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Let $\mathcal{A}$ be a rank 4 tensor and $X$ be a rank 2 tensor (i.e. a matrix). Define the Frobenius inner product (FIP) of real matrices $$ A : B = \operatorname{tr}(A^T B) $$ A side question: if $A$ and $B$ are complex matrices $A, B \in \mathbb{C}^{m \times n}$, then should we define the FIP as $$A : B = \operatorname{tr}(A^H B)$$ where $(\cdot)^H = (\overline{\cdot})^T$ is conjugate transpose?

I am particulary interested in the following rank 4 isotrpoic tensor $$ \mathcal{E}_{i,j,k,l} = \delta_{i,k}\delta_{j,l}, \qquad i,k=1,...,m, \quad j,l = 1,...,n $$ (I know there is a more general form of this tensor, see this question) and how it multiplies with matrices.

Let $\mathcal{A}, \mathcal{E} \in \mathbb{C}^{m \times n \times m \times n}$ and $X, Y, A, B \in \mathbb{C}^{m \times n}$ what are:

  • $\mathcal{A} X = ?$
  • $X \mathcal{A} Y^H = ?$
  • $\mathcal{A} : X = ?$
  • $\mathcal{A}^T = ?$ and $\mathcal{E}^T$? (if they are defined)

How they are defined, what tensor are they (e.g. what is their rank) and how they are calculated by components?

In particular, to prove:

  • $ X = \mathcal{E} : X = X : \mathcal{E}$
  • $ A X B = A \mathcal{E} B^H : X$

I would like to see definition both in general and by components (e.g. $(M \cdot P)_{ij} = \sum_k M_{ik} P_{kj}$).

This is related to this question: Differentiating a column with respect to a matrix but I think it is worth a specific question since this is quite general and can be useful to other tensor students.

Edit: From the question linked above, $$ \mathcal{A} X = \sum_l \mathcal{A}_{ijkl} X_{ln} $$ and $$ \mathcal{A} : X = \sum_k \sum_l \mathcal{A}_{ijkl} X_{kl} $$

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The trace/Frobenius Product for real matrices is (as you've noted) $$A:B = {\rm Tr}(A^TB)$$ For complex matrices, you can stick with this definition or switch to the Inner Product $$\langle A,B\rangle = {\rm Tr}(A^HB) \;\;\doteq\;\; A^*:B$$ I prefer the trace/Frobenius product because it has nicer alebraic properties
$\big({\rm where}\,\{\odot,\otimes\}$ denote Hadamard/Kronecker products$\big)$ $$\eqalign{ A:B &= B:A \quad&{\rm however} \quad \langle A,B\rangle\ne\langle B,A\rangle \\ (C\odot A):B &= C:(A\odot B) \quad&{\rm however}\quad \langle C\odot A,B\rangle\ne\langle C,A\odot B\rangle \\ (A\otimes B):(X\otimes Y) &= (A:X)\otimes(B:Y) \quad&{\rm however}\quad \langle A\otimes B,X\otimes Y\rangle\ne\langle A,X\rangle\otimes\langle B,Y\rangle \\\\ }$$ In terms of either product, the Frobenius Norm is very easy to compute $$ \big\|A\big\|^2_F \;=\; A:A^* \;=\; \langle A,A\rangle \\ $$ Extending multiple dot products to higher-order tensors is straightforward.
For example, a quadruple-dot product would be $$\eqalign{ {\cal P} &= {\cal A}::{\cal B} \\ {\cal P}_{ij....pq} &= \sum_k\sum_\ell\sum_m\sum_n {\cal A}_{ij..k\ell mn}{\cal B}_{k\ell mn..pq} \\ }$$ where the order of both $({\cal A,B})\ge 4$.
And once you become familiar with the Einstein convention, you can omit the $\Sigma$'s.

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  • $\begingroup$ So the expressions I wrote above are correct? $\endgroup$ Commented Apr 4, 2021 at 12:17

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