how to show a scalar product defined by trace of matrices is non-degenerate

Question

a) Let $V$ be a vector space of all $n \times n$ matrices over $\Bbb R $ , define the scalar product of two matrices $A$ and $B$ by $$\langle A,B\rangle = \text{tr}(AB)$$ where tr is trace. Show that this is a scalar product and non-degenerate.

b) If $A$ is a real symmetric matrix, show that $\text{tr}(AA) \ge 0$ and $\text{tr}(AA) > 0$ if $A \neq 0$. Thus defines a positive definite scalar product on the space of symmetric matrices.

$a)$ I don't have problem showing that it is scalar product. I don't know how to show that it is non degenerate. I know $\sum_{j=1}^n\sum_{i=1}^n a_{ji}b_{ij} = 0 , \forall b_{ij} \in \Bbb R$. Is it enough to show $A = 0$?

$b)$ I am not sure if this is correct $\sum_{j=1}^n\sum_{i=1}^n a_{ji}a_{ij} = \sum_{j=1}^n\sum_{i=1}^n a_{ji}^2 \ge 0 $. Is this correct?

Answer 1

For part (a), suppose the matrix $A$ satisfies $\operatorname{tr}(AB) = 0$ for all matrices $B$. Choose $B = e_{ij}$ for all $i, j$, where $e_{ij}$ is the matrix with 1 in the $i,j$ position, and 0 elsewhere.

I think part (b) is correct.