Let $B\in M_{2}(\mathbb{R})$ and let $T:M_{2}(\mathbb{R})\rightarrow M_{2}(\mathbb{R})$ defined by $T(A)=BA$. Find the rank of $T$ Let $B\in M_{2}(\mathbb{R})$ and let $T:M_{2}(\mathbb{R})\rightarrow M_{2}(\mathbb{R})$ defined by $T(A)=BA$. Find the rank of $T$
$\text{Im}(T)=\left\{ w\in M_{2}(\mathbb{R}):T(v)=w, \forall v\in M_{2}(\mathbb{R})\right\} $
$\dim(\text{Im}(T))= \text{rank} (T)$
Now, Let
$B=\begin{pmatrix}b_{1} & b_{2}\\
b_{3} & b_{4}
\end{pmatrix}$
$A=\begin{pmatrix}a_{1} & a_{2}\\
a_{3} & a_{4}
\end{pmatrix}$
Then, $BA=\begin{pmatrix}b_{1}a_{1}+b_{2}a_{3} & b_{1}a_{2}+b_{2}a_{4}\\
b_{3}a_{1}+b_{4}a_{3} & b_{3}a_{2}+b_{4}a_{4}
\end{pmatrix}=\begin{pmatrix}\alpha_{1} & \alpha_{2}\\
\alpha_{3} & \alpha_{4}
\end{pmatrix}$ with $\alpha_{i}\in M_{2}(\mathbb{R})$ and $1\leq i\leq4$
$\Rightarrow\begin{cases}
b_{1}a_{1}+b_{2}a_{3}=\alpha_{1}\\
b_{1}a_{2}+b_{2}a_{4}=\alpha_{2}\\
b_{3}a_{1}+b_{4}a_{3}=\alpha_{3}\\
b_{3}a_{2}+b_{4}a_{4}=\alpha_{4}
\end{cases}$
In this step i'm stuck, what am I supposed to clear to get the rank of $T$?

First we find the matrix $\ T$ in respect to standard basis of $\ M_{2x2} $
Let $\ B=\pmatrix{a&b\\c&d}$
$\ T(\pmatrix{1&0\\0&0})=\pmatrix{a&0\\c&0}$
$\ T(\pmatrix{0&1\\0&0})=\pmatrix{0&a\\0&c}$
$\ T(\pmatrix{0&0\\1&0})=\pmatrix{b&d\\0&0}$
$\ T(\pmatrix{0&0\\0&1})=\pmatrix{0&b\\0&d}$
Then to the respect to the standard basis we form a matrix $\ T$
$\ T=\pmatrix{a&0&b&0\\0&a&b&d\\c&0&0&0\\0&c&0&d}$ then in the respect to $\ a,b,c,d$ we find the $\ rankT$
For $\ a,b,c,d\neq0\ rankT=4$
Cases where they are $\ 0$ I leave up to you to calculate the rank of T
