Natural inner product on the Hom space.

Question

Let $P$ and $V$ be vector spaces of dimension $k$ and $n$, respectively. I want to know if it's possible to endow $\text{Hom}(P,V)$ with a natural inner product, but without using the (non-canonical) isomorphism between $V$ and $V^*$ and then considering something like $\langle{\phi},{\psi}\rangle = \text{tr}(\phi^{\top} \circ \psi)$.

What comes to my mind and what is more suitable to the setup I'm working on is something like this: assume that $P$ and $V$ have inner products $\langle \cdot, \cdot \rangle_P$ and $\langle \cdot, \cdot \rangle_V$, respectively. Then, we can identify $P \otimes V$ with $\text{Hom}(P,V)$ using the map $p \otimes v \mapsto \langle \cdot, p \rangle_P v$ and we can consider the canonical inner product on $P \otimes V$ given by $\langle p \otimes v, q \otimes w \rangle_{P \otimes V} := \langle p, q \rangle_P \langle v, w \rangle_P$. But how this translates to the Hom space?

I've tried this: choosing basis $\{p_1,\ldots,p_k\}$ and $\{v_1,\ldots,v_n\}$ (and I can't assume those basis are orthogonal) and defining $G$ as the Gram matrix of $V$, ie, $G = (\langle v_i, v_j \rangle_V)_{i,j}$, I've identified $\text{Hom}(P,V)$ as the space of matrices $n \times k$ written in those basis and then defined the inner product as $\text{tr}(A^{\top}GB)$. But this formula is wrong because on $P \otimes V$ we have $\langle{p_i \otimes v_r},{p_j \otimes v_s}\rangle_{P \otimes V} = \langle{p_i},{p_j}\rangle_P \langle{v_r},{v_s}\rangle_V$, but the inner product $\text{tr}(A^{\top}GB)$ on $\text{Hom}(P,V)$ computed on the corresponding matrices of the linear transformations $A := \langle \cdot,p_i \rangle_P v_r$ and $B := \langle \cdot,p_j \rangle_P v_s$ gives me $$\sum_{\lambda=1}^k \langle p_{\lambda},p_i \rangle_P \langle v_r,v_s \rangle_V \langle p_{\lambda},p_j \rangle_V$$ and this is not what I expected. Any idea how should I proceed?

Answer 1

$\def\hom{\operatorname{Hom}} \def\tr{\operatorname{tr}}$I will sum up what has already been discussed in the comments. Let $V,W$ be finite-dimensional inner product spaces over $k\in\{\mathbb{R},\mathbb{C}\}$. If $A:V\to W$ is $k$-linear, denote $A^*: W\to V$ to the adjoint operator (it is the unique $k$-linear map such that $\langle Av,w\rangle_W=\langle v,A^*w\rangle_V$ for all $v\in V,w\in W$). The canonical inner product in $\hom(V,W)$ is called the Frobenius inner product and is defined in the following way: for $A,B\in\hom(V,W)$, we set $$ \langle A,B\rangle=\tr(A^*B)=\tr(BA^*), $$ where $\tr$ denotes the trace (although computation of the trace requires picking a basis, the final result can be checked to be independent of the choice of basis).

Suppose $\{e_i\}$ and $\{c_i\}$ are respectively orthonormal bases of $V$ and of $W$, and denote $\rho_{ij}\in\hom(V,W)$ to the unique $k$-linear map such that $\rho_{ij}(e_k)=\delta_{ik}c_j$ (we will use Einstein summation convention). Then $\{\rho_{ij}\}$ is an orthonormal basis of $\hom(V,W)$. If $A=A^{ij}\rho_{ij}$, then $A^{ij}=c^j(Ae_i)$, where $\{c^i\}\subset W^*$ is the dual basis of $\{c_i\}$. Let $A,B\in\hom(V,W)$. Then

\begin{align*} \langle A,B\rangle &=\sum_{ij}(A^{*})^{ij}B^{ij}\\ &=\sum_{ij}\overline{c^j(Ae_i)}c^j(Be_i)\\ &=\sum_{ij}\langle c^j(Ae_i)c_j,c^j(Be_i)c_j\rangle_W\\ &=\sum_{ijk} \langle c^j(Ae_i)c_j,c^k(Be_i)c_k\rangle_W\\ &=\sum_i \langle Ae_i,Be_i\rangle_W, \end{align*} which gives a way of computing the inner product of $\hom(V,W)$ in terms of an orthonormal basis of $V$ and the inner product in $W$.

The generalization to infinite-dimensional Hilbert spaces is the Hilbert-Schmidt inner product.