To Find the Nullity of a Linear Transformation ... If $V(\Bbb R) $ be the vector space of $2\times2$ matrices and 
$$M=\begin{pmatrix}
1 & 2\\
0 & 3  \\
 \end{pmatrix}$$
If $T:V(\Bbb R)\to V(\Bbb R)$ be a linear transformation defined by $T(A)=AM-MA$ for every $A \epsilon V(\Bbb R)$.Then what is the dimension of kernel of $T$?

Here $T(I)=O$ where $I$ is the identity matrix and $O$ is the Zero matrix.Hence nullity of $T$ is $2$ since nullity of a  $2\times2$ Zero matrix is $2$. Is it true?

 A: Outline:
A matrix $A$ is in the kernel of $T$ if and only if $AM-MA=\begin{pmatrix}0&0\\0&0\end{pmatrix}$; or equivalently if and only if $AM=MA$.
So $A=\begin{pmatrix} a&b\\c&d \end{pmatrix}$ is in the kernel of $T$ if and only if
$$\begin{pmatrix} a&b\\c&d \end{pmatrix}\begin{pmatrix} 1&2\\0&3 \end{pmatrix}=\begin{pmatrix} 1&2\\0&3\end{pmatrix}\begin{pmatrix} a&b\\c&d \end{pmatrix}$$
Now multiply out each side of the above equation and equate entries of the resulting matrices.  This will give you conditions (equations) that $a, b, c, d$ must satisfy. Try to use these conditions to help solve the problem.
A: Great thing about Linear algebra is we can somewhat transform linear operator into matrix form using usual or standard basis, these type of problem can be attacked by transforming linear operator into matrix by substituting the standard basis for 2x2 matrix in the place of A and Using matrix representation method, we will get a 4x4 matrix, Row reduce that matrix to echelon form,we will get the dimension of Image space (Rank T), using the theorem dim V = Rank T + Nullity T. We can obtain the dimension of kernel space (Nullity). In our case dim V =4, Rank(T) = 2 and so Nullity = 2. 
A: More Simplest way:
$$ker(T)=\{A\in V(\mathbb R):T(A)=0\}$$
$$=\{A\in V(\mathbb R):AM=MA\}$$
Now, characteristic polynomial of $M$ is $(x-1)(x-3)$.
So, $dim(ket(T))=1^2+1^2=2$.( From this.)
A: Basis of $V(\mathbb R)$ is $$\left\{ \begin{pmatrix}
1 & 0\\
0 & 0  \\
 \end{pmatrix},\begin{pmatrix}
0 & 1\\
0 & 0  \\
 \end{pmatrix},\begin{pmatrix}
0 & 0\\
1 & 0  \\
 \end{pmatrix},\begin{pmatrix}
 0& 0\\
0 & 1  \\
 \end{pmatrix}\right\} $$
Find the matrix $T$.
Here, $$T=\left[\begin{matrix}0&0&-2&0\\2&2&0&-2\\0&0&-2&0\\0&0&2&0\end{matrix}\right]$$ Then find the  eigen values of $T$.
Number of zero eigen values of $T$ is  $N(T)$ , i.e. dimension of $ker(T).$
Clearly,  only one eigen value is zero ( as $rank(T)=2$). So, dimension of $ker(T)$ is $2$.  
