Dimension of $W_{2}$? Let $A = \begin{bmatrix} 1 & -1 & -5 & 1 & 4\\ -1 & 2 & 8 & -3 & -4\\ 3 & -1 & -9 & 0 & 4 \\ 2 & 2 & 2 & -5 & -10\\ 0&-3&-9&5&13\end{bmatrix}$
Now we define the subspace $W_{1},W_{2}$ of $A$ as follows - 
$W_{1} = \{X \in M_{5 \times 5}| AX = 0\}$
$W_{2} = \{Y \in M_{5 \times 5} | YA =0\}$
I can see that $W_{1}$ is the nullspace of $A$
using rank nullity theorem I got Nullity of $A$ as $2$ since we have the rank of matrix $A$ to be 3.
Now I am thinking about the dimension of $W_{2}$?
As from the comments and we know that row rank = column rank, hence dim$(W_{2}) = 2$
But Now I am thinking about the dimension of $W_{1} \cap W_{2}$ and $W_{1} + W_{2}$?
Any ideas?

Let us first look at 
\begin{align*}
W'&= \{X\in M_{5\times 1}; AX=0\}\\
W''&= \{Y\in M_{1\times 5}; YA=0\}
\end{align*}
In the other words we look at similar equations but with column/row vectors instead of matrices.
By a direct computation you can get that $\operatorname{rank}A=4$, which implies that $\dim(W')=\dim(W'')=1$. You can also compute that $W'$ is the span of the column vector $\vec a=(2,-3,1,0,0)^T$ and that $W''$ is the span of the row vector $\vec b^T=(5,0,-1,-1,2)$.
If we denote the columns of the matrix $X$ as $\vec c_1,\dots,\vec c_5$ then we have
$$AX = A\begin{pmatrix} \vec c_1 & \vec c_2 & \ldots & \vec{c_5} \end{pmatrix} = \begin{pmatrix} A\vec c_1 & A\vec c_2 & \ldots & A\vec{c_5} \end{pmatrix} = \begin{pmatrix} \vec 0 & \vec 0 & \ldots & \vec 0 \end{pmatrix}.$$
I.e., each of the columns fulfills the condition $A\vec c_i=\vec 0$. So we see that the matrices in $W_1$ are precisely those matrices where each column is a multiple of $\vec a$.
Similarly, we get for the rows of the matrix $X\in W''$ the condition $\vec r_i^TA=\vec 0^T$, and $W_2$ consists of those matrices where each row is multiple of $\vec b$.
We get that
\begin{align*}
W_1&=\{
\begin{pmatrix}
2a & 2b & 2c & 2d & 2e \\
-3a & -3b & -3c & -3d & -3e \\
a & b & c & d & e \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
\end{pmatrix}; a,b,c,d,e \in \mathbb R\}
\\
W_2&=\{
\begin{pmatrix}
5s & 0 & -s & -s & -2s \\
5t & 0 & -t & -t & -2t \\
5u & 0 & -u & -u & -2u \\
5v & 0 & -v & -v & -2v \\
5w & 0 & -w & -w & -2w \\
\end{pmatrix}; s,t,u,v,w \in \mathbb R\} 
\end{align*}
And we also see that $\dim(W_1)=\dim(W_2)=5$.
Now the matrices in the intersection $W_1\cap W_2$ are precisely the matrices which can be expressed in both ways.
$$\begin{pmatrix}
2a & 2b & 2c & 2d & 2e \\
-3a & -3b & -3c & -3d & -3e \\
a & b & c & d & e \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
\end{pmatrix}=
\begin{pmatrix}
5s & 0 & -s & -s & -2s \\
5t & 0 & -t & -t & -2t \\
5u & 0 & -u & -u & -2u \\
5v & 0 & -v & -v & -2v \\
5w & 0 & -w & -w & -2w \\
\end{pmatrix}
$$
Those are precisely the multiples of
$$
\begin{pmatrix}
10& 0 &-2 &-2 &-4 \\
-15& 0 & 3 & 3 & 6 \\
5 & 0 &-1 &-1 &-2 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
\end{pmatrix}
$$
This matrix generates $W_1\cap W_2$.
We see that $\dim(W_1\cap W_2)=1$. 
From the equation
$$\dim W_1+\dim W_2=\dim(W_1+W_2)+\dim(W_1\cap W_2)$$
we can calculate that  $\dim(W_1+W_2)=9$. 
