What is the dimension of the eigenspace corresponsing to eigenvalue lambda = 9? Another homework problem is asking me to find the dimension of the eigenspace corresponding to lambda = 9. From what I understand, I need to subtract original matrix A from 9 times a 4x4 identity matrix, but I'm seeking to understand what to do after that because my matrix has the number 9 in every column:
$$
    \begin{pmatrix}
    3-\lambda &amp; 0 &amp; 0 &amp; 0 \\
    3 &amp; 9-\lambda &amp; 0 &amp; 0 \\
    9 &amp; 0 &amp; 9-\lambda &amp; 0 \\
    3 &amp; 3 &amp; 9 &amp; 3-\lambda \\
    \end{pmatrix}
$$
What should I do now? Am I even on the right track?
Well...you're sort of on the right track. Plug in $9$ for $\lambda$ in your matrix and you get
$$
A - 9I = 
    \begin{bmatrix}
    -6 &amp; 0 &amp; 0 &amp; 0 \\
    3 &amp; 0 &amp; 0 &amp; 0 \\
    9 &amp; 0 &amp; 0 &amp; 0 \\
    3 &amp; 3 &amp; 9 &amp; -6 \\
    \end{bmatrix}
$$
And now you have to compute the dimension of the nullspace of that matrix (or you could compute the rank of that matrix and use the rank-nullity theorem).
$$(A-\lambda I)v=0$$
$$\begin{pmatrix}
    -6&amp; 0 &amp; 0 &amp; 0 \\
    3 &amp; 0 &amp; 0 &amp; 0 \\
    9 &amp; 0 &amp; 0 &amp; 0 \\
    3 &amp; 3 &amp; 9 &amp; -6 \\
    \end{pmatrix}\pmatrix {v_1 \\v_2 \\v_3 \\v_4}=\pmatrix {0\\0\\0\\0}$$
You deduce that:
$$
\begin{cases}
v_1=0 \\
v_2+3v_3-2v_4=0
\end{cases}
$$
So that we have :
$$v=\left(0,v_2,v_3,\dfrac 12 v_2+\dfrac 32 v_3\right)^t$$
$$v=c_1\left(0,2,0,1\right)^t+c_2\left(0,0,2,3\right)^t$$
You can deduce the dimenson of the eigenspace now.