Let $B\in \mathbb{R}^{n\times n} $ and $T:\mathbb{R}^{n\times > n}\rightarrow\mathbb{R}^{n\times n}$ linear transformation , $T(A)=BA$ for every $A\in \mathbb{R}^{n\times n}$ prove\disprove that if $T$ is not isomorphism then $\det(B)=0$
I thinking about that if $T$ is isomorphism then $T$ is invertible then $\det(BA)\ne0$ then if $T$ is not invertible then $\det(BA)=0$ and that happens only if $\det(B) = 0$ so the claim is correct
Does this prove it? If not, any hint how to solve this question?
If $T$ is an isomorphism, then there is a matrix $A\in\mathbb R^{n\times n}$ such that $T(A)=\operatorname{Id}$. This means that $BA=\operatorname{Id}$. But then$$1=\det\operatorname{Id}=\det(BA)=\det(B)\det(A)$$and therefore $\det(B)\neq0$.
And if $\det B\neq0$, then $T$ is an isomorphism because then, if $A\in\mathbb R^{n\times n}$,$$T(B^{-1}A=BB^{-1}A=A.$$