# Thread: Spectral Decomposition of a Matrix

1. ## Spectral Decomposition of a Matrix

A= $\begin{bmatrix} 7 & 3 & 3 &2 \\ 0 & 1 & 2 & -4 \\ -8 &
-4& -5 &0 \\ 2& 1 & 2 & 3\end{bmatrix}
$

Let T be the linear Operator in $R_n$ that has the given matrix A relative to the standard basis
$E_n$.
Find the Spectral Decomposition of T.

I found that characteristic polynomial is $(x-3)^2 (x+1)(x-1)$
So $T=P_1-P_2+3P_3$, where P are projections.

Trying to find E's such that $A=x_1E_1+x_2E_2+x_3E_3=E_1-E_2+3E_3$
E are Projections matrix relative to $E_3$

2. Originally Posted by firebio
A= $\begin{bmatrix} 7 & 3 & 3 &2 \\ 0 & 1 & 2 & -4 \\ -8 &
-4& -5 &0 \\ -2 & 1 & 2 & 3\end{bmatrix}
$

Let T be the linear Operator in $R_n$ that has the given matrix A relative to the standard basis $E_n$.
Find the Spectral Decomposition of T.

I found that characteristic polynomial is $(x-3)^2 (x+1)(x-1)$
So $T=P_1-P_2+3P_3$, where P are projections.

Trying to find E's such that $A=x_1E_1+x_2E_2+x_3E_3=E_1-E_2+3E_3$
E are Projections matrix relative to $E_3$
This problem is made harder by the fact that $(x-3)^2 (x+1)(x-1)$ is not the characteristic polynomial of the given matrix. I suspect that the matrix should be $\begin{bmatrix} 7 & 3 & 3 &2 \\ 0 & 1 & 2 & -4 \\ -8 &
-4& -5 &0 \\ 2 & 1 & 2 & 3\end{bmatrix}$
(with a +2 in the bottom left corner). That matrix at least appears to have the right eigenvalues.

The key fact that you need to solve the problem is that if $\mathbf{v} = (w,x,y,z)$ is a unit vector (I'll write vectors as rows rather than columns, to save space), then the projection onto the one-dimensional space spanned by $\mathbf{v}$ has matrix $\begin{bmatrix}w^2&wx&wy&wz\\ wx&x^2&xy&xz\\ wy&xy&y^2&yz\\ wz&xz&yz&z^2\end{bmatrix}$ (with respect to the standard basis).
For the matrix $\begin{bmatrix} 7 & 3 & 3 &2 \\ 0 & 1 & 2 & -4 \\ -8 &
-4& -5 &0 \\ 2 & 1 & 2 & 3\end{bmatrix}$
, a unit eigenvector corresponding to the eigenvalue 1 is $\tfrac1{\sqrt5}(1,-2,0,0)$. The projection $P_1$ onto the eigenspace of the eigenvalue x=1 is therefore $\frac15\!\begin{bmatrix}1&-2&0&0\\ -2&4&0&0\\ 0&0&0&0\\ 0&0&0&0\end{bmatrix}$.

The projection $P_{-1}$ onto the one-dimensional eigenspace of the eigenvalue x=–1 is calculated in a similar way.

For the two-dimensional eigenspace of the repeated eigenvalue x=3, there are various possible methods to find the projection $P_3$ onto it.

Method 1 (the slow but safe method). Find two orthonormal eigenvectors for the eigenvalue x=3, form the projections onto the one-dimensional spaces spanned by each of them, then add those two projections to get $P_3$.

Method 2 (the lazy way). Use the fact that sum of $P_1$, $P_{-1}$ and $P_3$ must be the identity matrix. Since you have already found $P_1$ and $P_{-1}$, you can just write down $P_3 = I-P_1-P_{-1}$.

Method 3 (the belt and braces method). Use Method 1 to find $P_3$. Then use Method 2 to check that your answer is right.

The advantage of Method 2 is that it is quick. The disadvantage is that it assumes that the eigenspace of the eigenvalue x=3 is two-dimensional. There is in principle no guarantee of this. If the eigenvalue x=3 only has one linearly independent eigenvector then the matrix A will not have a spectral decomposition at all.