How do you prove that $\langle Ax, Ay\rangle = \langle A^{*}Ax, y\rangle$ for an inner product space (IPS)? Lets just say $V = \mathbb{C^{n}}$ and $F = \mathbb{C}$ (Are there $F$ and $V$ where this isn't true?).
I assumed this would be a very simple proof using only the basic properties of the sesquilinear form and I am embarrassed that I can't get the details. Given an explicit definition for the IPS, e.g., the dot product, I can do this without issue, but for a general IPS, I can't figure out how to "move the matrix" so to speak.
Let $A_{i}$ denote the $i$th column of some matrix $A$, $x = \sum_{i=1}^{n} \alpha_{i}e_{i}$, and $y = \sum_{j=1}^{n} \beta_{j}e_{j}$. If I try this approach, I get something like
$$\langle Ax , By\rangle = \sum_{i} \overline{\beta_{i}}\langle Ax, A_{j}\rangle $$
which doesn't seem to go anywhere, however, I feel like expanding $x$ and $y$ into a linear combination of a standard basis is the only reasonable first step here.
Best Answer
If we are working in a general inner product space (over $\mathbb C$), then $A$ must be a linear transformation rather than a matrix. In this case, we can't define $A^*$ to be the conjugate transpose of $A$ (because $A$ is not a matrix). Usually we would define $A^*$ to be the unique linear transformation such that $\langle z, Ay\rangle = \langle A^*z, y \rangle$ for all $z, y$. From this definition, it follows immediately that $\langle Ax, Ay \rangle = \langle A^* A x, y \rangle$ for all $x, y$.
If $V = \mathbb C^n$ and $A$ is a matrix and $\langle \cdot, \cdot \rangle$ is the standard inner product on $\mathbb C^n$, and $A^*$ is defined to be the conjugate transpose of the matrix $A$, then we can use the following argument: \begin{align} \langle Ax, Ay \rangle &= (Ay)^* Ax \\ &= y^* A^* Ax \\ &= \langle A^* Ax, y \rangle. \end{align}