Find the spectral decomposition of $A$

Question

$$ A= \begin{pmatrix} -3 & 4\\ 4 & 3 \end{pmatrix} $$

So i am assuming that i must find the evalues and evectors of this matrix first, and that is exactly what i did.

The evalues are $5$ and $-5$, and the evectors are $(2,1)^T$ and $(1,-2)^T$

Now the spectral decomposition of $A$ is equal to $(Q^{-1})^\ast$ (diagonal matrix with corresponding eigenvalues) * Q

$Q$ is given by [evector1/||evector1|| , evector2/||evector2||]

and for Q i got the matrix

$$ Q= \begin{pmatrix} 2/\sqrt{5} &1/\sqrt{5} \\ 1/\sqrt{5} & -2/\sqrt{5} \end{pmatrix} $$

the inverse of Q is the matrix...

$$ \begin{pmatrix} 2 \sqrt{5}/5 & \sqrt{5}/5 \\ \sqrt{5}/5 & -2 \sqrt{5}/5 \end{pmatrix} $$

and the diagonal matrix with corresponding evalues is

$$ A= \begin{pmatrix} 5 & 0\\ 0 & -5 \end{pmatrix} $$

so now i found the spectral decomposition of $A$, but i really need someone to check my work.

Did i take the proper steps to get the right answer, did i make a mistake somewhere?

Answer 1

The needed computation is

$$\mathsf{A} = \mathsf{Q\Lambda}\mathsf{Q}^{-1}$$

Where $\Lambda$ is the eigenvalues matrix. And your eigenvalues are correct.

Hence you have to compute

$$\mathsf{AQ} = \mathsf{Q\Lambda}$$

Which gives you the solutions

$$a = 2c ~~~~~~~~~~~ d = -\frac{b}{2}$$

You can then choose easy values like $c = b = 1$ to get

$$Q = \begin{pmatrix} 2 & 1 \\ 1 & -\frac{1}{2} \end{pmatrix}$$

And easily

$$\mathsf{Q}^{-1} = \frac{1}{\text{det}\ \mathsf{Q}} \begin{pmatrix} -\frac{1}{2} & -1 \\ -1 & 2 \end{pmatrix}$$

Which you can compute alone.

Answer 2

\begin{align} \begin{bmatrix} -3 & 4 \\ 4 & 3\end{bmatrix}\begin{bmatrix} 2 \\ 1\end{bmatrix}= \begin{bmatrix} -2 \\ 11\end{bmatrix} \end{align}

The eigenvector is not correct. The correct eigenvecor should be $\begin{bmatrix} 1 & 2\end{bmatrix}^T$ since

\begin{align} \begin{bmatrix} -3 & 4 \\ 4 & 3\end{bmatrix}\begin{bmatrix} 1 \\ 2\end{bmatrix}= 5 \begin{bmatrix} 1 \\ 2\end{bmatrix} \end{align}

$\begin{bmatrix} 1 & -2\end{bmatrix}^T$ is not an eigenvector too.

\begin{align} \begin{bmatrix} -3 & 4 \\ 4 & 3\end{bmatrix}\begin{bmatrix} -2 \\ 1\end{bmatrix}= -5 \begin{bmatrix} -2 \\ 1\end{bmatrix} \end{align}

Also, at the end of the working, $A$ remains $A$, it doesn't become a diagonal matrix.

You should write $A$ as $QDQ^T$ if $Q$ is orthogonal.

Answer 3

I think of the spectral decomposition as writing $A$ as the sum of two matrices, each having rank 1. Let $A$ be given. Then compute the eigenvalues and eigenvectors of $A$. Then $$ A = \lambda_1P_1 + \lambda_2P_2 $$ where $P_i$ is an orthogonal projection onto the space spanned by the $i-th$ eigenvector $v_i$.

In your case, I get $v_1=[1,2]^T$ and $v_2=[-2, 1]$ from Matlab. Then we have: $$ \left[ \begin{array}{cc} -3 & 4 \\ 4 & 3\\ \end{array} \right] = 5\left[ \begin{array}{cc} 1/5 & 2/5 \\ 2/5 & 4/5\\ \end{array} \right] - 5\left[ \begin{array}{cc} 4/5 & -2/5 \\ -2/5 & 1/5\\ \end{array} \right] $$ Each $P_i$ is calculated from $v_iv_i^T$.