Multiplication of orthogonal matrix 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.
 A: 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 $AC \ne 0$.
A: 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
$$
AC = \begin{bmatrix}
1 & 1 \\
1 & 1 \\
\end{bmatrix}
$$
A: 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?
