Chapter 5 Study Guide

calendar_today August 9, 2026 article 665 words access_time 5 min replay Modified August 9, 2026

Table of Contents

Chapter 5 — Eigenvalues and Eigenvectors

Covers §5.1–5.3. Full problems and recorded answers in 2160 Compendium.

Weighting

44 questions, 6 true/false (13%). Highly procedural: find eigenvalues, find eigenvectors, assemble $P$ and $D$. The procedures chain together, so an error in §5.2 propagates through §5.3.

§5.1 Eigenvectors and Eigenvalues

Definition 1 (Eigenvector and eigenvalue).

A nonzero vector $\mathbf{x}$ is an eigenvector of $A$ if

$$A\mathbf{x}=\lambda\mathbf{x}$$

for some scalar $\lambda$, the corresponding eigenvalue.

$\mathbf{x}=\mathbf{0}$ is never an eigenvector — but $\lambda=0$ is a legitimate eigenvalue (it occurs exactly when $A$ is singular).

The two verification questions are different

“Is $\mathbf{v}$ an eigenvector of $A$?” — compute $A\mathbf{v}$ and check whether the result is a scalar multiple of $\mathbf{v}$. If $A\mathbf{v}=\lambda\mathbf{v}$, that $\lambda$ is the eigenvalue. Pure multiplication, no solving.

“Is $\lambda$ an eigenvalue of $A$?” — check whether $(A-\lambda I)\mathbf{x}=\mathbf{0}$ has a nontrivial solution, i.e. whether $A-\lambda I$ is singular. Equivalently $\det(A-\lambda I)=0$.

Confusing these wastes significant time — 5.1.4 and 5.1.6 are the first kind, 5.1.2, 5.1.7, 5.1.8 the second.

Eigenspace

The eigenspace for $\lambda$ is $\text{Nul}(A-\lambda I)$ — all eigenvectors for $\lambda$, together with $\mathbf{0}$.

To find a basis: form $A-\lambda I$, row reduce, solve the homogeneous system, and read off the parametric vector form. This is a §1.5 null-space computation wearing a new name.

Triangular matrices

The eigenvalues of a triangular matrix are exactly its diagonal entries — no characteristic polynomial needed.

Tested as — is $\lambda$ an eigenvalue 5.1.2, 5.1.7, 5.1.8 · is $\mathbf{v}$ an eigenvector 5.1.4, 5.1.6 · basis for an eigenspace 5.1.9–5.1.16 · eigenvalues directly 5.1.17

§5.2 The Characteristic Equation

Definition 2 (Characteristic polynomial).

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

The left side is the characteristic polynomial; its roots are the eigenvalues. For $n\times n$ it has degree $n$.

The $2\times2$ shortcut

For $A=\begin{bmatrix} a & b \\ c & d\end{bmatrix}$:

$$\det(A-\lambda I)=\lambda^{2}-(a+d)\lambda+(ad-bc)=\lambda^{2}-(\operatorname{tr}A)\lambda+\det A$$

Worth memorizing — it turns most $2\times2$ eigenvalue problems into a quadratic you can write down immediately.

Definition 3 (Algebraic vs. geometric multiplicity).

Algebraic multiplicity — the multiplicity of $\lambda$ as a root of the characteristic polynomial. Geometric multiplicity — $\dim$ of the eigenspace, i.e. $\dim\text{Nul}(A-\lambda I)$.

Always $1 \le \text{geometric} \le \text{algebraic}$. When they differ for some eigenvalue, the matrix is not diagonalizable — this is the bridge into §5.3 and what 5.2.18 is checking.

Listing "with multiplicity"

When asked to list eigenvalues repeated according to multiplicity, a double root must appear twice. The total count must equal $n$ (counting complex roots).

Tested as — characteristic polynomial and eigenvalues 5.2.1–5.2.6 · characteristic polynomial only 5.2.9–5.2.14 · eigenvalues with multiplicity 5.2.15, 5.2.16 · multiplicity comparison 5.2.18

§5.3 Diagonalization

Theorem 4 (The Diagonalization Theorem).

$A$ ($n\times n$) is diagonalizable $\iff$ $A$ has $n$ linearly independent eigenvectors. In that case $A=PDP^{-1}$, where

$P$ has the eigenvectors as its columns
$D$ is diagonal with the corresponding eigenvalues in the same order

The order must match: column $j$ of $P$ pairs with entry $(j,j)$ of $D$.

The procedure

Find the eigenvalues from $\det(A-\lambda I)=0$
Find a basis for each eigenspace
Count: if the eigenvectors total $n$, build $P$ and $D$; otherwise report not diagonalizable
Verify if time permits — check $AP=PD$, which avoids inverting $P$

Sufficient condition

If an $n\times n$ matrix has $n$ distinct eigenvalues, it is automatically diagonalizable — eigenvectors from distinct eigenvalues are always independent.

The converse fails: repeated eigenvalues may still diagonalize, provided each eigenspace is big enough to match its algebraic multiplicity. A $3\times3$ with two eigenvalues and both eigenspaces one-dimensional gives only $2 < 3$ independent eigenvectors — not diagonalizable (5.3.30).

Why diagonalize — computing powers

$$A^{k}=PD^{k}P^{-1}, \qquad D^{k}=\begin{bmatrix} \lambda_1^{k} & 0 \\ 0 & \lambda_2^{k}\end{bmatrix}$$

Raising a diagonal matrix to a power is entrywise, so $A^{4}$ or even $A^{k}$ for symbolic $k$ becomes tractable. This is the point of 5.3.1, 5.3.2, 5.3.3.

Tested as — diagonalize with eigenvalues given 5.3.11, 5.3.12, 5.3.15, 5.3.16 · diagonalize from scratch 5.3.17, 5.3.19 · compute $A^k$ from $PDP^{-1}$ 5.3.1–5.3.3 · diagonalizability true/false 5.3.21–5.3.30

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