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I can' t show the following equality. Could you give some hints? $$ A^T\big( \psi(t) \otimes I \big) Q \big(\psi^T(t) \otimes I\big) A= A^T\big( \psi(t) \psi^T(t) \otimes Q\big) A$$ where

$\otimes$ is the Kronecker product,

$Q$ is a positive semi-definite matrix, $A$ is a real matrix,

$I$ is the identity matrix.

$\psi(t)$ is the Legendre wavelet function,

$\psi^T(t)$ is the transpose of $\psi(t)$

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closed as off-topic by GNUSupporter 8964民主女神 地下教會, Rebellos, ArsenBerk, José Carlos Santos, Namaste Nov 7 '18 at 18:49

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  • $\begingroup$ Try evaluating one component at a time. $\endgroup$ – J.G. Nov 6 '18 at 11:44
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    $\begingroup$ How? Could you explain more clearly :) $\endgroup$ – HD239 Nov 6 '18 at 12:20
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From the definition of the Kronecker product, for all $A$ matrices, we have $$ A = I_1 \otimes A = A \otimes I_1,\tag{$\heartsuit$} $$ where $I_n$ is the $n \times n$ identity matrix. Specially $I_1 = {\begin{bmatrix}1\end{bmatrix}}$.

The mixed-product property of the Kronecker product states that if $A,B,C$ and $D$ are matrices of such size that we can form the matrix products $AC$ and $BD$, then $$ (A \otimes B)(C \otimes D) = (AC) \otimes (BD). $$ In general, if $A_1,\dots, A_p$ and $B_1,\dots,B_p$ are matrices of such size that the following matrix products exist, then we have $$ (A_1 \otimes B_1)(A_2 \otimes B_2)\cdots(A_p \otimes B_p) = (A_1A_2\cdots A_p) \otimes (B_1B_2 \cdots B_p).\tag{$\spadesuit$} $$

Since $\psi$ is a Legendre wavelet function, we know that $\psi(t) \in \mathbb{R}^{n \times 1}$ for all $t$ in the domain of $\psi$, for some $n \in \mathbb{N}$. First we decompose $Q \in \mathbb{R}^{n \times n}$ by using $(\heartsuit)$ as the following. $$ \left( \psi(t) \otimes I_n\right) Q \left(\psi^T(t) \otimes I_n\right) = \left( \psi(t) \otimes I_n \right) \left(I_1 \otimes Q\right) \left(\psi^T(t) \otimes I_n\right). $$ Then by using $(\spadesuit)$, we find that $$ \left( \psi(t) \otimes I_n\right) \left(I_1 \otimes Q\right) \left(\psi^T(t) \otimes I_n\right) = (\psi(t)\psi^T(t)) \otimes Q.\tag{$\diamondsuit$} $$ If $A \in \mathbb{R}^{n^2 \times n^2}$, then by multiplying both sides of $(\diamondsuit)$ by $A^T$ from left and by $A$ from right, we have the desired result.

$$ A^T\left( \psi(t) \otimes I_n\right) Q \left(\psi^T(t) \otimes I_n\right)A = A^T \left((\psi(t)\psi^T(t)) \otimes Q\right) A. $$

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  • $\begingroup$ many thanks. $[1]$ is $1$x$1$ matrix, isn't it? $\endgroup$ – HD239 Nov 6 '18 at 18:31
  • $\begingroup$ @HD239 Yes, it is. $\endgroup$ – user153012 Nov 6 '18 at 20:55

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