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

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?

-
You are missing a transpose in (a), it should be something like $tr(A^{t}B)$. $tr(AB)$ itself is degenerate. eg. $tr(A^2) = 0$ for any nilpotent matrix $A$ of degree 2. Your (b) is correct. –  achille hui Jan 31 '13 at 14:09
@achillehui the problem is from Serge Lang Linear Algebra chapter 6. Page no 112, you can find it here –  hasExams Jan 31 '13 at 14:23
You are missing a 'not' in your title. –  Chris Godsil Jan 31 '13 at 14:30
oh .. sorry .. i'll fix it. –  hasExams Jan 31 '13 at 14:30
@achillehui, the trace would be degenerate if there were a matrix $A \ne 0$ such that $\operatorname{tr}(AB) = 0$ for all matrices $B$. –  Andreas Caranti Jan 31 '13 at 14:32

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.
Part (b) is indeed correct, due to the fact that $A$ is assumed symmetric. –  kneidell Oct 10 '14 at 20:01