# 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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – GNUSupporter 8964民主女神 地下教會, Rebellos, ArsenBerk, José Carlos Santos, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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. $[1]$ is $1$x$1$ matrix, isn't it? – HD239 Nov 6 '18 at 18:31