1. Transformations/eigenvectors

I have the following question, sorry in advance for my poor mathematical notation, having some serious computer issues!
A a 3x3 matrix:
6 13 -8
2 5 -2
7 17 -9

I was given an eigenvector: v=(1 0 1)^T where its 3x1. I had to find the corresponding eigenvalue which is -2. Then had to show 3 is an eigenvalue and its corresponding eigenvector is (1/2 1/2 1)^T

I am now stuck at the following question:

The matrix A defines a linear transormationT: from real numbers of dimension 3 to the real numbers of dimension 3 by T(x)=Ax. It is known that T fixes a non-zero vector x, T(x)=x. Use this information to determine another eigenvalue of A

Any help would be gratefully received!

2. Eigenvalues are of course scaling factors by which a transformation may multiply a corresponding eigenvector. So in your case, you know that

if $A=\begin{pmatrix}6 &13 &-8 \\
2 &5 &-2 \\
7 & 17 &-9\end{pmatrix}$
, given $\lambda_1=-2$ and $\lambda_2=3$, then $v_1=\begin{pmatrix}1 \\0 \\1\end{pmatrix}$ and $v_2=\begin{pmatrix}1 \\1 \\2\end{pmatrix}$ behave as $Tv_i=\lambda_i$.

You now know there exists a vector $x=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}$ such that $Tx=1x$. Thus $x$ is an eigenvector with eigenvalue $1$. Since this does not match any of the previous eigenvalues, you know that this new $x$ is linearly independent of $v_1$ and $v_2$ (explain why). Then, you should be able to find the components of $x$ just as you found $v_2$.

3. Thats a great help, thank you. Why is it that we know for sure that this new x is linearly independent of v1 and v2?

4. Eigenvectors corresponding to distinct eigenvectors are always independent.

Suppose $v_1$ and $v_2$ are eigenvectors of T with corresponding eigenvalues $\lamba_1$ and $\lambda_2$ respectively, $\lambda_1\ne \lambda_2$, and that $av_1+ bv_2= 0$ for some scalars a and b. Applying T to both sides, $T(av_1+ bv_2)= aT(v_1)+ bT(v_2)= \lambda_1 av_1+ \lamba_2 bv_2= 0$. Multiplying the first equation by $\lambda_1$ gives $\lambda_1 av_1+ \lambda-1 bv_2= 0$ and subtracting those two equations, $(\lambda_2- \lambda_1) bv_2= 0$. Since $\lambda_2- \lambda_1$ is not zero, b= 0 and then similarly for a.