$f:M(n,\mathbb R) \to \mathbb R$ be , then $\exists ! C \in M(n,\mathbb R)$ such that $f(A)=Trace (AC) , \forall A \in M(n,\mathbb R)$?

Question

Let $f:M(n,\mathbb R) \to \mathbb R$ be a linear function , then does there exist a unique $C \in M(n,\mathbb R)$ such that $f(A)=Trace (AC) , \forall A \in M(n,\mathbb R)$ ?

Answer 1

The bilinear form $B(X, Y)=\operatorname{Tr}(XY^T)$ defines an inner product on the space $M(n , \mathbb{R})$ as you can easily check.

Thus (by https://en.wikipedia.org/wiki/Riesz_representation_theorem Riesz Representation Theorem) any linear functional $f\colon M(n, \mathbb{R})\to \mathbb{R}$ is given by the inner product with respect to a choice of a unique vector $Y_f$. That is, $$f(X)=B(X, Y_f)=\operatorname{Tr}(XY_f^T).$$

Denoting $Y_f=C$ you get the desired result.

Answer 2

Yes, this is true (assuming that $M(n, \mathbb R)$ is the space of all matrices $\mathbb R^{n \times n}$).

Let $E^{ij} \in M(n, \mathbb R)$ be the matrix with entries $(E^{ij})_{lm} := \delta_{il} \delta_{jm}$, i.e. with entry $1$ at position $(i, j)$ and $0$ everywhere else. The matrices $E^{ij}$ form a basis of $M(n, \mathbb R)$.

A linear function is described uniquely by the way it acts on a basis of the space, i.e. $f$ is determined uniquely through the values $$f_{ij} := f(E^{ij}).$$ For an arbitrary $C \in M(n, \mathbb R)$ the function $$tr(\cdot \, C)\colon M(n, \mathbb R) \to \mathbb R, \; A \mapsto tr(A C)$$ is also linear from $M(n, \mathbb R)$ to $\mathbb R$, i.e. also described by $tr(E^{ij} C)$. Now you can choose $C$ such that $tr(E^{ij} C) = f_{ij}$.