Thread: Orthogonal Matrix

1. Orthogonal Matrix

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

I have found the eigenvalues and vectors to be $\lambda=3,3,-3$ and eigenvectors $\left\{\begin{bmatrix}1\\1\\0\end{bmatrix},\begin{ bmatrix}1\\0\\1\end{bmatrix},\begin{bmatrix}-1\\1\\1\end{bmatrix}\right\}$.

Now I need to find an orthogonal matrix such that $P^tAP=D$.
D being my diagonal matrix.

I tried Gram-Schmidting my eigenspace but that didn't work. What do I need to do?

2. Re: Orthogonal Matrix

the problem lies with your eigenvectors for 3, as both are orthogonal to (-1,1,1). so apply gram-schmidt to just those 2.

for example if we take u1 = v1 = (1,1,0), then gram-schmidt gives:

u2 = v2 - [(u1.v1)/(u1.u1)]u1 = (1,0,1) - (1/2)(1,1,0) = (1/2,-1/2,1).

by construction, u2 is orthogonal to u1, and (1/2,-1/2,1).(-1,1,1) = -1/2 + -1/2 + 1 = 0.

3. Re: Orthogonal Matrix

That gives me the Diagonal matrix

$\begin{bmatrix}6&0&0\\0&\frac{9}{2}&0\\0&0&-9\end{bmatrix}$

If I am not mistaken, shouldn't D have entries corresponding to the eigenvalues?

4. Re: Orthogonal Matrix

normalize the eigenvectors.

5. Re: Orthogonal Matrix

Originally Posted by Deveno
normalize the eigenvectors.
Normalizing didn't help. Here is the output:

6. Re: Orthogonal Matrix

$\frac{1}{\sqrt{\frac{3}{2}}} = \frac{2}{\sqrt{6}}$

not $\frac{\sqrt{6}}{2}$.

re-check your matrix product with the proper matrix, because i get:

$P^TAP = \begin{bmatrix}3&0&0\\0&3&0\\0&0&-3\end{bmatrix}$

as expected.