Eigen space of $T_{A}$ where $T_{A}(v)=Av$ 
Let $T_{A}:\mathbb{C}^3\rightarrow \mathbb{C}^3$ ,$T_{A}(x,y,z)=A\begin{pmatrix}x\\ y\\ z\end{pmatrix}$ and $A=\begin{pmatrix}
1 & 5 & 0\\ 
0 & 1 & 0\\
0&0&3
\end{pmatrix}$.
I have found that the eigenvalues of $T$ are $1,3$.
When I tried finding the subspaces these eigenvalues create I couldnt understand how to continue.
What I tried:
finding 2 subspaces $W_1,W_3$ such that $W_1=\{v\in \mathbb{C}^3: T_A(v)=v\}$  $W_3=\{v\in \mathbb{C}^3: T_A(v)=3v\}$, I got 3 linear equations for each subspace, but I didnt understand the information it gave me. I need guidence for the rest.
 A: The space of solutions of $Av=\lambda v$ is the eigenspace of the
eigenvalue $\lambda$ so for example $W_{1}$ is the space of solutions
for $Av=1\cdot v=v$
$$
Av=v\iff(A-I)v=0
$$
$$
\begin{pmatrix}1-1 & 5 & 0\\
0 & 1-1 & 0\\
0 & 0 & 3-1
\end{pmatrix}\begin{pmatrix}a\\
b\\
c
\end{pmatrix}=\begin{pmatrix}0\\
0\\
0
\end{pmatrix}
$$
$$
\begin{pmatrix}0 & 5 & 0\\
0 & 0 & 0\\
0 & 0 & 2
\end{pmatrix}\begin{pmatrix}a\\
b\\
c
\end{pmatrix}=\begin{pmatrix}0\\
0\\
0
\end{pmatrix}
$$
which gives the equations 
$$
\begin{cases}
5b=0\\
0=0\\
2c=0
\end{cases}
$$
and so a general solution is $(a,0,0)=sp\{(1,0,0)\}$
So $W_{1}=sp\{(1,0,0)\}$.
Do similar work with $W_{3}$.
Note that in general that the sum of dimensions of the eigenspaces
need not be equal to $n$ (where $A\in M_{n}(\mathbb{F})$), this
happens iff $A$ is diagonalizable 
A: A linear vector space can be represented by a set of linear equations; I give you a couple of examples:
$W$ is the set of vectors for which $x+y-z=0$
If we know by other means that $W$ is a subspace of $\mathbb{R^3}$, its dimension would be 2 (a plane) because you put an arbitrary value to two variables and you can get the the third:
$x=0$
$y=1$
then $z=1$
You get the vector $(0,1,1)$, but because $\dim V=2$ we need at least another vector (independent to the first) to span $W$.
$x=1$
$y=0$
then $z=1$
and you get $(1,0,1)$ thus $W=Span((0,1,1),(1,0,1))$
Suppose now $W\subset\mathbb{R}^4$, we do the same thing, with the fourth variable (t) taking whatever value you want. So you can use "the same" already found vectors:
$(0,1,1,0)$ and $(1,0,1,0)$, now I need a third vector to span $W$ because its dimension is now 3 (there is this "hidden" t).
t can assume whatever value so a third vector could be: $(0,1,1,1)$.
So $W=span((0,1,1,0),(1,0,1,0),(0,1,1,1))$
The same thing applies when the space is described by more equations.
