1

Let $A$ and $B$ be two square matrices such that $\matrix {AB}=\matrix 0$, where $\matrix 0$ denotes the Null Matrix.

My textbook mentions that [without proof]: if $\matrix A\not= \matrix 0$, then $|\matrix B|=0$.

I am not sure if this is even correct:

Consider the scenario: $|\matrix B|=k$, where $k$ is a non-zero constant and $|\matrix A|=0$ with $\matrix A\not =\matrix 0$, then this above property predicts that the product $\matrix {AB}$ cannot be $\matrix 0$. Though I could not find counter-examples, this seems improbable.

  • Is this property true?
  • If not, can you provide counter-example of the kind I mentioned?
Truth-seek
  • 1,427
  • 6
    If $det(B)\neq 0$, then $B$ is invertible. $AB=0$ implies that $A=0$. Contradiction. – ntt Apr 27 '17 at 08:53
  • If B has an inverse you could right multiply to get A=0. Thus B can't have an inverse. – user121049 Apr 27 '17 at 08:54
  • In fact, for any $n$-by-$n$ matrix over a fixed field $B$, there exists an $m$-by-$n$ matrix $A$ such that $AB=0$ if and only if $\det(B)=0$. In general, for any $n$-by-$r$ matrix $B$, there exists an $m$-by-$n$ matrix $A$ such that $AB=0$ if and only if $\text{rank}(B)<n$. – Batominovski Apr 27 '17 at 09:13

3 Answers3

5

Note that saying $|B|=0$ is very different from saying $B=0$ (assuming that $|B|$ denotes the determinant of $B$).

If $\det B\ne0$, then $B$ is invertible, so from $AB=0$ you get $$ A=ABB^{-1}=0B^{-1}=0 $$

On the other hand it is easy to find nonzero matrices with zero product: $$ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} $$ but this doesn't contradict the above: indeed both matrices have zero determinant.

egreg
  • 238,574
2

We know that $dim(ker(A)) + dim(ker(B)) \geq dim(ker(AB))$.

(Sylvester rank inequality: $\operatorname{rank} A + \operatorname{rank}B \leq \operatorname{rank} AB + n$)

As $AB=0$, hence $dim(ker(AB)) = n$ (where A and B are $n \times n$ matrices.)

Hence, $dim(ker(A)) + dim(ker(B)) \geq n$.

As $A\neq0$, $dim(ker(A))<n$.

So, $dim(ker(B))>0$.

Thus, $B$ is not invertible. Hence, $|B| = 0$.

Naveen Kumar
  • 21
1

If $AB=0$ and $A\ne0$, then $aB=0$ for every nonzero row $a$ of $A$. Hence $B$ has linearly dependent rows (because $aB$ is a non-trivial linear combination of the rows of $B$) and its determinant must be zero.

Note that this proof is very basic: it works over any commutative ring (not just over fields) and the notion of determinant was devised to detect linear dependence in the first place.

user1551
  • 139,064