Finding matrices whose multiplication does not change the trace

Question

Let $C$ be an $n\times n$ matrix. Then find matrices $A$ of size $n \times n$ such that $trace(AC)=trace(C)$. One obvious choice for $A$ is the Identity matrix $I$, but are there other choices?

A similar question is, if $A$ and $C$ are $m\times m$ matrices and $B$ and $D$ are $n\times n$ matrices such that $trace(C)=trace(D)$ and $trace(AC)=trace(BD)$, then what is the explicit relationship between $A$ and $B$? Here also we can choose $A$ and $B$ to be Identity matrices, but are there other choices of matrices?

Answer 1

The expression $\mathrm{trace}(X^*C)$ can be thought of as an inner product between matrices.

So the equation $\mathrm{trace}(AC)=\mathrm{trace}(C)$ is equivalent to $\langle A^*-I,C\rangle=0$. So $A^*=I+D$ where $D$ is any matrix 'perpendicular' to $C$. One can create a basis for this space by starting with the elementary matrices $E_1,\ldots,E_{n^2}$ and modifying them: $$D_i=E_i-\frac{\langle E_i,C\rangle}{\langle C,C\rangle} C=E_i-\frac{\mathrm{trace}(E_i^*C)}{\mathrm{trace}(C^*C)}C$$

Hence the answer is $A=I+\sum_{i=1}^{n^2-1}\alpha_i\left(E_i^*-\frac{\mathrm{trace}(E_i^*C)}{\mathrm{trace}(C^*C)}C^*\right)$ for any $\alpha_i\in\mathbb{C}$.
[For $C=0$, then any matrix $A$ is valid.]

Answer 2

The answer to this question is "find it with this possible hint" Choose $A-I=B$ a non invertible matrix such that $BC=0$ then $det(A-I)=0$. This is answer to the first question. For 2nd question try this, might work. $$ A=C=\left[\begin{array}{ll} I_{n} & O\\ O & O \\ \end{array}\right] $$ and $B=D=I_{n}$