^T A \mathbf{v} = \overline{\mathbf{v}}^T (\lambda \mathbf{v}) = \lambda \overline{\mathbf{v}}^T \mathbf{v}

Since $A = A^T$ and $A$ has real entries, $\overline{A} = A = A^T$ So

\overline{\mathbf{v}}^T A \mathbf{v} = (A\overline{\mathbf{v}})^T \mathbf{v} = (\overline{A\mathbf{v}})^T \mathbf{v} = (\overline{\lambda}\,\overline{\mathbf{v}})^T \mathbf{v} = \overline{\lambda}\,\overline{\mathbf{v}}^T \mathbf{v}

Therefore $(\lambda - \overline{\lambda})\overline{\mathbf{v}}^T\mathbf{v} = 0$ . Since $\overline{\mathbf{v}}^T\mathbf{v} \gt 0$ We have $\lambda = \overline{\lambda}$ So $\lambda \in \mathbb{R}$ .

(2) and (3) By induction. For $n = 1$ the result is trivial. Assume it holds for $(n-1) \times (n-1)$ Symmetric matrices. Since all eigenvalues are real, $A$ has a real eigenvalue $\lambda_1$ with real Eigenvector $\mathbf{v}_1$ . Normalise: $\mathbf{q}_1 = \mathbf{v}_1 / \lVert \mathbf{v}_1 \rVert$ .

Let $W = \mathbf{q}_1^\perp = \{\mathbf{w} \in \mathbb{R}^n : \mathbf{q}_1^T \mathbf{w} = 0\}$ . For any $\mathbf{w} \in W$ :

\mathbf{q}_1^T (A\mathbf{w}) = (A\mathbf{q}_1)^T \mathbf{w} = (\lambda_1 \mathbf{q}_1)^T \mathbf{w} = \lambda_1 \cdot 0 = 0

So $A\mathbf{w} \in W$ . Therefore $A$ restricts to a symmetric linear map $A|_W : W \to W$ on an $(n-1)$ -dimensional space. By the inductive hypothesis, $W$ has an orthonormal basis $\{\mathbf{q}_2, \ldots, \mathbf{q}_n\}$ of eigenvectors of $A|_W$ .

Then $\{\mathbf{q}_1, \mathbf{q}_2, \ldots, \mathbf{q}_n\}$ is an orthonormal eigenbasis for $\mathbb{R}^n$ And $A = QDQ^T$ with $Q = [\mathbf{q}_1 \mid \cdots \mid \mathbf{q}_n]$ . $\blacksquare$

5.7 Worked Example: Full Diagonalisation

Problem. Find the eigenvalues, eigenvectors, and diagonalise

A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}

Solution

The characteristic polynomial is

\det(A - \lambda I) = \det\begin{pmatrix} 4 - \lambda & 1 \\ 2 & 3 - \lambda \end{pmatrix} = (4 - \lambda)(3 - \lambda) - 2

= \lambda^2 - 7\lambda + 10 = (\lambda - 5)(\lambda - 2)

So the eigenvalues are $\lambda_1 = 5$ and $\lambda_2 = 2$ .

For $\lambda_1 = 5$ : Solve $(A - 5I)\mathbf{v} = \mathbf{0}$ .

\begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix}\mathbf{v} = \mathbf{0} \implies -v_1 + v_2 = 0 \implies \mathbf{v} = t\begin{pmatrix} 1 \\ 1 \end{pmatrix}

For $\lambda_2 = 2$ : Solve $(A - 2I)\mathbf{v} = \mathbf{0}$ .

\begin{pmatrix} 2 & 1 \\ 2 & 1 \end{pmatrix}\mathbf{v} = \mathbf{0} \implies 2v_1 + v_2 = 0 \implies \mathbf{v} = t\begin{pmatrix} 1 \\ -2 \end{pmatrix}

Therefore $A = PDP^{-1}$ with

P = \begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix}, \quad D = \begin{pmatrix} 5 & 0 \\ 0 & 2 \end{pmatrix}

P^{-1} = \frac{1}{-3}\begin{pmatrix} -2 & -1 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} 2/3 & 1/3 \\ 1/3 & -1/3 \end{pmatrix}

Verification: $PDP^{-1} = \begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix}\begin{pmatrix} 5 & 0 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} 2/3 & 1/3 \\ 1/3 & -1/3 \end{pmatrix}$

$= \begin{pmatrix} 5 & 2 \\ 5 & -4 \end{pmatrix}\begin{pmatrix} 2/3 & 1/3 \\ 1/3 & -1/3 \end{pmatrix} = \begin{pmatrix} 10/3 + 2/3 & 5/3 - 2/3 \\ 10/3 - 4/3 & 5/3 + 4/3 \end{pmatrix} = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix} = A$ . $\blacksquare$

5.7a Intuition: What Do Eigenvalues and Eigenvectors Mean?

An eigenvector of a linear transformation $A$ is a direction that is preserved by $A$ : applying $A$ to an eigenvector only stretches or compresses it, without rotating it. The eigenvalue $\lambda$ measures how much the eigenvector is stretched: if $\lambda > 1$ , the eigenvector is elongated; if $0 < \lambda < 1$ , it is compressed; if $\lambda < 0$ , it is reflected and then scaled.

Geometrically, the eigenvectors of a matrix reveal the “natural axes” of the transformation. For a $2 \times 2$ matrix, the eigenvectors define the directions along which the transformation acts as simple scaling. In the eigenbasis, the matrix is diagonal: the transformation is just independent stretching along each axis.

This is why diagonalisation is powerful. Computing $A^{100}$ directly requires 100 matrix multiplications, but $D^{100}$ is trivial: just raise each diagonal entry to the 100th power. The change-of-basis matrix $P$ handles the translation between the standard basis and the eigenbasis.

Physical examples:

In mechanics, the eigenvectors of the inertia tensor are the principal axes of rotation.
In vibration analysis, the eigenvectors of the stiffness matrix are the normal modes, and the eigenvalues are the squared natural frequencies.
In Google’s PageRank, the eigenvector corresponding to eigenvalue 1 of the web matrix gives the importance ranking of all pages.

Why eigenvalues matter for stability: In the ODE $\mathbf{x}' = A\mathbf{x}$ , the solution is $\mathbf{x}(t) = e^{At}\mathbf{x}_0$ . If $A$ is diagonalisable, this becomes $\mathbf{x}(t) = P e^{Dt} P^{-1} \mathbf{x}_0$ . The behavior is governed by $e^{\lambda_i t}$ : if all $\mathrm{Re}(\lambda_i) < 0$ , the system decays to zero (stable); if any $\mathrm{Re}(\lambda_i) > 0$ , the system grows without bound (unstable).

5.7b Worked Example: 3x3 Diagonalisation

Problem. Diagonalise the matrix

A = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{pmatrix}

Solution

Since $A$ is upper triangular, the eigenvalues are the diagonal entries: $\lambda_1 = 2$ , $\lambda_2 = 3$ (with algebraic multiplicity 2).

For $\lambda_1 = 2$ : Solve $(A - 2I)\mathbf{v} = \mathbf{0}$ .

A - 2I = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \to \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}

Free variable: $x_1 = t$ . Eigenvector: $\mathbf{v}_1 = (1, 0, 0)^T$ .

For $\lambda_2 = 3$ : Solve $(A - 3I)\mathbf{v} = \mathbf{0}$ .

A - 3I = \begin{pmatrix} -1 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}

Free variables: $x_2 = s$ , $x_3 = t$ . Then $x_1 = s$ . Eigenvectors: $s(1, 1, 0)^T + t(0, 0, 1)^T$ .

The geometric multiplicity of $\lambda_2 = 3$ is 2, equal to its algebraic multiplicity. Therefore $A$ is diagonalisable with

P = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \quad D = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{pmatrix}

Key observation: When the geometric multiplicity equals the algebraic multiplicity for every eigenvalue, the matrix is diagonalisable. When they differ (as in the Jordan form example in Section 5.5), the matrix is not diagonalisable, and the Jordan normal form is the best alternative. $\blacksquare$

Problem. Use the Cayley—Hamilton theorem to compute $A^{10}$ for the same matrix $A$ above.

Solution

The characteristic polynomial is $p(\lambda) = \lambda^2 - 7\lambda + 10$ So by Cayley—Hamilton, $A^2 = 7A - 10I$ .

To find $A^{10}$ Divide $\lambda^{10}$ by $p(\lambda)$ :

\lambda^{10} = q(\lambda)(\lambda^2 - 7\lambda + 10) + r(\lambda)

Where $r(\lambda) = a\lambda + b$ has degree less than 2. Then $A^{10} = r(A) = aA + bI$ .

To find $a$ and $b$ Evaluate at the eigenvalues:

$\lambda^{10}\big|_{\lambda=5} = 5^{10} = 9765625 = 5a + b$

$\lambda^{10}\big|_{\lambda=2} = 2^{10} = 1024 = 2a + b$

Subtracting: $3a = 9765625 - 1024 = 9764601$ So $a = 3254867$ .

$b = 1024 - 2 \cdot 3254867 = 1024 - 6509734 = -6508710$ .

Therefore $A^{10} = 3254867 \cdot A - 6508710 \cdot I$ . $\blacksquare$

$A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ has Eigenvalue $\lambda = 1$ with algebraic multiplicity 2 but geometric multiplicity 1. It has only one Linearly independent eigenvector and is not diagonalisable.

5.8 Worked Example: Spectral Decomposition of a Symmetric Matrix

Problem. Orthogonally diagonalise the symmetric matrix

A = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix}

Solution

$\det(A - \lambda I) = \det\begin{pmatrix} 2-\lambda & 1 & 1 \\ 1 & 2-\lambda & 1 \\ 1 & 1 & 2-\lambda \end{pmatrix}$

$= (2-\lambda)[(2-\lambda)^2 - 1] - 1[(2-\lambda) - 1] + 1[1 - (2-\lambda)]$ $= (2-\lambda)(\lambda^2 - 4\lambda + 3) - (1-\lambda) + (\lambda - 1)$ $= (2-\lambda)(\lambda-1)(\lambda-3)$

Eigenvalues: $\lambda_1 = 1$ , $\lambda_2 = 2$ , $\lambda_3 = 3$ .

For $\lambda_1 = 1$ : $A - I = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix} \to \begin{pmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$ . Eigenspace: $\{(s, t, -s-t) : s, t \in \mathbb{R}\}$ . An orthonormal basis: $\mathbf{q}_1 = \frac{1}{\sqrt{2}}(1, -1, 0)$ , $\mathbf{q}_2 = \frac{1}{\sqrt{6}}(1, 1, -2)$ .

For $\lambda_2 = 2$ : $A - 2I = \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix}$ . Eigenspace: $\{(-t, -t, t) : t \in \mathbb{R}\}$ . Normalised: $\mathbf{q}_3 = \frac{1}{\sqrt{3}}(-1, -1, 1)$ .

For $\lambda_3 = 3$ : $A - 3I = \begin{pmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{pmatrix} \to \begin{pmatrix} 1 & -1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$ . Eigenspace: $\{(s+t, s, t) : s, t \in \mathbb{R}\}$ . Normalised: $\mathbf{q}_4 = \frac{1}{\sqrt{3}}(1, 1, 1)$ .

$A = QDQ^T$ where $Q = \frac{1}{\sqrt{6}}\begin{pmatrix} \sqrt{3} & 1 & -\sqrt{2} & \sqrt{2} \\ -\sqrt{3} & 1 & -\sqrt{2} & \sqrt{2} \\ 0 & -2 & \sqrt{2} & \sqrt{2} \end{pmatrix}$ and $D = \mathrm{diag}(1, 2, 3)$ .

$\blacksquare$

5.9 Common Pitfalls

Algebraic multiplicity $\geq$ geometric multiplicity, not the reverse. A matrix is diagonalisable if and only if equality holds for every eigenvalue.
The characteristic polynomial depends on the choice of eigenvalue variable convention. $\det(A - \lambda I)$ and $\det(\lambda I - A)$ differ by a factor of $(-1)^n$ but have the same roots.
Similarity preserves eigenvalues but not eigenvectors. If $A = PBP^{-1}$ and $B\mathbf{v} = \lambda\mathbf{v}$ then $A(P\mathbf{v}) = \lambda(P\mathbf{v})$ ; the eigenvectors transform by $P$ .

Counterexamples worth knowing by name

Claim	True?	Counterexample
Every square matrix is diagonalisable	no	$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ is nilpotent with one eigenvalue and only one eigenvector
Every real matrix has real eigenvalues	no	a rotation by $90^\circ$ has eigenvalues $\pm i$
Distinct eigenvalues imply diagonalisable	yes	but the converse fails: the identity matrix has one eigenvalue and is diagonalisable
A repeated eigenvalue implies a defective matrix	no	$I$ has eigenvalue $1$ with multiplicity $n$ and a full eigenspace
Algebraic and geometric multiplicity are equal	no	they are equal exactly when the matrix is diagonalisable; the Jordan block is the counterexample
$AB$ and $BA$ have different eigenvalues	no	they have the same characteristic polynomial, even when $A$ and $B$ are rectangular
The spectral theorem holds for all square matrices	no	it needs a normal matrix; symmetric, Hermitian, unitary and orthogonal are the standard special cases
Eigenvalues are continuous in the matrix entries	yes	but eigenvectors need not be, and that failure is what pseudospectra measure

The first row is the one to internalise, because everything else on the page is organised around it. The question “is this matrix diagonalisable?” is not about the eigenvalues — it is about whether there are enough eigenvectors, and eigenvalues only ever give you a necessary condition.

The last row is the numerical point. Eigenvalues are well-behaved; eigenvectors of a non-normal matrix can move discontinuously under an arbitrarily small perturbation, which is why stability analysis of non-normal systems is subtle.

Where the research frontier is

The QR algorithm. Computing eigenvalues is one of the ten algorithms with the greatest influence on science and engineering in the twentieth century, and it works by iterating a factorisation until the matrix converges to Schur form.
Random matrix theory. The distribution of eigenvalues of a large random matrix is universal — it does not depend on the distribution of the entries — and the semicircle law and Tracy-Widom distribution now appear in number theory and physics.
Pseudospectra. For non-normal matrices the spectrum is the wrong object: the $\epsilon$ -pseudospectrum describes what actually governs transient growth, and it can be enormous when the spectrum looks innocuous.
Spectral graph theory. The eigenvalues of a graph’s adjacency or Laplacian matrix determine connectivity, expansion and random-walk behaviour, which is the mathematical engine behind PageRank.
Generalised and polynomial eigenvalue problems. $Av = \lambda Bv$ arises in vibration analysis, and the theory is richer because $B$ may be singular.

Interleaving problems

Mix these with Systems of Linear Equations, Systems of Differential Equations, Spectral Theory, Representation Theory and Quantum Operators.

(Determinants.) Derive the characteristic polynomial of $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and show that its roots satisfy $\lambda^2 - (\operatorname{tr}A)\lambda + \det A = 0$ . That identity is the $2 \times 2$ case of Cayley-Hamilton.
(Differential equations.) Diagonalise the system $\dot{x} = Ax$ for $A = \begin{pmatrix} 0 & 1 \\ -2 & 3 \end{pmatrix}$ and write the general solution. The eigenvalues are the exponents — explain what a defective eigenvalue does to the solution, and why the answer involves $te^{\lambda t}$ .
(Spectral theory.) State the spectral theorem for self-adjoint operators and identify exactly which part of the finite-dimensional proof fails. That failure is why the theorem is a theorem.
(Representation theory.) Show that a representation of a finite group makes every group element diagonalisable, since some power of each element is the identity. Then explain why that fact gives the group algebra the structure the character theory needs.
(Quantum mechanics.) Explain why an observable must be a self-adjoint operator in terms of the spectral theorem: real eigenvalues, orthonormal eigenbasis, and somewhere for the Born-rule probabilities to live.
(Numerical analysis.) Describe the power method: iterating $x, Ax, A^2x, \ldots$ converges to the dominant eigenvector. Show why, and explain what happens when the two largest eigenvalues are equal in magnitude.

Cross-References

Matrices: The characteristic polynomial whose roots are eigenvalues is defined via the determinant of $A - \lambda I$ .
Singular Value Decomposition: The SVD relates singular values to eigenvalues of $A^T A$ and provides a diagonal-like decomposition for non-square matrices.
Inner Product Spaces: The spectral theorem for symmetric matrices guarantees an orthonormal basis of eigenvectors.

Mark this page as reviewed

Eigenvalues and Eigenvectors

Recall prompts

Motivation: eigenvectors are the directions the map does not rotate

5.1 Definitions

5.2 Characteristic Polynomial

5.3 Diagonalisation

5.4 Cayley—Hamilton Theorem

5.5 Jordan Normal Form

5.6 Spectral Theorem for Real Symmetric Matrices

5.7 Worked Example: Full Diagonalisation

5.7a Intuition: What Do Eigenvalues and Eigenvectors Mean?

5.7b Worked Example: 3x3 Diagonalisation

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