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Consider having a matrix whose structure is the following:

$$ A = \begin{pmatrix} a_{1,1} & a_{1,2} & a_{1,3} & 0 & 0 & 0 & 0 & 0 & 0\\ a_{2,1} & a_{2,2} & a_{2,3} & 0 & 0 & 0 & 0 & 0 & 0\\ a_{3,1} & a_{3,2} & a_{3,3} & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & a_{4,4} & a_{4,5} & a_{4,6} & 0 & 0 & 0\\ 0 & 0 & 0 & a_{5,4} & a_{5,5} & a_{5,6} & 0 & 0 & 0\\ 0 & 0 & 0 & a_{6,4} & a_{6,5} & a_{6,6} & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & a_{7,7} & a_{7,8} & a_{7,9}\\ 0 & 0 & 0 & 0 & 0 & 0 & a_{8,7} & a_{8,8} & a_{8,9}\\ 0 & 0 & 0 & 0 & 0 & 0 & a_{9,7} & a_{9,8} & a_{9,9}\\ \end{pmatrix} $$

Question.

What about its determinant $|A|$?.

Another question

I was wondering that maybe matrix $A$ can be expressed as a product of particular matrices to have such a structure... maybe using these matrices:

$$ A_1 = \begin{pmatrix} a_{1,1} & a_{1,2} & a_{1,3}\\ a_{2,1} & a_{2,2} & a_{2,3}\\ a_{3,1} & a_{3,2} & a_{3,3}\\ \end{pmatrix} $$

$$ A_2 = \begin{pmatrix} a_{4,4} & a_{4,5} & a_{4,6}\\ a_{5,4} & a_{5,5} & a_{5,6}\\ a_{6,4} & a_{6,5} & a_{6,6}\\ \end{pmatrix} $$

$$ A_2 = \begin{pmatrix} a_{7,7} & a_{7,8} & a_{7,9}\\ a_{8,7} & a_{8,8} & a_{8,9}\\ a_{9,7} & a_{9,8} & a_{9,9}\\ \end{pmatrix} $$

I can arrange $A$ as a compination of those: $A = f(A_1,A_2,A_3)$

Kronecker product

One possibility can be the Kronecker product:

$$ A= \begin{pmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0\\ \end{pmatrix} \otimes A_1 + \begin{pmatrix} 0 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0\\ \end{pmatrix} \otimes A_2 + \begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 1\\ \end{pmatrix} \cdot A_3 $$

But what about the determinant??? There are sums in this case which is not good...

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Could you cut down a bit on the $\huge\mathbf{Huge\ Text}$? –  Arturo Magidin May 23 '12 at 2:39
1  
Look at the definition of the determinant that defines it in terms of the a sum over the permutation group $S_n$. This will convince you of the idea. The result is here. –  John Engbers May 23 '12 at 2:42
    
@Arturo: I do not understand why you are getting mad. –  Andry May 23 '12 at 2:45
    
@John: Thanks a lot man... didn't see it when looked for it. Please post an answer, I will accept it :) –  Andry May 23 '12 at 2:46
    
@Andry: I'm not getting mad; I actually meant it to be humorous, making my point in what I thought was a somewhat amusing manner. I think you've been using this formatting (I know this is the second time I've removed the huge text in the last day or two), and I think it is disruptive and unnecessary; there are other ways of creating emphasis in the text without relying on changing the font size so drastically. –  Arturo Magidin May 23 '12 at 2:50

2 Answers 2

up vote 1 down vote accepted

As stated as a comment, the result is here. It really makes block diagonal matrices wonderful, hence finding canonical forms important.

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  1. The determinant of a block diagonal matrix is equal to the product of the determinants of the diagonal blocks. In your case, you have a block diagonal matrix of the form $$A=\left(\begin{array}{ccc} A_1 & 0 & 0\\ 0 & A_2 & 0\\ 0 & 0 & A_3 \end{array}\right)$$ so $\det(A) = \det(A_1)\det(A_2)\det(A_3)$.

  2. You can, though it is a bit ad-hoc. For example, note that if we let $$T_1 = \left(\begin{array}{ccccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right)$$ then $T_1^tA_1T_1$ is the block diagonal matrix $$\left(\begin{array}{ccc} A_1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{array}\right).$$ Likewise, if we let $$\begin{align*} T_2 &=\left(\begin{array}{ccccccccc} 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \end{array}\right)\\ T_3 &= \left(\begin{array}{ccccccccc} 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{array}\right) \end{align*}$$ then the resulting matrix is $$f(A_1,A_2,A_3) = T_1^tA_1T_1 + T_2^tA_2T_2 + T_3^t A_3 T_3.$$

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