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Mathematics-Online course: Linear Algebra - Matrices - Determinants

Determinants of Special Matrices

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For some special types of $ n\times n$-matrices $ A$ the determinant can be given immediately.

(i)
Triangular matrices: If $ a_{i,j}=0$ for $ i<j$ or $ i>j$, then we can calculate the determinant by

$\displaystyle \operatorname{det}A = a_{1,1}\cdots a_{n,n}\, . $

(ii)
Block-diagonal matrices: If matrix $ B$ has a blocked structure with $ A_{i,j}=0$, $ i\neq j$ and square diagonal blocks, then we have

$\displaystyle \operatorname{det}B = \prod\limits_{i=1}^k \operatorname{det}A_{i,i}\,. $

(iii)
For a unitary matrix $ U$ we have

$\displaystyle \vert\operatorname{det}U\vert=1\,. $

In particular, for orthogonal matrices $ (u_{i,j}\in \mathbb{R})$ we have

$\displaystyle \operatorname{det}U \in \{-1,1\}\,. $

(Authors: Burkhardt/Höllig/Hörner)
(i)
Since $ \operatorname{det} A=\operatorname{det} A^{\operatorname t}$ it is sufficient to consider upper triangle matrices. Let $ i$ be a permutation of $ (1,\ldots,n)$ wich is not the identity. Then, there exists at least one element $ i_k$ with $ i_k > k$, hence we have $ a_{i_k,k}=0$. Thus, by the expansion of the determinant in a sum over permutations, this sum reduces to one addend which corresponds to the identity permutation and, hence, is the product of the diagonal entries.
(ii)
At first, consider a decomposition in only two diagonal blocks of dimensions $ n_1$ and $ n_2$. Then, taking the sum over all permutations we see that we have to take into consideration only that permutations which permute the first $ n_1$ elements among each other and, consequently, permute the last $ n_2$ elements among each other as well. The addends corresponding to other permutations vanish. Thus, each remaining addend decomposes into a product of a permutation $ k$ of the first $ n_1$ elements and a product of a permutation $ l$ of the last $ n_2$ elements as follows:
$\displaystyle \operatorname{det}B$ $\displaystyle =$ $\displaystyle \sum\limits_{i\in S_n}\sigma(i) (a_{i_1,1}\cdots a_{i_{n_1},n_1}) (a_{i_{n_1+1},n_1+1}\cdots a_{i_{n},n})$  
  $\displaystyle =$ $\displaystyle \left( \sum\limits_{k\in S_{n_1}} \sigma(k) (a_{k_1,1}\cdots a_{k... ...ts_{l\in S_{n_2}} \sigma(l) (a_{n_1+l_1,n_1+1}\cdots a_{n_1+l_{n_2},n}) \right)$  
  $\displaystyle =$ $\displaystyle \operatorname{det}A_{1,1}\operatorname{det}A_{2,2}\,.$  

For more than two diagonal blocks the corresponding statements can be proved by induction.
(iii)
Since complex conjugation is a operation which commutes with the arithmetical operations of addition and multiplication, and since $ \operatorname{det}U = \operatorname{det}U^{\operatorname t}$, it follows from the multiplicativity of determinants that

$\displaystyle 1=\operatorname{det}E = \operatorname{det}\left(U^\ast U\right) =... ...{\operatorname{det}U}\operatorname{det}U = \vert\operatorname{det}U\vert\,. $

In the real case the determinant is real and, hence, $ 1$ and $ -1$ are the only values that can be attained.

(Authors: Burkhardt/Höllig/Hörner)
  automatically generated 4/21/2005