Question: let $M_2(\mathbb{R})$ be the set of $2×2$ real matrices. Let $A ∈M_2(\mathbb{R})$ be of trace $2$ and determinant $-3$. Consider the linear transformation $T:M_2(\mathbb{R}) → M_2(\mathbb{R})$ defined by $T(B) = AB$.
Then my professor has written that, $T$ is diagonalizable if and only if $A$ is diagonalizable. How they concluded this? Please help me.
Let $\{M_1,M_2,M_3,M_4\}$ be the canonical base of $M_2(\mathbb{R})$
$A$ is diagonalizable iff $\exists D,P\in M_2(\mathbb{R}), A=PDP^{-1}$ with $D$ diagonal (in this case you can show that the diagonal terms are $\lambda_1=3$ and $\lambda_2=-1$, but that is not needed) and $P$ invertible.
Let's now consider $b=\{N_1,N_2,N_3,N_4\}=\{PM_1,PM_2,PM_3,PM_4\}$. $b$ is a base of $M_2(\mathbb{R})$ because $P$ is invertible.
We have then $T(N_1)=AN_1=PDP^{-1}PM_1=PDM_1=\lambda_1N_1$ (just write $P$ and $M_1$ and do the product if you're not convinced) Similary, $T(N_2)=\lambda_2N_2; T(N_3)=\lambda_1N_3; T(N_4)=\lambda_2N_4$
Finally, $T$ is diagonalizable, having the same eigenvalues as $A$ with twice the multiplicity.