## Spectral Decomposition of a Matrix

To find the spectral decomposition of a matrix, first diagonalize if possible, then write the diagonal form as
$D= \sum_i \lambda_i E_i$
where
$E_i$
is the matrix with 1s on the leading diagonal in columns corresponding to the ith eigenvector of the matrix of eigenvectors,
$P$
.
Example:
$A= \left( \begin{array}{ccc} 4 & 2 & 2 \\ 2 & 4 & 2 \\ 2 & 2 & 4 \end{array} \right)$
.
The eigenvalues of
$A$
are the solutions to
$det(A- \lambda I)=0$
.
\begin{aligned} det(A- \lambda I) &=det(\left( \begin{array}{ccc} 4 & 2 & 2 \\ 2 & 4 & 2 \\ 2 & 2 & 4 \end{array} \right) - \lambda \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) ) \\ &= \left| \begin{array}{ccc} 4- \lambda & 2 & 2 \\ 2 & 4- \lambda & 2 \\ 2 & 2 & 4- \lambda \end{array} \right| \\ &= (4 -\lambda)((4- \lambda)^2-2^2)-2(2(4-\lambda) -2^2)+2(2(4-\lambda)-2^2) \\ &=(2- \lambda)^2(8- \lambda) = 0 \end{aligned}
.
Hence
$\lambda_1=2, \: \lambda_2 =8$
Now find the eigenvectors. Let
$\mathbf{v}=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}$

$\lambda_1=2$

$(A-2I) \mathbf{v}= \left( \begin{array}{ccc} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{array} \right)\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}= \begin{pmatrix}2x_1+2x_2+2x_3\\2x_1+2x_2+2x_3\\2x_1+2x_2+2x_3\end{pmatrix} = \begin{pmatrix}0\\0\\0\end{pmatrix}$

Set
$x_2=1, \: x_3=0$
then
$x_1=-1$
. An eigenvector is
$\begin{pmatrix}-1\\1\\0\end{pmatrix}$

Set
$x_2=0, \: x_3=1$
then
$x_1=-1$
. An eigenvector is
$\begin{pmatrix}-1\\0\\1\end{pmatrix}$

$\lambda_2=8$

$(A-2I) \mathbf{v}= \left( \begin{array}{ccc} -4 & 2 & 2 \\ 2 & -4 & 2 \\ 2 & 2 & -4 \end{array} \right)\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}= \begin{pmatrix}-4x_1+2x_2+2x_3\\2x_1-4x_2+2x_3\\2x_1+2x_2-4x_3\end{pmatrix} = \begin{pmatrix}0\\0\\0\end{pmatrix}$

A solution is
$x_1=x_2=x_3=1$
then the last eigenvector is
$\begin{pmatrix}1\\1\\1\end{pmatrix}$

The matrix of eigenvectors is
$\left| \begin{array}{ccc} -1 & -1 & 1 \\ 1 & 1 & 1 \\ 0 & w_2 & 1 \end{array} \right|$

Eigenvectors corresponding to
$\lambda_1 =2$
occupy the first two columns and the eigenvector corresponding to
$\lambda_2 =8$
occupies the third column.
Then
$D=2E_1 +8E_2$
where
$E_1 = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right), \: E_2 = \left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right)$

Then
$A=P(2E_1 +8E_2)P^{-1}=2 PE_1 P^{-1}+8 PE_2P^{-1}$