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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

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    $\begingroup$ Try from the other end. If $A$ is not invertible, can you show the existence of a $B\neq 0$ with $AB = 0$? $\endgroup$ Commented Sep 28, 2013 at 11:45
  • $\begingroup$ @DanielFischer :) Yes Yes.. I just now realized :) $\endgroup$
    – user87543
    Commented Sep 28, 2013 at 11:53

5 Answers 5

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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.

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  • $\begingroup$ please edit this as "another elegant proof" :) Nicely done.. Thank YOu :) :) $\endgroup$
    – user87543
    Commented Sep 28, 2013 at 12:01
  • $\begingroup$ This is a great proof, and shows the important of linear transformations as a tool for studying vector spaces (and matrices). $\endgroup$
    – Dan Rust
    Commented Sep 28, 2013 at 12:26
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Remember that $A$ is not invertible iff the equation $Ax=0$ has non-trivial solutions. Can you now construct a non-zero $B$ with $BA=0$?

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  • $\begingroup$ :) Thank you.... it is done now.... :) $\endgroup$
    – user87543
    Commented Sep 28, 2013 at 11:59
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Suppose $A$ isn't invertible. It follows that $(0,v)$ is an eigenpair of $A$, for some non-null $n\times 1$ vector $v$, that is, $Av=0_{n\times 1}$. Now consider the matrix $B$ whose columns are all equal to $v$.

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    $\begingroup$ :) Thank you.... it is done now.... :) $\endgroup$
    – user87543
    Commented Sep 28, 2013 at 11:52
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Suppose $A=(a_{ij})_{n \times n}$ is not invertible. Then the vectors $A_{i}=(a_{i1}, \cdots, a_{in})(1 \le i \le n)$ are not linearly independent. This means that there is a $k$ -dimensional vector space $L$ with $k<n$ such that all $A_i (1 \le i \le n)$ are in $L$. Let $b=(b_1,b_2, \cdots, b_n)$ be a non-zero vector being orthogonal to $L$. Let consider matrix $B$ whose each column coincides with $b$. Then $A \times B=0$ and we get a contradiction.

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Let V be a vector space of all square matrices of size n then we can define a operator T on V such that T(B) = AB and if AB not equal to zero for any non-zero matrix B this implies that the Operator T(B) = AB is non singular i.e. one - one . Here T is on finite dimensional Vector space so it must be onto hence identity matrix I must have pre-image in V i.e. there exist a non zero matrix M in V such that T(M) = AM = I.

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