Eigenvalue eigenvector (basis) I have a question regarding the use of eigenvectors as basis vectors.
For A = \begin{bmatrix}1 & -4 &7\\-4 & 4 & -4\\7& -4 & 1 \end{bmatrix}
"By expressing an arbitrary vector $r$ in terms of the eigenvectors or otherwise, show
that a non-zero vector $e$ exists such that
$Ar \cdot e = 0$
for all $r$."
I am not sure how I should tackle the question. I started off trying to convert $Ar$ into a diagonalized matrix. Is this the correct approach?
Thank you
 A: The characteristic polynomial $p_A(X)=\det(A-XI_3)$ of your matrix is
$$
p_A(X)=72X+6X^2-X^3
$$
which has roots $-6$, $0$ and $12$.
So the matrix is diagonalizable, hence it has a basis consisting of eigenvectors. (The fact that it is diagonalizable also directly follows from the fact that $A$ is symmetric.)
Just find an eigenvector relative to each eigenvalue and you have found a basis as required.
A: The problem as stated --
"By expressing an arbitrary vector r in terms of the eigenvectors or otherwise, show that a non-zero vector $e$ exists such that $Ar \cdot e=0$ for all $r$."
-- is false. For instance, if $A$ is the identity matrix, then there is no such vector $e$. I think that you have left off some important assumption about the matrix $A$. 
A: Say that your matrix $A$ has eigenvectors $x_1$, $x_2$ and $x_3$ with eigenvalues $\lambda_1$, $\lambda_2$ and $\lambda_3$ (resp.). Then you can express an arbitrary vector $r$ as $r=c_1x_1+c_2x_2+c_3x_3$ in the basis $\left\{x_1,x_2,x_3\right\}$. Therefore:
$$Ar\cdot e=A(c_1x_1+c_2x_2+c_3x_3)\cdot e=(c_1Ax_1+c_2Ax_2+c_3Ax_3)\cdot e=(c_1\lambda_1x_1+c_2\lambda_2x_2+c_3\lambda_3x_3)\cdot e$$
What can you say about $e$ when $Ar\cdot e=0$?
