Dimension of Eigenspace? Given a matrix A
$ A = \begin{bmatrix} 5 & 4 & -1 \\
    4 & 5 & -1 \\
    -4 & -4 & 2 \\  \end{bmatrix} 
$
I have to find out if A is diagonalizable or not. Also I have to write down the eigen spaces and their dimension.
For eigenvalue, $ \lambda = 1 $ , I found the following equation:
$ x_1 + x_2 - \frac{x_3}{4} = 0 $
Here, I have two free variables. $ x_2 $ and $ x_3 $. I'm not sure but I think the the number of free variables corresponds to the dimension of eigenspace and setting once $ x_2 = 0 $ and then $ x_3 = 0 $ will compute the eigenspace.  
Any detailed explanation would be appreciated.
 A: Set $x_3 = 0$ to get an eigenvector $v_1 = (1, -1, 0)^T$, then set $x_2 = 0$ to get another eigenvector $v_2 = (1, 0 , 4)^T$.  To find the third eigenvalue $t$, write the characteristic polynomial as $$\det(\lambda-AI)  = (\lambda-1)^2 (\lambda - t).$$  Substitute $\lambda = 0$ in the above equation to see that $-\det A = -t$, so
\begin{align}
t =& \det A = (5)(5)(2)+(4)(-1)(-4)+(-1)(4)(-4) \\
&- (-4)(5)(-1) - (-4)(-1)(5) - (2)(4)(4) \\
=& 50 + 16 + 16 - 20 - 20 - 32 = 10
\end{align}
$$(A-10I) \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix} -5 & 4 & -1 \\
    4 & -5 & -1 \\
    -4 & -4 & -8 \\  \end{bmatrix}
\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = 0, \tag{*} \label1$$
$v_3 = (x_1,x_2,x_3) = (1,1,-1)$ solves \eqref{1}, so it's a eigenvector corresponding to eigenvalue $10$.
The eigenspace $E_1 = \mathrm{span}\{v_1,v_2\}$ corresponding to $\lambda = 1$ has dimension $2$; the eigenspace $E_{10} = \mathrm{span}\{v_3\}$ corresponding to $\lambda = 10$ has dimension $1$.
A: Ummm If you can think of only one specific eigenvector for eigenvalue $1,$ with actual numbers, that will be good enough to start with. Call it $(u,v,w).$ It has a dot product of zero with $(4,4,-1.)$  We would like a second one. So, take second eigenvector $(4,4,-1) \times (u,v,w)$ using traditional cross product. 
