Here $A^*$ is the conjugate transpose. I've found other questions with nice answers on the $A^t$ version but with distinctly different proofs than the one I outline below. I'm curious if someone could enlighten me on the first equality below (in the main proof section) I think its something simple I'm missing but I can't figure it out.
Lemma: Let $A\in$ $m\times n$ Matrix with $x\in$ $F^n$ and $y\in$ $F^m$. Then
$\langle Ax,y\rangle_m=\langle x,A^*y\rangle_n$
Where $\langle\rangle$ denotes an inner product, with respect to $F^n$ and $F^m$ using the subscripts.
Main Proof:
The proof itself uses the dimension theorem after showing that $A^*$ and $A$ have the same null space.
The equality I'm struggling with is:
$\langle A^*Ax,x\rangle_n=\langle Ax, A^{**}x\rangle_m$
If someone could explain it would be greatly appreciated!
Thanks!
We already know that: $$\langle Cy,z\rangle_n = \langle y, C^* z\rangle_m$$ Now, let $C = A^*$, $y = Ax$ (which is a vector), and $z = x$. Then, we get that: $$\langle (A^*)(Ax),x\rangle_n = \langle Ax, (A^*)^* x\rangle_m$$ Getting rid of parenthesis where we can, we have that: $$\langle A^*Ax,x\rangle_n = \langle Ax,A^{**}x\rangle_m$$ as desired.