Modules › Eigenvalues and eigenvectors
Eigenvalues and eigenvectors
Characteristic polynomial, eigenspaces, diagonalization, powers of a matrix, symmetric matrices and orthogonal diagonalization.
Lecture 7 · Exercise Sheet 7 · Eigenvalues and eigenvectors
Textbook sections: Elementary Linear Algebra ch. 7 §7.1–7.3 · Strang ch. 6 §6.1–6.4
Textbook sections: Elementary Linear Algebra ch. 7 §7.1–7.3 · Strang ch. 6 §6.1–6.4
Eigenvalues and eigenvectors
Linear Algebra · Module 7
After this module you can
- Find eigenvalues and eigenspaces; decide whether a matrix is diagonalizable and find $P$ and $D$.
- Compute $A^k$ via $PD^kP^{-1}$.
- Orthogonally diagonalize a symmetric matrix and use the standard proofs.
Eigenvalues and eigenvectors
- $A\mathbf v=\lambda\mathbf v$, $\mathbf v\ne0$. Eigenvalues are the roots of $\det(A-\lambda I)=0$; eigenvectors for $\lambda$: $\ker(A-\lambda I)\setminus\{0\}$.
- $\sum\lambda_i=\operatorname{tr}A$ and $\prod\lambda_i=\det A$ (use them to check your work).
- Triangular matrices: the eigenvalues are the diagonal entries.
- Algebraic multiplicity (root multiplicity) $\ge$ geometric multiplicity $=\dim\ker(A-\lambda I)$.
Diagonalization: the recipe
- 1. $\det(A-\lambda I)$, factor it. 2. For each $\lambda$ find a basis of $\ker(A-\lambda I)$.
- 3. $A$ is diagonalizable iff the total number of independent eigenvectors is $n$ (geometric $=$ algebraic multiplicity for every $\lambda$).
- 4. $P$ = eigenvectors as columns, $D=\operatorname{diag}(\lambda_i)$ in the same order: $P^{-1}AP=D$.
- Not diagonalizable: $\begin{bmatrix}1&1&0\\1&2&1\\0&-1&1\end{bmatrix}$ has $\lambda=1$ twice but only a one-dimensional eigenspace.
Powers of a matrix
$$A=PDP^{-1}\ \Rightarrow\ A^k=PD^kP^{-1}$$
- $A=\begin{bmatrix}1&3\\2&0\end{bmatrix}$ has $\lambda=3,-2$: $(A^n)_{11}=\frac{3\cdot3^n+2(-2)^n}{5}$.
- Final 2022–23: $A=\begin{bmatrix}2&0&-2\\0&3&0\\0&0&3\end{bmatrix}$ gives $A^k$ with entries $2^k,\ 3^k$ and $2\cdot2^k-2\cdot3^k$.
- Check the formula at $k=1$ (it must return $A$).
Symmetric matrices
- Spectral theorem: a real symmetric matrix is orthogonally diagonalizable: $P^TAP=D$ with $P$ orthogonal ($P^{-1}=P^T$).
- Eigenvectors for different eigenvalues are already orthogonal; inside a repeated eigenvalue orthogonalise with Gram–Schmidt, then normalise.
- Example: $\begin{bmatrix}2&-2&2\\-2&5&1\\2&1&5\end{bmatrix}$ has $\lambda=0,6,6$.
- Then $A^k=PD^kP^T$.
Standard proofs from the finals
| Statement | Proof idea |
|---|---|
| $A$ invertible, $Av=av$ ($a\ne0$) $\Rightarrow$ $A^{-1}v=\frac1av$ | apply $A^{-1}/a$ to $Av=av$ |
| $A^2=2A$ $\Rightarrow$ eigenvalues $\in\{0,2\}$ | $\lambda^2v=2\lambda v$, so $\lambda(\lambda-2)=0$ |
| no $T:\mathbb{R}^5\to\mathbb{R}^2$ with $\dim\ker=2$ | rank $=5-2=3>2=\dim\mathbb{R}^2$ |
Write the hypothesis, the one key computation, and the conclusion.
Typical mistakes
Order of $P$ and $D$The $i$-th column of $P$ must go with the $i$-th diagonal entry of $D$.
Repeated eigenvalueCount independent eigenvectors before claiming diagonalizable.
Eigenvector $\mathbf 0$Eigenvectors are non-zero by definition.
Practice next
- Chapter quizzes Ch. 7 (3 sets)
- Practice bank, filtered to this chapter, then written mode
- Lab: Eigenvalues, eigenvectors, diagonalization
- Then the mock exams (Exams page)
Worked examples from the course problems
Worked example 1 · Eigenvalues (Exercise Sheet 7, Ex. 3)
For $A=\begin{bmatrix}2 & -1 & 1 \\ -2 & 3 & -2 \\ -1 & 1 & 0\end{bmatrix}$, the eigenvalue $\lambda=1$ has algebraic multiplicity 2. What is the dimension of its eigenspace?
Show the answer and the solution
Answer: $2$
$\dim\ker(A-1I)=n-\operatorname{rank}(A-1I)=3-1=2$.
Worked example 2 · Powers via diagonalization (Exercise Sheet 7, Ex. 8)
For $A=\begin{bmatrix}1 & 3 \\ 2 & 0\end{bmatrix}$ (eigenvalues $3$ and $-2$), the $(1,1)$ entry of $A^5$ is
Show the answer and the solution
Answer: $133$
$A=PDP^{-1}$ gives $A^n=PD^nP^{-1}$; here $(A^n)_{11}=\frac{2 \left(-2\right)^{n}}{5} + \frac{3 \cdot 3^{n}}{5}$. For $n=5$: $A^5=\begin{bmatrix}133 & 165 \\ 110 & 78\end{bmatrix}$, so the entry is $133$.
Worked example 3 · Eigenvectors (Exercise Sheet 7, Ex. 3)
For $A=\begin{bmatrix}2 & -1 & 1 \\ -2 & 3 & -2 \\ -1 & 1 & 0\end{bmatrix}$, which vector is an eigenvector for $\lambda=3$?
Show the answer and the solution
Answer: $(-1, 2, 1)$
Solve $(A-3I)x=0$. Then $Ax=\lambda x$: $A(-1, 2, 1)=(-3, 6, 3)=3\cdot(-1, 2, 1)$.
Worked example 4 · Diagonalization (Linear Algebra Summary (Gen 16), VII.2 (sufficient condition))
An $n\times n$ matrix has $n$ distinct eigenvalues. What follows?
Show the answer and the solution
Answer: it is diagonalizable; the converse is false (e.g. $I_3$ is diagonalizable with the single eigenvalue $1$)
Eigenvectors of distinct eigenvalues are linearly independent, so $n$ distinct eigenvalues give $n$ independent eigenvectors: a sufficient condition. It is not necessary: $I_3$ has one repeated eigenvalue but is diagonalizable, whereas $\begin{bmatrix}1&1\\0&1\end{bmatrix}$ (repeated eigenvalue, only one eigenvector direction) is not.
Practice
Quiz set A Quiz set B Quiz set C Practice bank (this chapter) Written mode Lab: Eigenvalues, eigenvectors, diagonalization