# No invertible matrix

I'm trying to show:

If $A$ is a $n\times n$ matrix which is not invertible, then exists a matrix $n\times n$, $B$ such that: $AB=0$ with $B\neq 0$.

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It isn't true for all $A$. Specifically, it isn't true if $A$ is invertible. Are you trying to show that if $A$ is not invertible, then there is such a matrix $B$? – Brian M. Scott Apr 30 '12 at 15:48
I'm sorry, $A$ is not invertible. You right. Thanks – Hiperion Apr 30 '12 at 15:52

If $A$ is not invertible then there exists a non-zero column vector $v$ such that $Av=0$. Take $B$ to be formed of $n$ columns all equal to the vector $v$.

If $A$ is invertible then $AB=0$ implies $B=0$ because you can multiply the equality by $A^{-1}$.

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+1 because this is the best method (I just wanted to post something out of the ordinary.) – rschwieb Apr 30 '12 at 15:54
@Beni Bogosel, of course, thanks! – Hiperion Apr 30 '12 at 15:58

One short (but advanced) method to show that non-invertible matrices are zero divisors is to remember that $A$ is invertible iff its characteristic polynomial has nonzero constant term, and that matrices satisfy their minimal polynomials.

Then for a non-invertible matrix $A$ with minimal polynomial $p(x)$, $p(A)=A*q(A)=0$, where it is possible to factor out an $A$ because there is no constant term. Thus $q(A)$ is a nonzero matrix multiplying with $A$ to make zero.

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This is a very nice answer. :) – Beni Bogosel Apr 30 '12 at 15:54
You should take the minimal polynomial and not the caracteristic one as $q(A)$ might already be $0$. Take some nilpotent matrix $A$ with $A^2=0$ and $n\geq 3$. – Olivier Bégassat Apr 30 '12 at 16:00
@rschwieb, thank you very much! – Hiperion Apr 30 '12 at 16:00
@OlivierBégassat Indeed, thanks for catching the slip. – rschwieb Apr 30 '12 at 16:57

Since the most obvious answers are taken, let me show here another method.

Suppose that $AB\ne0$ for all $B\ne0$. Then the map $X\mapsto AX$, is linear and injective (because if $AX=0$ then $X=0$ by the assumption). Now, since the set of $n\times n$ matrices is a finite dimensional vector space, an injective map of the space into itself is necessarily surjective. This means in particular that there exists $B$ such that $AB=I$, i.e. $A$ is invertible.

The contrapositive of the previous paragraph then holds, and this is the required fact "$A$ non-invertible implies that there exists $B\ne0$ with $AB=0$.

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