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...
A: Here is a direct proof that uses only the property of $A^{-1}$ and hence is true in more general settings....
$$
A B  = 0 \Rightarrow A^{-1} (AB) = 0 \Rightarrow \left(A^{-1} A \right) B = 0 \Rightarrow I B = 0 \Rightarrow B=0$$
A: Since $A^{-1}$ exists we have
$$AB={\bf0}\quad\Rightarrow\quad A^{-1}AB=A^{-1}{\bf0}\quad\Rightarrow\quad
  IB={\bf0}\quad\Rightarrow\quad B={\bf0}\ .$$
A: 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 $AB = 0$.
A: 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 $\{0\}$
