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Suppose we have $n\times n$ matrices $A, B$, and there is no information about singularity of $A, B$.

Also $C$ is a $n\times n$ projection matrix, i.e. singular matrix and it is not the zero metrix.

If $AC=BC$, can we say that $A=B$?

All comments would be appreciated.

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    Only if C is invertable. Then $ACC^{-1} = BCC^{-1}$. But a fundamental point of matrices is that not all matrices are invertable. – 2017-02-21

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Take $$A = \begin{pmatrix}0 & 0 \\ 0 & 0\end{pmatrix}, B = \begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}, C = \begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix}.$$

Another example: $$A = \begin{pmatrix}0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{pmatrix}, B = \begin{pmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{pmatrix}, C = \begin{pmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix}.$$

To make this more general: Let $C$ be a projection with nullspace at least 2 dimensional. Then Take for $A$ and $B$ projections on different subspaces of the nullspace.

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    Thank you for your comment. But, $A$ is also non-zero matrix. – 2017-02-21
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    @beginner there are many counter examples., try a bit longer to find one yourself. – 2017-02-21
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    in that case can you spot an A that is not a zero matrix – 2017-02-21
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    It is greatly appreciated!! Thank you so much for all comments. – 2017-02-21
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Here's one source of matrices $A$ and $B$: reflections in $\Bbb R^2$ can be achieved as half-rotations (that is, $180^\circ$ rotations) in $\Bbb R^3$ (and more generally, reflections in $\Bbb R^n$ can be achieved as rotations in $\Bbb R^{n+1}$).

As an example, consider the reflection (in $\Bbb R^2$) across the $y$-axis (the line $x = 0$). We can find two $3 \times 3$ matrices that do the same thing to the $xy$-plane $\{(x, y, 0):x, y \in \Bbb R \}$.

Try and find the matrix $A$ that gives the reflection across the $yz$-plane, and the matrix $B$ that gives the half-rotation in the $xz$-plane (so about the $y$-axis).

Then, if $C$ is the matrix that projects the point $(x, y, z)$ to $(x, y, 0)$, we have $AC = BC$, since both $A$ and $B$ perform the same action on the plane $xy$-plane.

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You should think in terms of linear maps.

If $a,b,c$ are endomorphisms of $\mathbb{R}^n$, canonically associated to the matrices $A,B,C$, then $AC=BC$ is equivalent to $(a-b)\circ c=0$.

Hence if $a,b$ are chosen in such a way that $\mathrm{im}(c)\subset\ker(a-b)$, we will have $(a-b)\circ c=0$ and hence $AC=BC$.