Exercises: Eigenvalues and Diagonalization

Practice problems and worked solutions for eigenvalues, eigenvectors, diagonalization, and matrix powers.
Modified

April 26, 2026

Keywords

exercises, eigenvalues, eigenvectors, diagonalization

1 Scope and Goals

This set trains four things:

  • finding eigenvalues and eigenvectors in small examples
  • checking whether diagonalization is available
  • reading matrix powers through eigen-modes
  • preparing for spectral applications and SVD

2 Prerequisites

  • matrix multiplication
  • basis and independence
  • diagonal matrices and powers

3 Warm-Up Problems

  1. Find the eigenvalues of

    \[ \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}. \]

  2. Find one eigenvector of

    \[ \begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix} \]

    for each eigenvalue.

4 Core Problems

  1. Diagonalize

    \[ A = \begin{bmatrix} 4 & 0 \\ 0 & 1 \end{bmatrix} \]

    and compute \(A^5\).

  2. Let

    \[ P = \begin{bmatrix} 0.8 & 0.2 \\ 0.2 & 0.8 \end{bmatrix}. \]

    Find its eigenvalues and explain which mode decays.

  3. Explain why a matrix with \(n\) distinct eigenvalues is diagonalizable.

5 Proof Problems

  1. Prove that if \(Av = \lambda v\) and \(c \neq 0\), then \(cv\) is also an eigenvector with eigenvalue \(\lambda\).
  2. Show that if \(A = S \Lambda S^{-1}\), then \(A^k = S \Lambda^k S^{-1}\) for every positive integer \(k\).

6 Computational or Applied Problems

  1. Open Computation Lab: Matrix Powers and Spectral Modes. Change the mixing parameter and describe how the second eigenvalue controls the rate of convergence.
  2. In software, compute the eigenvalues of a symmetric matrix and compare them with the singular values of the same matrix when it is positive semidefinite.

7 Hints

  1. For diagonal matrices, the eigenvalues are already visible.
  2. For matrix powers, diagonalize first and then power the diagonal entries.
  3. Distinct eigenvalues give independent eigenvectors.

8 Full Solutions

8.1 Solution 1

The eigenvalues are \(2\) and \(3\).

8.2 Solution 2

For \(\lambda=2\), solve

\[ \begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix} v = 0, \]

so one eigenvector is \((1,0)^\top\).

For \(\lambda=3\), solve

\[ \begin{bmatrix} -1 & 1 \\ 0 & 0 \end{bmatrix} v = 0, \]

so one eigenvector is \((1,1)^\top\).

8.3 Solution 3

The matrix is already diagonal, so

\[ A = I \begin{bmatrix} 4 & 0 \\ 0 & 1 \end{bmatrix} I^{-1}. \]

Therefore

\[ A^5 = \begin{bmatrix} 4^5 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1024 & 0 \\ 0 & 1 \end{bmatrix}. \]

8.4 Solution 4

The eigenvalues are \(1\) and \(0.6\).

The consensus mode \((1,1)^\top\) persists, while the disagreement mode \((1,-1)^\top\) decays because \(0.6^k \to 0\).

8.5 Solution 5

Distinct eigenvalues give linearly independent eigenvectors. So an \(n \times n\) matrix with \(n\) distinct eigenvalues has \(n\) independent eigenvectors, which is exactly what diagonalization needs.

9 Common Errors

  • treating the zero vector as an eigenvector
  • assuming repeated eigenvalues automatically prevent diagonalization
  • forgetting that eigenvectors are determined only up to nonzero scaling

10 What To Do Next

Open SVD and Low-Rank Approximation next. That page brings spectral thinking back through the symmetric matrix \(A^\top A\) and turns it into a stable approximation tool.

11 Sources and Further Reading

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