# Diagonalization of a matrix

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Diagonalization is the process of finding a corresponding diagonal matrix (a matrix in which the only non-zero components are on the diagonal line from $A_{1,1}$ to $A_{n,n}$ for an $n\times n$ matrix) for a given diagonalizable matrix. A matrix is diagonalizable if and only if the matrix of eigenvectors is invertable (that is, the determinant does not equal zero). If a matrix is not diagonalizable, is is called a defective matrix.

The diagonal $D$ of a matrix is equal to $P^{-1}AP$ such that $P$ is the matrix of eigenvectors ($P=[v_1,\ldots,v_n]$). Diagonal matrices are very useful, as computing determinants, products and sums of matrices, and powers becomes much simpler. For example, given the matrix $A$ , $A^n=PD^nP^{-1}$ .

## Computation of the diagonal matrix

Given $AP=PD$ , $D$ can be found be making a diagonal matrix of the eigenvalues of $A$ . $P$ will be equal to the matrix of corresponding eigenvectors. For example, say we have the matrix

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

To find the eigenvalues, we must first find the characteristic polynomial, which will be equal to

$\begin{vmatrix}\lambda I-\begin{bmatrix}1&2\\-1&4\end{bmatrix}\end{vmatrix} =\begin{vmatrix}\lambda\begin{bmatrix}1&0\\0&1\end{bmatrix}-\begin{bmatrix}1&2\\-1&4\end{bmatrix}\end{vmatrix}=0$
$\begin{vmatrix}\lambda-1&-2\\1&\lambda-4\end{vmatrix}=(\lambda^2-5\lambda+4)+2=\lambda^2-5\lambda+6=(\lambda-2)(\lambda-3)=0$
$\lambda=2,3$

Therefore $D$ will be equal to

$\begin{bmatrix}2&0\\0&3\end{bmatrix}$

$P$ will be the matrix of eigenvectors corresponding to the above diagonal matrix. The eigenvectors will be the non-trivial solution to

$2I-\begin{bmatrix}1&2\\-1&4\end{bmatrix} =\begin{bmatrix}1&-2\\1&-2\end{bmatrix}=0_{2,1}$
$\begin{bmatrix}x\\y\end{bmatrix}=t\begin{bmatrix}2\\1\end{bmatrix}$
$\vec{v}_1=\begin{bmatrix}2\\1\end{bmatrix}$

$3I-\begin{bmatrix}1&2\\-1&4\end{bmatrix} =\begin{bmatrix}2&-2\\1&-1\end{bmatrix}=0_{2,1}$
$\begin{bmatrix}x\\y\end{bmatrix}=t\begin{bmatrix}1\\1\end{bmatrix}$
$\vec{v}_2=\begin{bmatrix}1\\1\end{bmatrix}$
$P=\begin{bmatrix}\vec{v}_1&\vec{v}_2\end{bmatrix} =\begin{bmatrix}2&1\\1&1\end{bmatrix}$
$P^{-1}=\frac1{|P|}\text{adj}\begin{bmatrix}2&1\\1&1\end{bmatrix} =\begin{bmatrix}1&-1\\-1&2\end{bmatrix}$

Therefore,

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

This is useful to us because, among other things, we can use this to find large powers of $A$ .

\begin{align}A^5&=\begin{bmatrix}1&2\\-1&4\end{bmatrix}^5=PD^5P^{-1} =\begin{bmatrix}2&1\\1&1\end{bmatrix}\begin{bmatrix}2^5&0\\0&3^5\end{bmatrix}\begin{bmatrix}1&-1\\-1&2\end{bmatrix}\\ &=\begin{bmatrix}2&1\\1&1\end{bmatrix}\begin{bmatrix}32&0\\0&243\end{bmatrix}\begin{bmatrix}1&-1\\-1&2\end{bmatrix}=\begin{bmatrix}-179&422\\-211&454\end{bmatrix}\end{align}