Page 236 - Linear Algebra Done Right

into lists of consecutive integers and in each list move the first term to
the end of that list. For example, taking n = 9, the permutation
(2, 3, 1, 5, 6, 7, 4, 9, 8)
10.20 Determinant of a Matrix 227
is obtained from (1, 2, 3), (4, 5, 6, 7), (8, 9) by moving the first term of
each of these lists to the end, producing (2, 3, 1), (5, 6, 7, 4), (9, 8), and
then putting these together to form 10.20. Let T ∈L(V) be the operator
such that

10.21 Tv k = c k v p k

for k = 1,...,n. We want to find a formula for det T. This generalizes
our earlier example because if (p 1 ,...,p n ) happens to be the permuta-
tion (2, 3,...,n, 1), then the operator T whose matrix equals 10.18 is
the same as the operator T defined by 10.21.
With respect to the basis (v 1 ,...,v n ), the matrix of the operator T
defined by 10.21 is a block diagonal matrix
 
A 1 0
 . 
A =  . .  ,
 
0 A M
where each block is a square matrix of the form 10.18. The eigenvalues
of T equal the union of the eigenvalues of A 1 ,...,A M (see Exercise 3 in
Chapter 9). Recalling that the determinant of an operator on a complex
vector space is the product of the eigenvalues, we see that our definition
of the determinant of a square matrix should force

det A = (det A 1 )...(det A M ).

However, we already know how to compute the determinant of each A j ,
which has the same form as 10.18 (of course with a different value of n).
Putting all this together, we see that we should have

det A = (−1) n 1 −1 ...(−1) n M −1 c 1 ...c n ,

where A j has size n j -by-n j . The number (−1) n 1 −1 ...(−1) n M −1 is called
the sign of the permutation (p 1 ,...,p n ), denoted sign(p 1 ,...,p n ) (this
is a temporary definition that we will change to an equivalent definition
later, when we define the sign of an arbitrary permutation).