Diagonalization

Let $ A$ 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

$\displaystyle (A - \lambda_i I) \mathbf{v}_i = \mathbf{0},$   $\displaystyle \mbox{ or, equivalently $A \mathbf{v}_i = \lambda_i \mathbf{v}_i$ for $i = 1,\ldots,n$. }$ (12)

Then we can define a diagonal matrix $ D$ and a matrix $ P$ 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 $\displaystyle \quad P = [\mathbf{v}_1  \ldots \mathbf{v}_n] .$ (13)

Together we can summarize the equations of eigenvectors as

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

Furthermore, if $ \mathbf{v}_1,\ldots,\mathbf{v}_n$ are linearly independent, the matrix $ P$ is invertible, and therefore, we obtain

$\displaystyle A = P D P^{-1} .$ (14)

The matrix $ A$ is said to be diagonalizable if $ A$ has the above expression.



Subsections

Department of Mathematics
Last modified: 2005-10-21