# Basic Linear Alg Question..Sufficient?

Q: Prove that if $A$ is invertible and $AB = 0$, then $B = 0$.

A: if $A = 0$, $\nexists A^{-1} \mid AA^{-1} = I$. but it's given that $\exists A^{-1}$. Thus $B=0$.

This just seems too easy to be a sufficient answer... is it?

In my opinion I don't think it is because you can multiply a nonzero matrix that is not invertible and a nonzero matrix to get the zero matrix. So I'd think you'd need to involve the determinant here...

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"if $A=0$", why do you assume that? $A$ is invertible. –  Zircht Feb 7 '14 at 2:23
@NotMe well that's the issue, the only two matrices that are not invertible are $0$ and a matrix such that $det(A) = 0$. So im 1 for 2 here –  Anthony Peter Feb 7 '14 at 2:25
Not very sufficient proof. Nice try though! –  NasuSama Feb 7 '14 at 2:26
@NasuSama Like I said 1/2. I don't want to get too complicated by using a $n \times n$ determinant. –  Anthony Peter Feb 7 '14 at 2:27
It is not true that if $AB=0$ and $A\not=0$ then $B=0$. You need to make use of the fact that A is invertible. –  Rainbow Feb 7 '14 at 2:30

Here is a direct proof that uses only the property of $A^{-1}$ and hence is true in more general settings....

$$A B = 0 \Rightarrow A^{-1} (AB) = 0 \Rightarrow \left(A^{-1} A \right) B = 0 \Rightarrow I B = 0 \Rightarrow B=0$$

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I wasn't sure if it was sufficient. But I guess it is. –  Anthony Peter Feb 7 '14 at 2:48

Elementary proof: We have to show that $By=0$ for all $y$. $$0=0y=ABy = A(By)$$ shows that $By$ is in the kernel of $A$, which is $\{0\}$

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Since $A^{-1}$ exists we have $$AB={\bf0}\quad\Rightarrow\quad A^{-1}AB=A^{-1}{\bf0}\quad\Rightarrow\quad IB={\bf0}\quad\Rightarrow\quad B={\bf0}\ .$$

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This was my orignial proof –  Anthony Peter Feb 7 '14 at 2:47

First, we need to have $A$ to be a square matrix. Invertibility does not apply for non-square matrices. This is why true statements, like theorems, need to have careful conditions to avoid the possibility of the "counter-proof" or counter-example.

Let's start the proof...

We assume that $A$ is invertible and $AB = 0$. We want to show that $B = 0$.

Since $A$ is invertible, we know that $\det(A) \neq 0$. Given that $AB = 0$, we can set both sides by $\det$ to get

\begin{aligned} \det(AB) &= \det(0) = 0 \end{aligned}

Since $\det(AB) = \det(A)\det(B)$,

$$\det(A)\det(B) = 0$$

Thus, since $\det(A) \neq 0$, $\det(B) = 0$. Let's see if you can figure out the last part by yourself. Note: $B$ is not really zero since $B$ is an arbitrary matrix such that $AB = 0$.

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$det(B)=0$ does not imply $B=0$.. –  Rainbow Feb 7 '14 at 2:34
Thanks for your thought! ;) –  NasuSama Feb 7 '14 at 2:35
It is actually true in this case that $B=0$. You can multiply by $A^{-1}$ on the left for both sides and you get $B=0$. –  Rainbow Feb 7 '14 at 2:37
This reasoning has nothing to do with the question. –  Artem Feb 7 '14 at 2:50