0
$\begingroup$

Prove that if $AB=0$ then either $A$ or $B$ is null matrix or both are singular.
I am unable to submit my working as i have no udea how to proceed. I can only think of taking mod and saying $|AB|=0$ . I am able to understand the null matrix part as all elements would be $0$ in $AB$. Can someone help me in the second part?

  • 0
    Please use elementary methods as I have very little knowledge of the topic. – 2017-01-05
  • 0
    1. The question's title is not clear enough. 2. Please use Mathjax. – 2017-01-05
  • 0
    Possible duplicate of [Showing AB=0 does not imply either A,B=0, but that singular](http://math.stackexchange.com/questions/783593/showing-ab-0-does-not-imply-either-a-b-0-but-that-singular) – 2017-01-05
  • 0
    It didn't properly address the condition that both can be singular( or it may have, maybe i didn't understand) – 2017-01-05

2 Answers 2

1

We can assume that neither $A$ and $B$ is the null matrix. What we need to prove is that

If $AB = 0$, then both $A$ and $B$ is singular.

This is the same as proving

If one of $A$ and $B$ is not singular, then $AB\neq 0$.

This should be easier to prove.

Let's say that $A$ is not singular.

Then,

  • take some $x$ such that $Bx\neq 0$ (why does this $x$ exist?).
  • What is $ABx$? (is it $0$?)
  • Can you therefore conclude that $AB\neq 0$?

Now, assume that $B$ is not singular. Since you know there exists some $y$ such that $Ay\neq 0$ (again, why?), can you find an $x$ such that $(AB)x\neq 0$?

Hint: Remember, $(AB)x = A(Bx)$.

1

It is equivalent to proving that, if $AB=0$ and one of $A,B$ is non-singular, the other matrix is $0$.

Suppose $A$ is non-singular. Then it is invertible and left-multiplying both sides by $A^{-1}$ yields $$0=A^{-1}(AB)=(A^{-1}A)B=IB=B.$$ Similarly, if $B$ is non-singular, you right-multiply by $B^{-1}$ to ge $A=0$.