Thanks.
Although using my definition for the standard inner product I had to use \langle a , b \rangle = a^t \bar{y}, but it all worked out in the end.
Is there a non-ugly proof of the following identity:
\langle Ax,y \rangle = \langle x,A^*y \rangle
where A is an nxn matrix over, say, \mathbb{C}, A* is its conjugate transpose, and \langle \cdot , \cdot \rangle is the standard inner product on \mathbb{C} ^n.
I'm 19 and I get allowance. (I have a part-time job during school, and ideally a full-time job otherwise.) Where I'm from, it's perfectly natural for your parents to take care of you financially until you get your degree. Then it's your turn to work your ass off so they can retire with luxury...