Lecture 7: Eigenvalues and Diagonalization

---
title: "Lecture 7: Eigenvalues and Diagonalization"
subtitle: Linear Algebra II · Autumn term 2026
author: Department of Mathematics
number-sections: true
---

# Eigenvalues and eigenvectors

Let $A$ be an $n \times n$ matrix. A non-zero vector $\mathbf{v}$ is an **eigenvector** of $A$ with **eigenvalue** $\lambda$ if

$$
A\mathbf{v} = \lambda \mathbf{v}.
$$ {#eq:eig}

Rewriting [@eq:eig] as $(A - \lambda I)\mathbf{v} = \mathbf{0}$ shows that a non-zero solution exists exactly when $A - \lambda I$ is singular, that is, when

$$
\det(A - \lambda I) = 0.
$$ {#eq:char}

The left-hand side of [@eq:char] is a polynomial of degree $n$ in $\lambda$, the **characteristic polynomial** of $A$.

::: {.tip title="Example 7.1"}
For $A = \left(\begin{smallmatrix} 4 & 1 \\ 2 & 3 \end{smallmatrix}\right)$,
$\det(A - \lambda I) = (4-\lambda)(3-\lambda) - 2 = \lambda^2 - 7\lambda + 10 = (\lambda - 2)(\lambda - 5)$,
so the eigenvalues are $\lambda_1 = 2$ and $\lambda_2 = 5$.
:::

# Diagonalization

A matrix $A$ is **diagonalizable** if there is an invertible matrix $P$ and a diagonal matrix $D$ with

$$
A = P D P^{-1}, \qquad D = \begin{pmatrix} \lambda_1 & & \\ & \ddots & \\ & & \lambda_n \end{pmatrix}.
$$ {#eq:diag}

> [!NOTE]
> **Theorem 7.2.** An $n \times n$ matrix is diagonalizable if and only if it has $n$ linearly independent eigenvectors. In that case the columns of $P$ in [@eq:diag] are those eigenvectors.

Diagonalization makes powers easy to compute: $A^k = P D^k P^{-1}$, and $D^k$ only needs the $k$-th powers of the eigenvalues.[^power]

[^power]: This is how the closed form of the Fibonacci numbers is usually derived.

# Summary

| Property                    | Condition                                  |
|-----------------------------|--------------------------------------------|
| $\lambda$ is an eigenvalue  | $\det(A - \lambda I) = 0$                  |
| $A$ is invertible           | $0$ is not an eigenvalue                   |
| $A$ is diagonalizable       | $n$ linearly independent eigenvectors      |
| $\operatorname{tr} A$       | $\lambda_1 + \lambda_2 + \cdots + \lambda_n$ |
| $\det A$                    | $\lambda_1 \lambda_2 \cdots \lambda_n$     |

: Facts to remember

# Exercises

1. Find the eigenvalues and eigenvectors of $\left(\begin{smallmatrix} 2 & 0 \\ 1 & 3 \end{smallmatrix}\right)$.
2. Show that $\left(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix}\right)$ is not diagonalizable.
3. Use diagonalization to compute $A^{10}$ for the matrix of Example 7.1.