Matrix Diagonalization

A Basis That Separates the Coordinate Directions

Diagonalization rewrites a linear transformation in a $basis of eigenvectors$ . In that basis, each coordinate is multiplied by its own eigenvalue, so the coordinates stop mixing with one another.

The main goal is to find a basis in which the linear transformation $y = A \cdot x$ is represented by the diagonal matrix $\Lambda = S^{-1} \cdot A \cdot S$ . If the eigenbasis can be chosen orthonormally, then $S^{-1} = S^T$ in the real case and $S^{-1} = S^H$ in the complex case.

Definition of Diagonalization

A matrix $A \in \mathbb{K}^{n \times n}$ is called $diagonalizable$ if it is similar to some diagonal matrix $\Lambda \in \mathbb{K}^{n \times n}$ , that is, if there exists an invertible matrix $S \in \mathbb{K}^{n \times n}$ such that:

\Lambda = S^{-1} \cdot A \cdot S

Equivalently, $A = S\Lambda S^{-1}$ . Repeated application then has the form:

For every nonnegative integer $m$ :

A^m = S\Lambda^m S^{-1}

Because $\Lambda^m$ is obtained by raising each diagonal entry to the power $m$ , diagonalization turns a matrix-power problem into scalar arithmetic.

Basic Conditions for Diagonalization

When can a matrix $A \in \mathbb{K}^{n \times n}$ be diagonalized? The answer is when we can find a basis of $\mathbb{K}^n$ that consists entirely of eigenvectors $v_1, \ldots, v_n \in \mathbb{K}^n$ of $A$ with corresponding eigenvalues $\lambda_1, \ldots, \lambda_n \in \mathbb{K}$ .

The diagonal matrix $\Lambda$ is:

\begin{aligned} \Lambda &= \begin{pmatrix} \lambda_1 & & \\ & \ddots & \\ & & \lambda_n \end{pmatrix} \\ &= \text{diag}(\lambda_1, \ldots, \lambda_n) \end{aligned}

The matrix $S$ has the eigenvectors as its columns:

S = (v_1 \quad \ldots \quad v_n)

If $A$ is diagonalizable, then the columns $v_1, \ldots, v_n$ of $S$ form a $basis of eigenvectors$ . From $\Lambda = S^{-1} \cdot A \cdot S$ we obtain $A \cdot S = S \cdot \Lambda$ and thus $A \cdot v_i = \lambda_i \cdot v_i$ for $i = 1, \ldots, n$ .

Conversely, if $v_1, \ldots, v_n$ is a basis of eigenvectors, then $S$ is invertible and from $A \cdot v_i = \lambda_i \cdot v_i$ for $i = 1, \ldots, n$ we obtain $A \cdot S = S \cdot \Lambda$ and thus $\Lambda = S^{-1} \cdot A \cdot S$ .

A Matrix That Cannot Be Diagonalized

Consider the matrix:

A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}

This matrix has eigenvalue $\lambda = 1$ with algebraic multiplicity $\mu_A(1) = 2$ . Its eigenspace is the kernel of $A - I$ :

E_A(1) = \ker(A - I)

The span shown above contains all scalar multiples of the displayed vector and has dimension $1$ . Thus the eigenspace supplies $only one independent eigenvector$ , but a basis of $\mathbb{K}^2$ needs two. Therefore $A$ is not diagonalizable.

Matrix	Algebraic multiplicity of 1	Dimension of the eigenspace	Diagonalizable?
Identity $I_2$	2	2	Yes
Matrix $A$ above	2	1	No

Requirements for Matrix Diagonalization

If a matrix $A \in \mathbb{K}^{n \times n}$ is diagonalizable, then the characteristic polynomial $\chi_A(t)$ of $A$ over $\mathbb{K}$ factors into linear factors:

\chi_A(t) = (\lambda_1 - t) \cdots (\lambda_n - t)

The matrix $A$ has $n$ eigenvalues $\lambda_i \in \mathbb{K}$ , counted with multiplicity, and they need not be distinct.

$The chosen field matters$ . A real matrix can fail to be diagonalizable over $\mathbb{R}$ yet become diagonalizable over $\mathbb{C}$ after its characteristic polynomial splits there.

Distinct eigenvalues remove the need to compare algebraic and geometric multiplicities. Suppose the characteristic polynomial $\chi_A(t)$ of $A \in \mathbb{K}^{n \times n}$ factors over $\mathbb{K}$ as follows.

\chi_A(t) = (\lambda_1 - t) \cdots (\lambda_n - t)

If the eigenvalues are pairwise distinct, so $\lambda_i \neq \lambda_j$ for $i \neq j$ with $i, j \in \{1, \ldots, n\}$ , then $A$ is diagonalizable.

Eigenvectors for pairwise distinct eigenvalues of $A$ are linearly independent. With $n$ such eigenvalues, they form a basis of $\mathbb{K}^n$ .

Repeated eigenvalues require an additional check. Each eigenvalue has algebraic multiplicity $\mu_A(\lambda_i)$ and geometric multiplicity $\dim E_A(\lambda_i)$ , with the relationship:

\dim E_A(\lambda_i) \leq \mu_A(\lambda_i)

Diagonalization Characterization Theorem

For a matrix $A \in \mathbb{K}^{n \times n}$ , the following statements are equivalent:

$A$ is diagonalizable.
Both of the following conditions are satisfied. First, the characteristic polynomial of $A$ must factor into linear factors:

$\begin{aligned} &\chi_A(t) \\ &=(\lambda_1-t)^{\mu_A(\lambda_1)}\cdots(\lambda_k-t)^{\mu_A(\lambda_k)} \end{aligned}$

with pairwise distinct eigenvalues $\lambda_1, \ldots, \lambda_k \in \mathbb{K}$ of $A$ . Second, for all eigenvalues of $A$ , $the algebraic multiplicity must equal the geometric multiplicity$ :

$\begin{aligned} &\mu_A(\lambda_i) = \dim E_A(\lambda_i) \\ &(i = 1, \ldots, k) \end{aligned}$
The direct sum of all eigenspaces is the entire vector space:

$E_A(\lambda_1) \oplus \cdots \oplus E_A(\lambda_k) = \mathbb{K}^n$

This means there exists a basis of $\mathbb{K}^n$ consisting of eigenvectors of $A$ .

For each $i = 1, \ldots, k$ , let $v_1^{(i)}, \ldots, v_{d_i}^{(i)}$ be a basis of the eigenspace $E_A(\lambda_i)$ . Then:

\begin{aligned} &v_1^{(1)},\ldots,v_{d_1}^{(1)}, \\ &v_1^{(2)},\ldots,v_{d_2}^{(2)},\ldots, \\ &v_1^{(k)},\ldots,v_{d_k}^{(k)} \end{aligned}

These vectors form a basis of $\mathbb{K}^n$ consisting of eigenvectors of $A$ . Therefore, $A$ is diagonalizable.