Theorems

## Theorem 2

An nxn matrix A is diagonalizable if and only if A has n linearly independent eigenvectors

Proof

Let A be diagonalizable
Thus, $A = PDP^{-1}$ for some invertible matrix P and diagonal matrix D

$P = \begin{bmatrix} \vec{v}_1&\vec{v}_2&\vec{v}_3,...,\vec{v}_n \end{bmatrix}$ and $D = \begin{bmatrix} d_1&0&0&...&0\\0&d_2&0&...&0\\0&0&d_3&...&0\\0&0&0&...&0\\0&0&0&...&d_n \end{bmatrix}$

Where P is linearly independent since P is invertible

Since $A = PDP^{-1}$

(1)
\begin{equation} AP = PDP^{-1}P = PD \end{equation}
(2)
\begin{align} A\begin{bmatrix} \vec{v}_1&\vec{v}_2&\vec{v}_3,...,\vec{v}_n \end{bmatrix} = \begin{bmatrix} \vec{v}_1&\vec{v}_2&\vec{v}_3,...,\vec{v}_n \end{bmatrix}D \end{align}
(3)
\begin{bmatrix} A\vec{v}_1&A\vec{v}_2&A\vec{v}_3,...,A\vec{v}_n \end{bmatrix} = \begin{bmatrix} d_1\vec{v}_1&d_2\vec{v}_2&d_3\vec{v}_3,...,d_n\vec{v}_n \end{bmatrix}

$A\vec{v}_1 = d_1\vec{v}_1$
$A\vec{v}_2 = d_2\vec{v}_2$

Therefore, $\begin{bmatrix} \vec{v}_1&\vec{v}_2&\vec{v}_3,...,\vec{v}_n \end{bmatrix}$ are eigenvectors

Since these are also the columns of P, we know they are linearly independent.

A has n linearly independent eigenvectors
let $\begin{bmatrix} \vec{v}_1&\vec{v}_2&\vec{v}_3,...,\vec{v}_n \end{bmatrix}$ be linearly independent eigenvectors
let $\lambda_1, \lambda_2, \lambda_3...\lambda_n$ be the associated eigenvalues
let $P = \begin{bmatrix} \vec{v}_1&\vec{v}_2&\vec{v}_3,...,\vec{v}_n \end{bmatrix}$

P is invertible since its components are linearly independent
D is the diagonal matrix with the eigenvalues as the entries on the primary diagonal

Then

(4)
\begin{align} AP = A \begin{bmatrix} \vec{v}_1&\vec{v}_2&\vec{v}_3,...,\vec{v}_n \end{bmatrix} = \begin{bmatrix} A\vec{v}_1&A\vec{v}_2&A\vec{v}_3,...,A\vec{v}_n \end{bmatrix} \end{align}

Since $\begin{bmatrix} \vec{v}_1&\vec{v}_2&\vec{v}_3,...,\vec{v}_n \end{bmatrix}$ are eigenvectors,

$A\vec{v}_1 = \lambda_1\vec{v}_1$
$A\vec{v}_2 = \lambda_2\vec{v}_2$

So it can be shown that $AP = PD$, and multiplying both sides on the right by $P^{-1}$, we find that $A = PDP^{-1}$ and is therefore diagonalizable.

page revision: 0, last edited: 18 Apr 2015 20:00