I have
$$A = \begin{bmatrix} 5 & -2 & 1 \\ -2 & 2 & -2 \\ -1 & -2 & 5 \end{bmatrix}$$
which has eigenvalues $\lambda_1 = \lambda_2 = 6$ and $\lambda_3 = 0$.
I want to find the eigenvectors for these eigenvalues.
I've tried to turn it into equations and trying to solve them (for $\lambda_1 = \lambda_2 = 6$)
$$ -x -2y – z = 0 \\ -2x – 4y – 2z = 0 \\ -x -2y – z = 0$$
But no matter how I try to approach this, I have no idea how to get my eigenvectors. I know from the answer that the eigenvector for $\lambda_1$ & $\lambda_2$ should be:
and eigenvector for $\lambda_3$$ should be:
But I have no idea how to arrive to that answer. Can someone explain (every step)?
Best Answer
I am assuming you meant $$A = \begin{bmatrix} 5 & -2 & 1 \\ -2 & 2 & -2 \\ -1 & -2 & 5 \end{bmatrix}.$$
Let's find the eigenvector corresponding to $\lambda = 6$. So we wish to solve $A\vec{v} = 6\vec{v}$. Suppose $$\vec{v} = \begin{bmatrix} x \\ y \\ z\end{bmatrix}.$$
Then $$A\vec{v} = \begin{bmatrix} 5x -2y + z \\ -2x + 2y -2z \\ -x -2y + 5z\end{bmatrix} = \begin{bmatrix} 6x \\ 6y \\ 6z \end{bmatrix}$$.
Thus we get the three equations: $$ -x -2y + z = 0 \\ -2x -4y - 2z = 0 \\ -x - 2y - z = 0.$$
Can you find the general solution from here?
EDIT: Since in the question you were stuck at exactly this point, I'll complete this part for you.
Note that adding the first and third equation gives $x + 2y = 0$, so $x = -2y$. Since the second equation is identical to the third (upto a factor of $2$), it is redundant. So any solution of the form $(-2y, y, z)$ will solve this system. Thus, an arbitrary eigenvector will be of the form $$ \vec{v} = \begin{bmatrix} -2s \\ s \\ t \end{bmatrix} = s\begin{bmatrix} -2 \\ 1 \\ 0\end{bmatrix} + t \begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix}.$$ Does this help?