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My ultimate goal is to find a way to compute the determinant of the following block symmetric matrix:

$$ \underset{np\times np}H=\begin{bmatrix} \frac{1}{n}A+\frac{2(n-1)}{n^3s^2}I_p & -\frac{2}{n^3s^2}I_p & \dots & -\frac{2}{n^3s^2}I_p \\ -\frac{2}{n^3s^2}I_p & \frac{1}{n}A+\frac{2(n-1)}{n^3s^2}I_p & \dots & -\frac{2}{n^3s^2}I_p \\ \dots & \dots & \dots & \dots\\ -\frac{2}{n^3s^2}I_p & -\frac{2}{n^3s^2}I_p & \cdots & \frac{1}{n}A+\frac{2(n-1)}{n^3s^2}I_p\\ \end{bmatrix}, $$

where $A$ is a $p\times p$ symmetric matrix and $n,s\in \mathbb{R}$. As a first step, I would like to express this matrix with a unique expression envolving Kronecker product. Any idea of how to achieve such expression? Or any suggestion to compute the determinant of $H$ in another way?

coolsv
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1 Answers1

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We can write $$ H = \frac 1n I_n \otimes A + M \otimes I_p $$ where the entries of $M$ are the coefficients to your $I_p$. That is, $$ M = \pmatrix{ \frac{2(n-1)}{n^3s^2} & -\frac{2}{n^3s^2} & \cdots & -\frac{2}{n^3s^2} \\ -\frac{2}{n^3s^2} & \frac{2(n-1)}{n^3s^2} & \cdots & -\frac{2}{n^3s^2} \\ \vdots & \vdots & \ddots & \vdots\\ -\frac{2}{n^3s^2} & -\frac{2}{n^3s^2} & \cdots & \frac{2(n-1)}{n^3s^2}\\ } $$ You could say that your $H$ is the Kronecker sum $M \oplus \frac{A}{n}$.

One approach to finding the determinant is as follows: you can find the eigenvalues of $M$ with one of the methods outlined here. If $\lambda_1,\dots,\lambda_n$ are the eigenvalues of $M$ and $\mu_1,\dots,\mu_p$ are the eigenvalues of $\frac An$, then the eigenvalues of $H$ will be $\lambda_i + \mu_j$ for every pair $i,j$ with $1 \leq i \leq n$ and $1 \leq j \leq p$. So, the determinant of $H$ is the product of the eigenvalues of $H$, that is the product of all such sums.

Here is a reference for the Kronecker sum. See section 2.1.3 in particular.

Ben Grossmann
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