3

I am reading the book Matrix Variate Distribution by A. K. Gupta and D. K. Nagar. In the first chapter (Definition 1.2.8), they define a matrix $B_{p}$ ($p \in \mathbb{N}^{\ast}$) as follows :

Definition 1.2.8 : The matrix $B_{p}$ of order $p^{2} \times \frac{1}{2}p(p+1)$, with typical element $$ (B_{p})_{ij,gh} = \frac{1}{2}\big( \delta_{ig}\delta_{jh} + \delta_{ih}\delta_{jg} \big) \quad i \leq p, \; j \leq p, \; g \leq h \leq p, $$ where $\delta_{rs}$ is the Kronecker's delta, is called the transition matrix.

I am not very comfortable with the notation $(B_{p})_{ij,gh}$. Here is how I understand this notation :

let $\varphi \, : \, \lbrace 1,\ldots,p^{2} \rbrace \, \rightarrow \, \lbrace 1,\ldots,p \rbrace \times \lbrace 1,\ldots,p \rbrace$ a one-to-one and onto mapping. Let's assume that $p$ is even and let $\psi \, : \, \lbrace 1,\ldots,\frac{1}{2}p(p+1) \rbrace \, \rightarrow \, \lbrace 1,\ldots,\frac{p}{2} \rbrace \times \lbrace 1,\ldots,p+1 \rbrace$ a one-to-one and onto mapping. For all $1 \leq i \leq p^{2}$ and all $1 \leq j \leq \frac{1}{2}p(p+1)$, let $\varphi_{1}(i)$ (resp. $\varphi_{2}(i)$) be the first (resp. second) coordinate of $\varphi(i)$. Let $\psi_{1}(j)$ (resp. $\psi_{2}(j)$) be the first (resp. second) coordinate of $\psi(j)$. Therefore, I would define $B_{p}$ as follows :

$$ \forall 1 \leq i \leq p, \; \forall 1 \leq j\leq \frac{1}{2}p(p+1), \; (B_{p})_{i,j} = \frac{1}{2}\big( \delta_{\varphi_{1}(i)\psi_{1}(j)}\delta_{\varphi_{2}(i)\psi_{2}(j)} + \delta_{\varphi_{1}(i)\psi_{2}(j)}\delta_{\varphi_{2}(i)\psi_{1}(j)} \big). $$

Would this be correct ? Is there another way to understand this definition 1.2.8 ?

Odile
  • 1,171
  • 8
  • 11
  • Your rewriting mainly mangles things. Recall that for every finite sets $S$ and $T$, a matrix $B$ indexed by $S\times T$ with values in $\mathbb R$, say, is nothing but a function $\beta:S\times T\to\mathbb R$ where one uses the notation $B_{s,t}$ for $\beta(s,t)$. In your case, $S={1,\ldots,p}^2$ and $T={(i,j)\mid1\leqslant i\leqslant j\leqslant p}$. Your idea would have some benefits if, for some reason, one wanted to order canonically $S$ and $T$ but one does not, does one? – Did Feb 01 '16 at 12:39

0 Answers0