Prove the following matrix commute with every matrix

Question

This question is from A Second Course in Linear Algebra. Some appropriate terminologies are defined as follows:

Definition 1. Let $V$ be a finite-dimensional complex inner product space over and $\phi:V\to\mathbb{C}$ a linear functional. The Riesz vector of $\phi$ is a vector $w\in V$ such that for every $v\in V$, $$\phi(v)=\langle v,w\rangle,$$ whose existence and uniqueness follows from the Riesz representation theorem and basic properties of inner products.

Definition 2. A matrix $C\in M_n(\mathbb{C})$ is a commutator if $C=AB-BA$ for some matrices $A,B\in M_n(\mathbb{C})$. By Shoda’s theorem, $C$ is a commutator if and only if $\operatorname{tr}(C)=0$.

Now the question goes as follows:

Question. Consider $M_n(\mathbb{C})$ as an inner product space under the Frobenius inner product $\langle A,B\rangle_F=\operatorname{tr}(B^*A)$. Let $\phi: M_n(\mathbb{C})\to\mathbb{C}$ be a linear functional such that $\phi(C)=0$ for every commutator $C$. Suppose that $Y$ is the Riesz vector of $\phi$. Prove that $Y$ commutes with every matrix in $M_n(\mathbb{C})$.

Once this is proven, it is trivial that $Y$ is a scalar matrix whence $\phi$ becomes a multiple of the trace functional. However I am stuck with proving this. I was thinking of $\phi(AY-YA)=0$, but the dimension of the kernel space of $\phi$ is positive, so we cannot infer $AY-YA=0$ from it.

Any suggestions will be appreciated.

Updates. Thanks a lot to those various solutions either posted or in the comments below. I agree it is somewhat easier to prove $Y$ to be a scalar matrix directly, nevertheless I am still wondering if there is any way of proving $Y$ commutes with every matrix without showing it to be scalar, as the book indicates such approach might exist by its statements. (I am not doing any homework, but simply using this book for reviewing some knowledge in matrix theory, so different solutions or ideas are highly welcomed.)

Answer 1

Here is a solution that might be more in the spirit of what the book author was after:

By assumption, $$ \phi(C)=\mathrm{tr}(Y^\ast C) $$ for all $C\in M_n(\mathbb C)$. In particular, if $C=A^\ast B-BA^\ast$, then $$ 0=\phi(C)=\mathrm{tr}(Y^\ast(A^\ast B-BA^\ast))=\mathrm{tr}((Y^\ast A^\ast-A^\ast Y^\ast)B)=\langle A Y-Y A,B\rangle_F. $$ Taking $B=A Y-Y A$, you get $\langle A Y-Y A,A Y-Y A\rangle_F=0$, hence $AY-YA=0$.

Answer 2

Let $U$ denote the kernel of the trace functional. As $U$ has codimension $1$ we have $U^\bot = [I]$. Thus $Y= \lambda I + C$ for some scalar $\lambda$ and some $C \in U$. Now $0=\phi(C)=\langle C,Y \rangle =\overline{\lambda}\langle C,I \rangle + \langle C,C \rangle = \|C\|^2$. So, $Y= \lambda I$.