Not sure regarding this one. I've been give 2 claims, one that I solved and this one. This is the question:
a) Let $V=M_{nxn}(\mathbb R)$ and $a,b \in V$. Does $(a,b)=\text {tr}\,(ba)$ define an inner product on $ V$? (I don't know how to approach it even)
b) $V=R^2$ and $u=(x_1,x_2)$, $v=(y_1,y_2)$. $u,v \in V$. Does $(u,v)=x_1+y_1$ define an internal product on $V$? (this claim is false, since $(u,v)=x_1y_1+x_2y_2$)
My main problem is with a), as I don't know how to approach it because Idon't know how to show the relation between $\text {tr}\, (ba)$ to $(a,b)=\text {tr}\,(ba)$ being an inner product space.
(if I was wrong in b and it can occur, please note that).
Thank you very much for your help.
The formula in a) defines a pre-inner product (i.e., a bilinear form), but there exist nonzero matrices with $(a,a)=0$. On the other hand, $$(a,b)=\text {Tr}\, (b^Ta) $$ does define an inner product.
Your argument for b) is not right. In this case the formula doesn't even give you a bilinear form. For instance, $$((3,0), (1,1))=4\ne7= ((3,0),(1,0))+((3,0), (0,1)). $$Or, even easier, $$((2,0), (1,0))=3\ne 4=2 ((1,0), (1,0)). $$