2
$\begingroup$

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

which can be interpreted as the Euclidean inner product on ${\bf R}^{np}$. All inner products on ${\bf R}^{np}$ can be written as $$\langle a, b\rangle_P:=a^TPb$$ for $P\in {\bf R}^{np\times np}$ (symmetric) positive-definite. The best I could do in attempting to extend the trace inner product on ${\bf R}^{n\times p}$ is something of the form \begin{align} \langle A, B\rangle_{XY}:=\sum_{i=1}^N\text{tr}((X_iAY_i)^T(X_iBY_i))&=\sum_{i=1}^N\big((X_i\otimes Y_i^T)\text{vec}(A)\big)^T(X_i\otimes Y_i^T)\text{vec}(B)\\ &=\text{vec}(A)^T\big(\sum_{i=1}^N(X_i\otimes Y_i^T)^T(X_i\otimes Y_i^T)\big)\text{vec}(B) \end{align}

for $X_i\in{\bf R}^{n\times n}$ and $Y_i\in{\bf R}^{p\times p}$ all non-singular, $\text{vec}(\cdot)$ the row-ordered vectorization of a matrix and $\otimes$ the Kronecker product. Thus the sum $$\sum_{i=1}^N(X_i\otimes Y_i^T)^T(X_i\otimes Y_i^T)$$ is the sum of positive-definite matrices and is thus positive-definite, and it shares a striking resemblance to the decomposition of a positive-definite matrix $$P=U^T\Lambda U=(\sqrt{\Lambda}U)^T(\sqrt{\Lambda}U)=\sum_{i=1}^{np}\lambda_iu_iu_i^T$$

The number of free parameters for $\lambda_iu_iu_i^T$ is $np$, while for $(X_i\otimes Y_i^T)^T(X_i\otimes Y_i^T)$ it's $n^2+p^2$, and thus it seems reasonable to assume that we have the bound $N\leq np$.

Can anyone confirm that $\sum_{i=1}^N(X_i\otimes Y_i^T)^T(X_i\otimes Y_i^T)$ can be made to form any positive-definite matrix? Also is there more to this story? Can things be simplified at all from what I have here?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.