Determining the eigenvectors from a reduced-row echelon matrix I am given the following matrix:
$$A = P\left(\begin{matrix} -5 & -6 & 3\\3 & 4 & -3\\0 & 0 & -2\end{matrix}\right)$$
After finding the following eigenvalues by finding the characteristic polynomial I get: 
$\lambda_1 = \lambda_2 = -2$ 
and $\lambda_3 = 1$
After finding the I now need to find the eigenvectors for $\lambda_1$ and $\lambda_2$. After putting matrix into reduced-row echelon form for $\lambda_1$:
$$\left(\begin{matrix} 1 & 2 & -1\\0 & 0 & 0\\0 & 0 & 0\end{matrix}\right)$$
I now want to find the eigenvector from this, but am I bit puzzled how to find it an then find the basis for the eigenspace (I know this involves putting it into vector form, but for some reason I found the steps to translating-to-vector-form really confusing and still do).
A step-by-step explanation on this point would be very helpful for a linear algebra newbie.
 A: Hint: If $A v_i = \lambda_i v_i$ then $(A - \lambda_i I) v_i = 0.$ You can solve fox $v_i$ by Gaussian elimination on the augmented matrix
$$\left( A-\lambda_i I \quad {\bf 0} \right).$$
For $\lambda = -2,$ we have
$$
A - (-2) I =  
\begin{pmatrix}
-3 & -6 & 3\\
3 & 6 & -3 \\
0 & 0 & -1
\end{pmatrix}
$$
And the reduced echelon form of $(A-\lambda_i I )v = 0$ is
$$
\begin{pmatrix}
1 & 2 & 0\\
0 & 0 & 1\\
0 & 0 & 0 
\end{pmatrix}
\begin{pmatrix}
x \\y \\z
\end{pmatrix}
= 0. \tag{1}
$$
This gives you a solution $z = 0, y = t, x = -2.$ i.e., the set of vectors satisfying $Av = -2v$ are of the form $t-$multiple of
$$
\begin{pmatrix}
-2 \\ 1 \\0
\end{pmatrix}
t.
$$
The's the basis.
A: For $\lambda=-2$, let $\bf{x}$$=(x_1,x_2,x_3)$ be an eigenvector corresponding to $-2$. Then, $A\bf{x}$ $=-2\bf{x}$, and multplying the matrix $A$ by $\bf{x}$ you obtain the following system of equations: $$-5x_1-6x_2+3x_3=-2x_1$$$$3x_1+4x_2-3x_3=-2x_2$$$$0x_1+0x_2-2x_3=-2x_3$$From the last equation it is clear that $x_3=0$. The last equation does not provide any information, so we go ahead and look at the other equations. We note that the second and first equation are the same equation since if you multiply one by $-1$ you get the other. Ergo you have that $$3x_1+6x_2-3x_3=0$$$$\Rightarrow x_1+2x_2-x_3=0$$and we want to write this parametrically. We have that the degree of freedom is $2$, so you have that solutions are of the form $(-2t+v,t,v)$ for any $t$ and $v$. Hence, you can for instance take $t=0$ and $v=1$ and obtain $(1,0,1)$ and then you can take $t=1$ and $v=0$ and get $(-2,1,0)$. This two vectors form a basis for $E_{-2}$ the eigenspace corresponding to the value $-2$. For the other eigenvalue, I will let you figure it out. 
Let me know if you have any questions.  
