Linear functional on $M_n(\mathbb{C})$

Question

Any linear functional on $M_n(\mathbb{C})$ is of the form $X⟼Tr(AX)$ for some $A∈M_{n}(\mathbb{C})$. Can this statement be proved with the help of the Riesz representation theorem and also by any other method?

Answer 1

The bilinear form $\varphi : \mathcal{M}_n(\mathbb{C}) \times \mathcal{M}_n(\mathbb{C}) \rightarrow \mathbb{C}$ defined by $$\varphi(A,B)=\mathrm{Tr}(AB)$$

is non-degenerated, hence it induces a canonical isomorphism $ \mathcal{M}_n(\mathbb{C}) \rightarrow \left( \mathcal{M}_n(\mathbb{C})\right)^*$.

Answer 2

If you want to use Riesz theorem, indeed $$\langle A, B\rangle=\mathrm{Tr}(A^*B)$$

is an inner product on $M_n(\mathbb C)$, the Frobenius inner product. So any linear form $h(X)$ on $M_n(\mathbb C)$ can be written as

$$h(X)=\mathrm{Tr}(A^*X).$$

Riesz representation theorem is however as mentioned in a comment an over killer as the result is quite obvious if you just write what $\mathrm{Tr}(A^*B)$ is for any matrix $A$.