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Suppose there are three $n \times n$ matrices $A$, $B$ and $C$. Now, I know that $B$ is an orthogonal matrix and that $ABC=0$. Does this mean $AC=0$? How can I prove it?

Thank you very much.

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    Use $B=\begin{bmatrix}0&1\\1&0\end{bmatrix}$ and look for a counterexample. – 2017-02-09

3 Answers 3

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If $ABC=0$, this means $\text{Im}(BC)\subset \ker(A)$.

Since $B$ is orthogonal, it is invertible, and it just means that $\text{Im}(BC)$ and $\text{Im}(C)$ have the same dimension.

EDIT: Expanding on the answer. The observation above indicates what, in a counter-example, one should be looking for. Our matrices must satisfy

$(1):$ $A,C\neq 0$. Otherwise, $AC=0$.
$(2):$ $\dim\big(\ker(A)\big)>0$. Otherwise, $ABC\neq 0$.
$(3):$ $\dim\big(\text{Im}(C)\big)\leq \dim(\ker(A))$. Otherwise, $ABC\neq 0$.
$(4):$ $\text{Im}(C)\not\subset\ker(A)$. Otherwise, $AC=0$.
$(5):$ $B\big(\text{Im}(C)\big)\subset\ker(A)$, so that $ABC=0$.

The basic idea is that the image of $C$ can fit inside the kernel of $A$ (in terms of 'size', or dimension to be precise), but is not a natural fit. It needs $B$ to reposition it first.

If we're cooking an example for small $n$, say $n=2$, we see that we're pretty constrained. We must have that $\dim\big(\text{Im}(C)\big)=\dim\big(\ker(A)\big)=1$. Can you think of an example?

Hint: Let $A=C$ be the projection along the $x$-axis. What could $B$ be?

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    Why the downvote? – 2017-02-09
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    I didn't downvote, but this doesn't address the question. The fact that Im(C) has the same dimension as Im(BC) doesn't have any relation to $Im(C) \subset ker(A)$, which is what the question asks. – 2017-02-09
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    Eh, I think there are different ways to see it. It was more of a hint towards what a counter-example must satisfy. Regardless, I have expanded it. And in fact, that $B$ being orthogonal is mostly inconsequential to the question at large. – 2017-02-09
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Take $$ A = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} $$ $$ B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} $$ $$ C = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} $$ Then $ABC = 0$ but $AC \ne 0$.

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Consider $$ A = \begin{bmatrix} 1 & 0 \\ 1 & 0 \\ \end{bmatrix},\quad B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}, \quad C = \begin{bmatrix} 1 & 1 \\ 0 & 0 \\ \end{bmatrix}. $$ We have $$ ABC = 0 $$ but $$ AC = \begin{bmatrix} 1 & 1 \\ 1 & 1 \\ \end{bmatrix} $$