Sarah Morash's question at Yahoo Answers about eigenvalues

[Fernando Revilla]

Here is the question:

ey! So I have a question on an assignment asking to orthogonally diagonalize the matrix:
a 0 b
0 a 0
b 0 a
I know the steps on how to do this, but am having a hard time trying to figure out how to factor this correctly to get all of the eigenvalues at the beginning. I can factor it to a point, but then cannot seem to figure out how to solve for the eigenvalues.

If anyone could help, that would be great!

Here is a link to the question:

Help finding the eigenvalues of a matrix? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[Fernando Revilla]

Hello Sarah,

Denote by $$A$$ to the given matrix. Let's find the corresponding eigenvalues.

$$\det (A-\lambda I)=\begin{vmatrix}{a-\lambda}&{0}&{b}\\{0}&{a-\lambda}&{0}\\{b}&{0}&{a-\lambda}\end{vmatrix}=(a-\lambda)\begin{vmatrix}{a-\lambda}&{b}\\{b}&{a-\lambda}\end{vmatrix}$$

Now we use the transformations: $$F_2\to F_2-F_1$$ and $$C_1\to C_1+C_2$$:

$$\begin{vmatrix}{a-\lambda}&{b}\\{b}&{a-\lambda}\end{vmatrix}=\begin{vmatrix}{a-\lambda}&{b}\\{b-a+\lambda}&{a-b-\lambda}\end{vmatrix}=\begin{vmatrix}{a+b-\lambda}&{b}\\{0}&{a-b-\lambda}\end{vmatrix}$$

As a consequence:

$$\det (A-\lambda I)=(a-\lambda)(a+b-\lambda)(a-b-\lambda)$$

and the eigenvalues are

$$\lambda_1=a,\lambda_2=a+b,\lambda_3=a-b$$