I have the matrix \begin{bmatrix}1&0&0\\2&2&-1\\0&1&0\end{bmatrix}
I know that the only eigenvalue is 1 with multiplicity 3
I solved for the first eigenvalue and got \begin{bmatrix}0\\1\\1\end{bmatrix}
How do I find the other two? I know they are \begin{bmatrix}0\\1\\0\end{bmatrix} and \begin{bmatrix}1/2\\0\\0\end{bmatrix}
but when I do $(A-\lambda I)v_2 = v_1$, I get the system of equations $2x + y -z = 1$,
$y -z =1$. I don't see how that gives the second eigenvector.
Thanks
Best Answer
All you need is just to solve the system. The system of equations $$ \begin{cases} 2x+y-z=1,\\ \qquad y-z=1 \end{cases} $$ has the solution $$ \begin{bmatrix} x\\y\\z \end{bmatrix}=\begin{bmatrix} 0\\1+t\\t \end{bmatrix},\quad t\in\Bbb R. $$ Now take $t=0$.
The second vector is obtained similarly: solve $$ \begin{cases} 2x+y-z=1,\\ \qquad y-z=0 \end{cases} $$ to get $$ \begin{bmatrix} x\\y\\z \end{bmatrix}=\begin{bmatrix} 1/2\\-t\\t \end{bmatrix},\quad t\in\Bbb R $$ and set $t=0$.
P.S. Those other two vectors are not eigenvectors. They are called generalized eigenvectors.