Linear functionals on M_n(R) are traces against a matrix

Idea. In a finite-dimensional inner product space, every linear functional is "taking the inner product with a fixed vector" (Riesz representation). Here the inner product is $\operatorname{tr}(B^TA)$.

Step 1: existence, by hand. Write $A=\sum_{i,j}a_{ij}E_{ij}$ where $E_{ij}$ is the matrix with a $1$ in position $(i,j)$ and zeros elsewhere. By linearity,
$$\varphi(A)=\sum_{i,j}a_{ij}\varphi(E_{ij})=\sum_{i,j}a_{ij}b_{ij}=\operatorname{tr}(B^TA),\qquad b_{ij}:=\varphi(E_{ij}).$$

Step 2: uniqueness. If $\operatorname{tr}(B^TA)=\operatorname{tr}(B'^TA)$ for all $A$, then $\langle B-B',A\rangle=0$ for all $A$, in particular for $A=B-B'$, which gives $\|B-B'\|_F^2=0$.

Step 3: continuity. By Cauchy-Schwarz,
$$|\varphi(A)|=|\langle B,A\rangle_F|\le\|B\|_F\|A\|_F,$$
so $\varphi$ is Lipschitz with constant $\|B\|_F$, hence continuous (for $\|\cdot\|_F$, and therefore for every norm by equivalence of norms).

Examples. $\operatorname{tr}$ corresponds to $B=I$; the $(i,j)$ entry corresponds to $B=E_{ij}$; $A\mapsto x^TAy$ corresponds to $B=xy^T$.

Remark. The same argument shows that every linear map between finite-dimensional normed spaces is continuous. The kernel of a non-zero functional is a hyperplane, which is therefore closed (and has empty interior), a fact we use in the topology of subspaces.