Matrix Algebra

Diagonalization

Matrix representation. Let be an $n\times n$ matrix, and let $\mathbf{v}_1,\ldots,\mathbf{v}_n$ be the eigenvectors corresponding to the eigenvalues $\lambda_1,\ldots,\lambda_n$ satisfying the matrix equations

$\displaystyle (A - \lambda_i I) \mathbf{v}_i = \mathbf{0},$ or, equivalently $A \mathbf{v}_i = \lambda_i \mathbf{v}_i$ for $i=1,\ldots,n$ . $\displaystyle$

Then we can define a diagonal matrix

and a matrix

respectively by

$\displaystyle D = \begin{bmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \hdotsfor{4} \\ 0 & 0 & \cdots & \lambda_n \end{bmatrix}$ and $\displaystyle \quad P = [\mathbf{v}_1 \ldots \mathbf{v}_n] .$

Together we can summarize the matrix equations of eigenvector as

$\displaystyle A P = P D = [\lambda_1\mathbf{v}_1 \:\ldots\: \lambda_n\mathbf{v}_n] .$

Diagonalization. If $\mathbf{v}_1,\ldots,\mathbf{v}_n$ are linearly independent, the matrix is invertible, and therefore, we obtain

$\displaystyle A = P D P^{-1} .$

The matrix

is said to be diagonalizable if

has the above expression.

EXAMPLE 4. Diagonalize each of the following matrices if possible.

$A = \begin{bmatrix} 1 & 3 & 3 \\ -3 &-5 &-3 \\ 3 & 3 & 1 \end{bmatrix}$
$A = \begin{bmatrix} 2 & 4 & 3 \\ -4 &-6 &-3 \\ 3 & 3 & 1 \end{bmatrix}$
$A = \begin{bmatrix} 5 & 0 & 0 & 0 \\ 0 & 5 & 0 & 0 \\ 1 & 4 &-3 & 0 \\ -1 &-2 & 0 &-3 \end{bmatrix}$

Matlab/Octave. To find whether is diagonalizable, we can use the command [P,D] = eig(A) and the function det(P). If det(P) returns 0, or the magnitude of $10^{-8}$ (for example, 1.2608e-08), is not invertible numerically, and therefore, is not diagonalizable.