I'm having trouble with row reduction on the following matrix when I try to find the eigenvector for $\lambda$=1.
$\frac{1}{7} \times \begin{bmatrix} 2 & 6 & 3 \\ 6 & -3 & 2 \\ 3 & 2 & -6\\ \end{bmatrix}$ For $\lambda$ = 1 this becomes $\frac{1}{7} \times \begin{bmatrix} 1 & 6 & 3 \\ 6 & -4 & 2 \\ 3 & 2 & -7\\ \end{bmatrix}$
But when I row reduce this I get
$\frac{1}{7} \times \begin{bmatrix} 1 & 6 & 3 \\ 0 & -40 & -16 \\ 0 & -16 & -16\\ \end{bmatrix}$
The eigenvector that I should be getting is (3,2,1). Any idea where I'm going wrong?
The characteristic polynomial for the matrix $A=\begin{bmatrix}2&6&3\\6&-3&2\\3&2&-6\end{bmatrix}$ is $\chi(\lambda)=\det(A-\lambda I)$ which one finds to be equal to:
$-\lambda^3-7\lambda^2+49\lambda+343$. After some algebraic manipulation, you will find it can be written as:
$\chi(\lambda)=-(\lambda+7)(\lambda+7)(\lambda-7)$
That implies that for $A$, the eigenvalues are $7,-7,-7$
The actual matrix we are interested in is $(\frac{1}{7} A)$. By properties of matrices, the eigenvectors of $(\frac{1}{7} A)$ are the same as the eigenvectors of $A$ and the corresponding eigenvalues of $(\frac{1}{7} A)$ are $\frac{1}{7}$ times the eigenvalues of $A$. That is to say, the eigenvalues of $(\frac{1}{7} A)$ are $1,-1,-1$
We try to find the eigenvector of $1$ for $(\frac{1}{7}A)$. This is the same as trying to find the eigenvector of $7$ for $A$.
So, we row reduce $A-7I$ (not $A-I$). I.e. row reduce $\begin{bmatrix}2-7&6&3\\6&-3-7&2\\3&2&-6-7\end{bmatrix}=\begin{bmatrix}-5&6&3\\6&-10&2\\3&2&-13\end{bmatrix}$
Completing the rowreduction process, one sees it row reduces to $\begin{bmatrix}1&0&-3\\0&1&-2\\0&0&0\end{bmatrix}$ implying an eigenvector is $\begin{bmatrix}3\\2\\1\end{bmatrix}$ (and so the eigenspace for the eigenvalue is the span of this vector)