# General expression for determinant of a block-diagonal matrix

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
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

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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First write $$\left[ \begin{array}{cccc} A_1 \hspace{-5pt} &&& \\ & A_2 \hspace{-5pt} && \\[-3pt] && \ddots \hspace{-5pt} & \\ &&& A_k \end{array} \right] = \left[ \begin{array}{cccc} A_1 \hspace{-5pt} &&& \\ & \text{I}_{n_2} \hspace{-5pt} && \\[-3pt] && \ddots \hspace{-5pt} & \\ &&& \text{I}_{n_k} \end{array} \right] \left[ \begin{array}{cccc} \text{I}_{n_1} \hspace{-5pt} &&& \\ & A_2 \hspace{-5pt} && \\[-3pt] && \ddots \hspace{-5pt} & \\ &&& \text{I}_{n_k} \end{array} \right] \dots \left[ \begin{array}{cccc} \text{I}_{n_1} \hspace{-5pt} &&& \\ & \text{I}_{n_2} \hspace{-5pt} && \\[-3pt] && \ddots \hspace{-5pt} & \\ &&& A_k \end{array} \right]$$ Also, $$\det \left( \left[ \begin{array}{ccccc} \text{I}_{n_1} \hspace{-5pt} &&&& \\[-3pt] & \ddots \hspace{-5pt} &&& \\ && A_j \hspace{-5pt} && \\[-3pt] &&& \ddots \hspace{-5pt} & \\ &&&& \text{I}_{n_k} \end{array} \right] \right) = \det (A_j)$$ which can be seen by using the cofactor formula and repeatedly expanding along a row or column with all 0's and one 1

$$\implies \det \left( \left[ \begin{array}{cccc} A_1 \hspace{-5pt} &&& \\ & A_2 \hspace{-5pt} && \\[-3pt] && \ddots \hspace{-5pt} & \\ &&& A_k \end{array} \right] \right) = \det (A_1) \det (A_2) \cdots \det (A_k)$$

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