Page 235 - Linear Algebra Done Right

226


0
c n


0
 c 1



0
 ;


10.18 Chapter 10. Trace and Determinant
c 2

 . . 
 . . . . 
 
c n−1 0
here all entries of the matrix are 0 except for the upper-right corner
and along the line just below the diagonal. Let’s find the determinant
of T. Note that
2
(v 1 ,Tv 1 ,T v 1 ,...,T n−1 v 1 ) = (v 1 ,c 1 v 2 ,c 1 c 2 v 3 ,...,c 1 ...c n−1 v n ).
Thus (v 1 ,Tv 1 ,...,T n−1 v 1 ) is linearly independent (the c’s are all non-
zero). Hence if p is a nonzero polynomial with degree at most n − 1,
then p(T)v 1 = 0. In other words, the minimal polynomial of T cannot
n
have degree less than n. As you should verify, T v j = c 1 ...c n v j for
n
each j, and hence T n = c 1 ...c n I. Thus z − c 1 ...c n is the minimal
n
Recall that if the polynomial of T. Because n = dim V, we see that z − c 1 ...c n is also
minimal polynomial of the characteristic polynomial of T. Multiplying the constant term of
n
an operator T ∈L(V) this polynomial by (−1) , we get
has degree dim V, then
the characteristic 10.19 det T = (−1) n−1 c 1 ...c n .
polynomial of T equals
If some c j equals 0, then clearly T is not invertible, so det T = 0 and
the minimal polynomial
the same formula holds. Thus in order to have det T = det M(T),we
of T. Computing the n−1
minimal polynomial is will have to make the determinant of 10.18 equal to (−1) c 1 ...c n .
often an efficient However, we do not yet have enough evidence to make a reasonable
method of finding the guess about the proper definition of the determinant of an arbitrary
characteristic square matrix.
polynomial. To compute the determinants of a more complicated class of op-
erators, we introduce the notion of permutation. A permutation of
(1,...,n) is a list (m 1 ,...,m n ) that contains each of the numbers
1,...,n exactly once. The set of all permutations of (1,...,n) is de-
noted perm n. For example, (2, 3,...,n, 1) ∈ perm n. You should think
of an element of perm n as a rearrangement of the first n integers.
For simplicity we will work with matrices with complex entries (at
this stage we are providing only motivation—formal proofs will come
later). Let c 1 ,...,c n ∈ C and let (v 1 ,...,v n ) be a basis of V, which
we are assuming is a complex vector space. Consider a permutation
(p 1 ,...,p n ) ∈ perm n that can be obtained as follows: break (1,...,n)