Prove that [B]ij =  Let B = A*A. That is, the product of a matrix and its transpose. Prove that the $[B]_{ij} = < u_{i}, u_{j}>$.
Now, upon looking at this, I decide to test it out first to see if I believe it. I started with $ A = [2,1,-2],[0,-1,3],[3,1,3],[1,1,0]$. I then found the transpose and multiplied them together. When I did this, I noticed a pattern. On the diagonal line would be some number, but then on the opposite of each side would be the same numbers. Sorry, am I making sense? Anyway, I tested out to see if each entry was the inner product of some row and column and it's true. I just don't know where to start with a formal proof of this. Can someone help?
 A: You did not specify what are the $u_i$. But one can guess that they are the columns of $A$, in which case the equation you are asking about is just an immediate translation of the definition of matrix multiplication.
The symmetry you discovered is just the fact that $\langle u_i,u_j\rangle=\langle u_i,u_j\rangle$, the symmetry of the inner product. This holds for real inner product spaces; for complex inner products you would find conjugate symmetry.
A: Yes, it's true in general and really quite easy to see.  Suppose the $u_i$ are the columns of the matrix $A$.  Then they are the rows of the matrix $A^T$ (I am using the somewhat more conventional notation $A^T$ for the transpose of $A$, rather than $A^*$).  Since $B_{ij}$, the $ij$-th entry of $B$, is the inner product of the $i$-th row of $A^T$ with the $j$-th column of $A$ (which follows immediately from the definition of matrix multiplication), the result $B_{ij} = \langle u_i, u_j \rangle$ follows directly.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
