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We know that the Frobenius inner product $A \cdot B = tr(A^\top B)$ between two $n\times n$ matrices $A$ and $B$ can be represented as $$ A \cdot B = \langle\vec A, \vec B\rangle = tr \vec A (\vec B)^\top$$ where on the right side $\langle \cdot, \rangle$ is the ordinary Euclidean scalar product and where $\vec X$ denotes the vectorization of $X$. I am curious about (and would have use of) a slightly more general case.

In particular, I have matrices $Q,A,B$ where I want to use positive definite matrix $Q$ to "skew" the inner product, i.e., I am interest in the rewriting the quantity $$ tr (Q A^\top B) $$ in terms of $\vec A$ and $\vec B$ as is possible when $Q=I$.

My question is thus: Can we write $$tr (QA^\top B) = \langle \vec A, \vec B \rangle_Q $$ where $\langle\cdot, \cdot \rangle_Q$ is a skewed euclidean inner product on $\mathbb{R} ^{n^2}$, depending on $Q$?

My guess would be that this is indeed possible and should look something like $$tr (QA^\top B) = tr \left( (I \otimes Q) (\vec A \otimes (\vec B)^\top))\right) $$ but I am certainly not sure and don't really know how to proceed.

I would very much appreciate help in finding a rewrite of the quantity $tr (QA^\top B)$which is similar to this.


Edit for clarity:

The question could be reformulated as: Is there a euclidean inner product $\langle , \rangle _Q$ weighted by some matrix (say $M=M(Q)$), such that $tr( QA^\top B) = \langle \vec A, \vec B\rangle_Q=\langle M(Q) \vec A,\vec B\rangle$? In the end what I am after is to express $tr QA^\top B$ in terms of the trace of the outer product of $\vec A$ and $\vec B$.


Progress Update:

I was able to find a reference ( https://www.ime.unicamp.br/~cnaber/Kronecker.pdf) online which states that

$$ tr (ABC ) = (\vec A^\top)^\top (I\otimes B) \vec C$$

which is more or less equivalent to what I am asking. However, as the reference does include a proof, I would still like this to be resolved.

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    $\begingroup$ How do you mean skew? As in a prescribed weight for position of each element? If so, then I think this won't work but you will need to work in vectorized space. But maybe you are happy to only be able to skew different rows differently? Then I think it will work. $\endgroup$ Oct 25, 2019 at 11:58
  • $\begingroup$ By skew I am referring to the premultplication by the matrix $Q$ in the trace. $\endgroup$ Oct 25, 2019 at 12:01
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    $\begingroup$ Ok, you can for sure get something more general with $vec(A)^T Q vec(B)$, where $Q$ is big diagonal matrix with one diagonal entry per matrix entry in each $A$ and $B$ but maybe you don't need it. Yes Kronecker product with identity matrix shall also help you get what you want. I think there is a good wikipedia on it. en.wikipedia.org/wiki/Kronecker_product#Matrix_equations $\endgroup$ Oct 25, 2019 at 13:29

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For ease of typing, let's use $A:B$ instead of $A\cdot B$ to denote the Frobenius inner product,
and $a$ instead of $\vec A$ to denote vectorization, i.e. $$\eqalign{ A:B &= {\rm Tr}(A^TB) = &{\rm Tr}(B^TA) = B:A \\ A:B &= a:b &({\rm Frobenius\,product\,notation})\\ A:B &= a^Tb &({\rm Matrix\,product\,notation})\\ }$$ Using this we can develop various expressions for the following scalar quantity $$\eqalign{ {\rm Tr}(QA^TB) &= {\rm Tr}(A^TBQ) \\ &= A:BQ \\ &= {\rm vec}(A):{\rm vec}(IBQ) \\ &= a:(Q^T\otimes I)\,b \\ &= a^T(Q^T\otimes I)\,b \\ &= b^T(Q\otimes I)\,a \\ }$$

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