Trace of the multiplication of two matrix 1) Let $F$ be a field and $f$ be a linear functional on $M_{n}(F)$. Then there is a nonzero matrix $T$ in $M_{n}(F)$ such that $f(M)=\mathrm{tr}(TM)$ for all $M\subset M_{n}(F)$.
Every one help me. Why?
2) Let $F$ be a  field. Consider the a mapping on the algebra of all linear transformations on $F^n$ as follows:
$$(T,A)=trace(TA).$$ Clearly, this is an inner product. Then this is non-singular. 
Why? 
What's the ralation between two  parts of questions?
 A: Let $E_{ij}$ be the matrix with $1$ in the $(i,j)$ position and
$0$ elsewhere. Then $Tr(E_{i,j}A)=a_{j,i}$, the $(j,i)$ entry.
Thus the linear functional $\phi_{j,i}:A\mapsto a_{j,i}$ has a 
representation as a trace map $A\mapsto Tr(TA)$.
All linear functionals on $M_n(F)$ are linear combinations of the $\phi_{j,i}$.
A: How would you express an arbitrary linear combination of the entries of $M$?  You'd need $n^2$ coefficients, which might be conveniently encapsulated by a matrix $T \in M_n(F)$.  You might try multiplying out $TM$ to see what this gives you.  Doing so you'd notice that the sum of the diagonal elements, $\operatorname{tr}(TM)$, gives you the desired linear combination.
A: If you do out some linear algebra in the elementary matrices $E_{ij}$, you can find that the bilinear form on $M_n(F)$:
$$ X,Y \mapsto  \text{Trace}(X^tY) $$
is actually the entrywise dot product on $F^{n \times n}$. Here $E_{ij}$ is the matrix with a $1$ is the $ij$th entry and zeros elsewhere, and $X^t$ denotes the transpose of $X$.
