Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $B\neq 0$.

Thanks for your help.

share|cite|improve this question
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
up vote 6 down vote accepted

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 $A^{-1}$.

share|cite|improve this answer
+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.

share|cite|improve this answer
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 $AB=0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.