A question about Kronecker Product [closed]

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 $$\psi(t)$$

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

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• Try evaluating one component at a time. – J.G. Nov 6 '18 at 11:44
• How? Could you explain more clearly :) – HD239 Nov 6 '18 at 12:20

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.
$$A^T\left( \psi(t) \otimes I_n\right) Q \left(\psi^T(t) \otimes I_n\right)A = A^T \left((\psi(t)\psi^T(t)) \otimes Q\right) A.$$
• many thanks. $$ is $1$x$1$ matrix, isn't it? – HD239 Nov 6 '18 at 18:31