Extending the Frobenius inner product to all matrix inner products

Question

So in ${\bf R}^{n\times p}$ we have the Frobenius inner product given by $$\langle A, B\rangle=\text{tr}(A^TB)$$

which can be interpreted as the Euclidean inner product on ${\bf R}^{np}$. My understanding is that all inner products on ${\bf R}^{np}$ can be written as $$a^TPb$$ for $P$ positive-definite. The best I could do in attempting to extend the Frobenius inner product on ${\bf R}^{n\times p}$ is something of the form $$\langle A, B\rangle=\sum_{i=1}^N\text{tr}((X_iAY_i)^T(X_iBY_i))$$ for $X_i\in{\bf R}^{m_i\times n}$ and $Y_i\in{\bf R}^{p\times q_i}$ all full rank. However I would like to know if this covers all inner products on ${\bf R}^{np}$, or if maybe it's more complex than necessary due to redundancies.

I can find the corresponding $P$ matrix for any specific matrix inner product by taking the standard basis for ${\bf R}^{n\times p}$ and forming the matrix

\begin{bmatrix} \langle E_1,E_1\rangle & \langle E_1,E_2\rangle & \dots & \langle E_1,E_{np}\rangle \\ \langle E_2,E_1\rangle & \langle E_2,E_2\rangle & & \vdots \\ \vdots & & \ddots \\ \langle E_{np},E_1\rangle & \dots & \dots & \langle E_{np},E_{np}\rangle \end{bmatrix}

but I don't know if the general form for a matrix inner product I gave above covers all positive-definite matrices $P$.

Update:

newer version of this question on MathOverflow: https://mathoverflow.net/questions/229675/extending-the-trace-inner-product-to-all-matrix-real-inner-products

Answer 1

You can see an inner product as an operation $f(a,b)=\left<a,b\right>$, i.e., it is a bilinear function that (i) returns a non-negative number, (ii) satisfies the relationship $f(a,b)=f(b,a)$.

For vectors $a,b\in\mathbb R^n$, all bilinear functions that satisfy these properties can be written as $$ f(a,b) = \sum_{i,j=1}^n a_i P_{ij} b_j $$ where $P$ is symmetric and positive definite. For matrices $a,b\in\mathbb R^{n\times p}$, all such functions can be written as $$ f(a,b) = \sum_{i,k=1}^n \sum_{j,l=1}^p a_{ij} P_{ijkl} b_{kl} $$ where now $P$ is a tensor of rank 4 that is symmetric in the sense that $P_{ijkl}=P_{klij}$ and positive definite in the sense that $f(a,a)> 0$ for all $a\neq 0$.

Your question boils down to whether every $P$ that satisfies such conditions can be written a form that results from the vectors $X_i,Y_i$. I believe the answer to this is no. This is simply so because the (for simplicity assuming $n=p$) symmetric $P$ has (asymptotically) $n^4/2$ degrees of freedom, whereas the $n$ vectors $X_i,Y_i$ only have $2n^2$ degrees of freedom. In other words, I don't think that for sufficiently large $n$, your approach has sufficiently many degrees of freedom.