1. ## Eigenvector

I have
1 1 0
1 1 0
0 0 -1
And i need to find the eigenvalue and eigenvector.
I found the eigenvalue 0, 2 and -1.
then i took B+I
2 1 0
1 2 0
0 0 0
that after beeing reduced is
1 0 0
0 1 0
0 0 0
How can i find the eigenvector from this? its the general solution?

2. ## Re: Eigenvector

First, your eigenvalues, -1, 0, and 2, are correct. Of course the definition of "eigenvalue", $\displaystyle \lambda$, (of matrix B) is that there exist a non-zero vector v such that $\displaystyle Bv= \lambda v$ which is the same as $\displaystyle Bv- \lambda v= (B- \lambda I)v= 0$.

Assuming the matrix you give is B (you never actually say that) then, for eigenvalue -1, yes, $\displaystyle B- \lambda I= B+ I= \begin{bmatrix}2 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 0 \end{bmatrix}$ and you want to find non-zero $\displaystyle \begin{bmatrix} x \\ y \\ z\end{bmatrix}$ such that

$\displaystyle \begin{bmatrix}2 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} x \\ y \\ z\end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 0\end{bmatrix}$

Of course that is the same as solving the system of equations 2x+ y= 0, x+ 2y= 0, 0= 0.
From the first equation, y= -2x so the second equation can be written x+ 2(-2x)= 5x= 0 which gives x= 0 and then y= 0. But we have no equation z must satisfy so we can take z to be any non-zero number, say z= 1. That gives eigenvector $\displaystyle \begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}$, or any non-zero multiple, as eigenvector corresponding to eigenvalue -1.
Doing it your way, row-reducing the matrix, would be the same as using the "augmented matrix" to solve the system of equations. Of course, since the last column, the "augmented" part, is all 0s, you don't need to write that:
$\displaystyle \begin{bmatrix}2 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 0\end{bmatrix}$
That reduces, as you say, to $\displaystyle \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\0 & 0 & 0\end{bmatrix}$
with the fourth, "augmented" column, still 0. So it corresponds to the equations x= 0, y= 0, 0= 0. The fact that the last column is all 0s tells us z can be anything (non-zero to give a non-zero vector, of course).

Now, try eigenvalues 0 and 2.