Let $A$ be an $n \times n$ symmetric matrix. If for all singular $n \times n$ matrices $C$, $AC=0$. Then does $A=0$? I think this is true as I tried to come up with a counterexamples but couldn't.
Is there a way to prove it? I can't think of a proof.
Thanks.
 A: If $n = 1$, the only singular matrix is $C = 0$ but then any matrix $A$ will satisfy $AC = 0$ and not only $A = 0$.
For $n > 1$, denote by $e_1,\dots,e_n$ the standard basis of $\mathbb{F}^n$. Choose $1 \leq i \leq n$ and consider a matrix $C$ whose columns are the same and equal to $e_i$. This is a singular matrix (it has rank $1 < n$) and $AC$ is a matrix whose columns are $Ae_i, \dots, Ae_i$. By assumption, $Ae_i = 0$ and since this is true for all $1 \leq i \leq n$, we get $A = 0$. Note that we haven't used the fact that $A$ is symmetric.
A: For $n=1$, $\boldsymbol{c}=\boldsymbol0$ hence $\boldsymbol{a} = \boldsymbol0$.
For $n=2$, have you tried $$\boldsymbol C = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix},\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix},\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix},\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}?$$
Can we generalize this to the more general case?
A: There is a Basis $C_1,C_2,..., C_{n^2}$ of $M_{n \times n}$ such that each $C_j$ is singular. For the identity matrix $I$ we have
$I=\sum_{k=0}^{n^2}s_kC_k$ for some scalars $s_k$.
Then $A=\sum_{k=0}^{n^2}s_kAC_k=0$
(symmetry was not needed)
