Diagonalization

Overview

Diagonalization is the process of finding a diagonal matrix $D$ similar to $A$. A diagonalizable matrix can be written as $A = PDP^{-1}$, where $P$ contains eigenvectors and $D$ contains eigenvalues.

Definition

A square matrix $A$ is diagonalizable if there exists an invertible matrix $P$ such that:

$P^{-1}AP = D$

where $D$ is diagonal. Equivalently:

$A = PDP^{-1}$

The Matrices P and D

• P: columns are linearly independent eigenvectors of $A$
• D: diagonal entries are the corresponding eigenvalues

$P = \begin{bmatrix} \mathbf{v}_1 & \mathbf{v}_2 & \cdots & \mathbf{v}_n \end{bmatrix}$

$D = \begin{bmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_n \end{bmatrix}$

Diagonalization Process

Step 1: Find Eigenvalues

Solve $\det(A - \lambda I) = 0$

Step 2: Find Eigenvectors

For each eigenvalue $\lambda_i$, solve $(A - \lambda_i I)\mathbf{v} = \mathbf{0}$

Step 3: Form P and D

• $P$: eigenvectors as columns (order matches $D$)
• $D$: corresponding eigenvalues on diagonal

Step 4: Verify (Optional)

Check that $P^{-1}AP = D$

Example

Diagonalize:

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

Eigenvalues:

$p(\lambda) = \lambda^2 - 7\lambda + 10 = (\lambda-5)(\lambda-2)$

$\lambda_1 = 5$, $\lambda_2 = 2$

Eigenvectors:

For $\lambda = 5$: $\mathbf{v}_1 = \begin{bmatrix} 2 \\ 1 \end{bmatrix}$

For $\lambda = 2$: $\mathbf{v}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$

Matrices:

$P = \begin{bmatrix} 2 & 1 \\ 1 & -1 \end{bmatrix}, \quad D = \begin{bmatrix} 5 & 0 \\ 0 & 2 \end{bmatrix}$

$P^{-1} = \frac{1}{-3}\begin{bmatrix} -1 & -1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 1/3 & 1/3 \\ 1/3 & -2/3 \end{bmatrix}$

Verify: $P^{-1}AP = D$ ✓

Conditions for Diagonalizability

$A$ is diagonalizable if and only if:

1. $A$ has $n$ linearly independent eigenvectors
2. For each eigenvalue: geometric = algebraic multiplicity
3. Sum of geometric multiplicities = $n$

When NOT Diagonalizable

Defective Matrix

$A = \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix} \quad p(\lambda) = (\lambda-2)^2$

Only one eigenvector $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ for $\lambda = 2$

Not diagonalizable (defective)

Applications of Diagonalization

Computing Matrix Powers

If $A = PDP^{-1}$:

$A^n = PD^n P^{-1}$

where $D^n$ is easy to compute (raise diagonal entries to power $n$).

Example: Matrix Square

$A^2 = (PDP^{-1})(PDP^{-1}) = PD(P^{-1}P)DP^{-1} = PD^2 P^{-1}$

Solving Systems of ODEs

For $\frac{d\mathbf{x}}{dt} = A\mathbf{x}$:

$\mathbf{x}(t) = Pe^{Dt}P^{-1}\mathbf{x}_0$

Special Cases

Symmetric Matrices

Real symmetric matrices are always diagonalizable:

• Eigenvalues are real
• Eigenvectors can be chosen orthonormal
• $A = QDQ^T$ where $Q$ is orthogonal

Distinct Eigenvalues

If all $n$ eigenvalues are distinct, $A$ is diagonalizable.

Summary Table

Condition	Diagonalizable?
$n$ distinct eigenvalues	Yes
Symmetric matrix	Yes
All geometric = algebraic mult.	Yes
Has $n$ lin. indep. eigenvectors	Yes
Defective (fewer eigenvectors)	No