Question is to check that :
If $A$ is an $n\times n$ matrix over a field $F$ and $AB\neq 0$ for any non zero matrix $B_{n\times n}$ over $F$ then, $A$ is invertible.
This does make some sense to me but i am not sure how to prove this.
As $AB\neq 0$ for any $B$ , in particular, we have $A.A\neq 0$ i.e., $A^2\neq 0$
for similar reasons we see that $A^n\neq 0$ for any positive integer $n$
So, $A$ is not nilpotent… I see that this is just nilpotent…
I am stuck to prove that $A$ is invertible.
I do have some thoughts inbetween but nothing gives me simple way to conclude final result.
please help me to see this by giving some hints (I am sure this must be very easy)
Thank you
Best Answer
Yet another proof. Consider the linear map $L_A : M_n(F)\to M_n(F)$ defined by $$\forall B\in M_n(F)\;:\; L_A(B)=AB\, .$$ The assumption means that $\ker (L_A)=\{ 0\}$, i.e. $L_A$ is 1-1. Since $M_n(F)$ is finite-dimensional, it follows that $L_A$ is invertible. In particular, one can find $B\in M_n(F)$ such that $AB=Id$; so $A$ is invertible.