# Orthonormal Set of Eigenvectors of A

• Dec 10th 2011, 06:13 PM
divinelogos
Orthonormal Set of Eigenvectors of A
Problem:
http://img46.imageshack.us/img46/5769/capture2br.jpg

Part (a): This part seems simple. I just find the eigenvectors of the matrix A, use the GS process, and then normalize to get an orthonormal set. Is this correct?

Part (b): I'm not sure how to do this part. Is "S" referring to [S]B? That is, do I use the eigenvectors of A=[S]B? I'm pretty lost here...

Part (c): Not sure how to do this without completing part b.

Any help will be appreciate and I'll click the thank you button for you! Thanks! :)
• Dec 10th 2011, 08:22 PM
Drexel28
Re: Orthonormal Set of Eigenvectors of A
Quote:

Originally Posted by divinelogos
Problem:
http://img46.imageshack.us/img46/5769/capture2br.jpg

Part (a): This part seems simple. I just find the eigenvectors of the matrix A, use the GS process, and then normalize to get an orthonormal set. Is this correct?

Part (b): I'm not sure how to do this part. Is "S" referring to [S]B? That is, do I use the eigenvectors of A=[S]B? I'm pretty lost here...

Part (c): Not sure how to do this without completing part b.

Any help will be appreciate and I'll click the thank you button for you! Thanks! :)

I'm sorry, is this for a takehome exam?
• Dec 10th 2011, 09:43 PM
divinelogos
Re: Orthonormal Set of Eigenvectors of A
Quote:

Originally Posted by Drexel28
I'm sorry, is this for a takehome exam?

Nope. It's a study guide for the final. I can send you the whole thing if you need to verify that. Thanks! :)
• Dec 11th 2011, 12:43 AM
FernandoRevilla
Re: Orthonormal Set of Eigenvectors of A
(a) The eigenvalues of $\displaystyle A$ are $\displaystyle \lambda=1$ (double) and $\displaystyle \lambda=1$ (simple). We have $\displaystyle \ker (A-I) \equiv \begin{Bmatrix}-x_1+x_3=0\\x_1-x_3=0\end{matrix}$ and an orthogonal basis is $\displaystyle \{(1,0,1)^t,(0,1,0)^t\}$ .

Now, $\displaystyle \ker (A+I) \equiv \begin{Bmatrix}x_1+x_3=0\\2x_2=0\\x_1+x_3=0\end{ma trix}$ and a basis is $\displaystyle \{(1,0,-1)^t\}$ .

Hence, $\displaystyle B=\{(1/\sqrt{2},0,1/\sqrt{2})^t,(0,1,0)^t,(1/\sqrt{2},0,-1/\sqrt{2})^t\}$ is an orthonormal basis of eigenvectors of $\displaystyle A$ .

(b) $\displaystyle (1/\sqrt{2},0,1/\sqrt{2})^t\equiv (1/\sqrt{2})\vec{u_1}+(1/\sqrt{2})\vec{u_3}=\ldots$

Hope you can continue.