Eigenvalues and Eigenvectors

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 $\lambda$.

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).

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 $F^n$. $\blacksquare$

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 $p(A) = 0$. $\blacksquare$

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

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

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

1. All eigenvalues of $A$ are real.
2. $A$ has $n$ linearly independent orthonormal eigenvectors.
3. $A$ 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 $Q = [\mathbf{q}_1 \mid \cdots \mid \mathbf{q}_n]$. $\blacksquare$

Problem. Find the eigenvalues, eigenvectors, and diagonalise

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

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

:::caution Common Pitfall Not every matrix is diagonalisable. For example, $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ has Eigenvalue $\lambda = 1$ with algebraic multiplicity 2 but geometric multiplicity 1. It has only one Linearly independent eigenvector and is not diagonalisable.

Problem. Orthogonally diagonalise the symmetric matrix

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

• 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 $P$.

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