need to find the rank of $T$ $B=\begin{pmatrix}2&-2\\-2&4\end{pmatrix}$ , $T:M_2(\mathbb{R})\to M_2(\mathbb{R})$ defined by $T(A)=BA$, we need to find the rank of $T$, I have calculated by hand and got rank is $4$, could any one tell me if there is any process with out calculating that?
 A: We have $\det B = 4$, so $B$ is invertible.  Given $A \in M_2(\mathbb{R})$, $B^{-1}A \to A$, so $T$ is onto $\implies \operatorname{rank}T = 4$.
A: Yet another approach:
$B$ is symmetric and hence we can choose orthogonal eigenvectors $v_1,v_2$.
It is straightforward to check that the dyads $v_i v_j^T$ are linearly independent, and $T (v_i v_j^T) = \lambda_i v_i v_j^T$, where $\lambda_i$ is the eigenvalue of $B$ corresponding to $v_i$.
Since $\lambda_i \neq 0$, we see that all of the above dyads are in the range of $T$, hence it has rank 4.
Alternatively, we note from the above that the eigenvalues of $T$ are $\lambda_1, \lambda_1, \lambda_2, \lambda_2$,and so $\det T = (\lambda_1)^2 (\lambda_2)^2 = (\det B)^2 = 16$. Hence $T$ is invertible, and so has rank $4$.
A: Yes. $B$ is invertible. Hence $T$ is invertible and has full rank, i.e. $\operatorname{rank}(T)=4$.
A: $T$ is a linear map of spaces of the same dimension, and it is injective (an inverse is given by multiplying by $B^{-1}$ on the left) and therefore an isomorphism. So it is surjective, hence rank 4.
