If $BNA=N$ it can happen that $B$ and $A$ identities? Consider two matrices $A\in GL_m(K)$, $B\in GL_n(K)$ such that for any $N\in M_{n,m}(K)$ is true that
$$BNA=N$$
How can I prove that $B$ and $A$ are identities?
 A: This is not true. For example, if we take $n = m$, $A = 2I$, and $B = \frac{1}{2}I$, then $BNA = N$ for all $N$.
There's nothing special about the assumption $n = m$ here. For example, if $m = 2$ and $n = 1$, $$\left(\matrix{2&0\\0&2}\right)\left(\matrix{a\\b}\right)\left(\matrix{1/2}\right) = \left(\matrix{a\\b}\right).$$
A: It is clear that neither $A$ nor $B$ can be zero.
Assuming summation notation, and choose $N$ so that only the $m,n$-th element is $1$ and others all $0$, we have
$$B_{ij}\delta_{jm}\delta_{nl}A_{kl}=\delta_{im}\delta_{ln}$$
$$B_{im}A_{nl}=\delta_{im}\delta_{jm}$$
Now set $i\ne m$, we have
$$B_{im}A_{nl}=0 \,\, \forall n,l$$
which implies
$$B_{im}=0$$
and hence $B$ is diagonal.
Similarly, we can argue that $A$ is diagonal.
For $i=m,l=n$, we have
$$B_{ii} A_{ll}=1 \,\, \forall i,l$$
This can hold only when $A=cI_m$ and $B=\frac{1}{c}I_n$ for some $c\ne 0$.
A: By vectorizing we obtain:
$$(B^T\otimes A - I) \times vec(N) = 0$$
This condition must be true for any $N$, thus $B^T\otimes A = I$. This is possible iff $A = \sigma I_m$, $B = \frac{1}{\sigma}I_n$ for any $\sigma \neq 0$.
