Isomorphism with matrices I have the following linear map. 
$T: M_{n*m}(F) \to M_{m*n}(F)$ such that $X \to AXB$
where $X \in M_{n*m}(F)$ and $A$ and $B$ are fixed matrices in $M_{m*n}(F)$
I am supposed to show that if $m \neq n$, then T is not an isomorphism. 
I have been thinking that when you try to prove injectivity, I have to use that inverses of $A$ and $B$ only exists when $m = n$. 
This makes me think that this has to do with whether matrices are invertible or not, but I am not sure how to take this further. How should I approach further? 
Thanks
 A: We can write this as the composition of two maps in two different ways:
$$ M_{n,m} \stackrel{A\cdot -}{\to} M_{m,m} \stackrel{-\cdot B}{\to} M_{m,n}$$
$$ M_{n,m} \stackrel{-\cdot B}{\to} M_{n,n} \stackrel{A\cdot -}{\to} M_{m,n}$$
If $m < n$, the first map of the second line is a linear map from a vector space of strictly larger dimension to a vector space of strictly smaller dimension, and therefore must have a nontrivial kernel. Therefore the composite map has a nontrivial kernel, and cannot be injective.
If instead $n < m$, the first map of the first line is again a linear map from a vector space of strictly larger dimension to a vector space of strictly smaller dimension, and the same logic applies.
A: Suppose $m>n$; then $A$ (which is $m\times n$ has rank at most $n$; hence its null space is nonzero; let $x\ne0$ be such that $Ax=0$ and form the matrix
$$
X=[\underbrace{\begin{matrix}x&x&\dots&x\end{matrix}}_{\text{$m$ columns}}\,]
$$
Then $X\ne0$ and $AXB=0$.
For the case $m<n$ consider $B$ and do similarly (you may want to consider the null space of $B^T$).
