1

The determinant of the following $14 \times 14$ matrix

$$\begin{bmatrix} D_1 & D_2 \\ D_3 & D_4 \end{bmatrix}$$

where the $D_i$ blocks are $7 \times 7$ diagonal matrices given by

$$D_1 = \mbox{diag} (a_1, a_2, \ldots a_7)$$

$$D_2 = \mbox{diag} (a_8, a_9, \ldots a_{14})$$

$$D_3 = \mbox{diag} (a_8, a_{14}, a_{13},\ldots a_9)$$

$$D_4 = \mbox{diag} (a_1, a_7, a_6, \ldots a_2)$$

is

$$(a_1^2-a_8^2)(a_2a_7-a_9a_{14})^2(a_3a_6-a_{10}a_{13})^2(a_4a_5-a_{11}a_{12})^2$$

Am I correct?

Rodrigo de Azevedo
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PAMG
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2 Answers2

6

A classical result is the following one.

Being given a $2 \times 2$ block matrix with square blocks:

$$\mathbf{S} := \begin{pmatrix} A & B\\ C & D \end{pmatrix},$$

if $DC=CD$ [which is the case here : any diagonal matrix commutes with any other], then :

$$\det(S)=\det(AD-BC)$$

Can you take it from here ?

Reference: (Determinant of block matrix with commuting blocks)

Rodrigo de Azevedo
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Jean Marie
  • 88,997
2

Let

$$S= \begin{pmatrix} D_1 & D_2\\ D_3 & D_4 \end{pmatrix}$$

Here is a completely different way to consider this issue by using a certain simultaneous re-ordering operation on rows and columns of $S$ giving rise to a matrix $B$ which is block diagonal (see formula (2) with diagonal blocks that will be denoted $B_k$).

This process can be expressed under the equivalent matrix form :

$$B=P^{-1}SP\tag{*}$$

where $P$ is a certain shuffling permutation matrix that will be given below in (1) corresponding to the new row ordering :

$$[0,7,1,8,2,9,3,10,4,11,5,12,6,13]\tag{0}$$

whereas $P^{-1}$ (which is also equal to $P^T$) operates in an identical way on columns.

As $\det(B)=\det(S)=\prod_k \det(B_k)$, we obtain the desired result.

More precicely, let $S$ be the initial matrix :

$$S=\left(\begin{smallmatrix} a & & & & & & &|& h & & & & & & \\ & b & & & & & &|& & i & & & & & \\ & & c & & & & &|& & & j & & & & \\ & & & d & & & &|& & & & k & & & \\ & & & & e & & &|& & & & & l & & \\ & & & & & f & &|& & & & & & m & \\ & & & & & & g &|& & & & & & & n \\ \hline h & & & & & & &|& a & & & & & & \\ & n & & & & & &|& & g & & & & & \\ & & m & & & & &|& & & f & & & & \\ & & & l & & & &|& & & & e & & & \\ & & & & k & & &|& & & & & d & & \\ & & & & & j & &|& & & & & & c & \\ & & & & & & i &|& & & & & & & b \end{smallmatrix}\right)$$

Let

$$P = \left(\begin{smallmatrix} 1 & & & & & & & & & & & & & \\ & & & & & & & 1 & & & & & & \\ & 1 & & & & & & & & & & & & \\ & & & & & & & & 1 & & & & & \\ & & 1 & & & & & & & & & & & \\ & & & & & & & & & 1 & & & & \\ & & & 1 & & & & & & & & & & \\ & & & & & & & & & & 1 & & & \\ & & & & 1 & & & & & & & & & \\ & & & & & & & & & & & 1 & & \\ & & & & & 1 & & & & & & & & \\ & & & & & & & & & & & & 1 & \\ & & & & & & 1 & & & & & & & \\ & & & & & & & & & & & & & 1 \end{smallmatrix}\right)\tag{1}$$

Then : $$B=P^{-1}SP= \left(\begin{smallmatrix}a & h & & & & & & & & & & & & \\ h & a & & & & & & & & & & & & \\ & & b & i & & & & & & & & & & \\ & & n & g & & & & & & & & & & \\ & & & & c & j & & & & & & & & \\ & & & & m & f & & & & & & & & \\ & & & & & & d & k & & & & & & \\ & & & & & & l & e & & & & & & \\ & & & & & & & & e & l & & & & \\ & & & & & & & & k & d & & & & \\ & & & & & & & & & & f & m & & \\ & & & & & & & & & & j & c & & \\ & & & & & & & & & & & & g & n \\ & & & & & & & & & & & & i & b \end{smallmatrix}\right)\tag{2}$$

whose determinant is the product of the determinants of the diagonal blocks, i.e.,

$$(a^2-h^2)(d e - k l)^2 (c f - j m)^2 (b g - i n)^2$$

as desired.

Here is the "Sage" program that I have written in order to ease and/or check my calculations.

[Let us recall that Sage is accessible to everybody, being a public domain software]

 var('a b c d e f g h i j k l m n')

D1=diagonal_matrix(7,[a,b,c,d,e,f,g])
 D2=diagonal_matrix(7,[h,i,j,k,l,m,n])
 D3=diagonal_matrix(7,[h,n,m,l,k,j,i])
 D4=diagonal_matrix(7,[a,g,f,e,d,c,b])
 S=block_matrix([[D1,D2],[D3,D4]])
 show(S)


L=[0,7,1,8,2,9,3,10,4,11,5,12,6,13]


B=S.matrix_from_rows(L).matrix_from_columns(L)
 show(B)
 dd=B.determinant().factor()
 show(dd)


A less efficient way to obtain matrix B


#P=P.matrix_from_columns(L)
 #B=(P.T)SP; # formula (*)

Important remark : ordering given by (0) isn't unique, i.e., matrix $P$ giving a diagonal-by-blocks structure using conjugation (formula (*)) isn't at all unique; here is another one

$$[7, 0, 8, 1, 9, 2, 10, 3, 11, 4, 12, 5, 13, 6]$$

which has been obtained by using the reverse Cuthill McKee algorithm which "returns a permutation vector p such that S(p,p) tends to have its diagonal elements closer to the diagonal than S" (extracted from Matlab documentation of its "symrcm" routine).

Jean Marie
  • 88,997