Prove or disprove that Y = AX-C

Let $A$ be an $m \times n$ matrix such that $\mathrm{rank}(A) = n \le m$. Prove, or disprove using a counter example:

Every $m\times n$ matrix $Y$ has a decomposition $Y = AX-C$, where $X$ and $A^TC$ are nonnegative semidefinite symmetric matrices.

I have a feeling this is false, but I can't come up with a good counter example to show this. I've been trying to find a $Y$ such that $X$ would have to be non semidefinite to construct this $Y$.

Is it possible that the statement is true?

This is a problem from a book on semidefinite optimization.

Any insights would be greatly appreciated.

Thanks

• @MichaelC.Grant: I don't see anything wrong with fixing a matrix and asking if every matrix has a decomposition in terms of the first one and and some others. Is that what was intended, user? – Eric Stucky May 16 '13 at 2:37
• My suspicion is that something is missing from the statement. Care to point us to the reference you're talking about? I do semidefinite programming as part of my work, I might be familiar with the text. – Michael Grant May 16 '13 at 2:51

Let $A$ be the $n\times n$ identity matrix. Then the decomposition reduces to $X-C$, where both $X$ and $C$ are symmetric. Clearly no asymmetric $Y$ has such a decomposition.