Page 180 - Linear Algebra Done Right

The Characteristic Polynomial
Theorem: Let T ∈L(V) and λ ∈ F . Then for every basis of V
8.10
with respect to which T has an upper-triangular matrix, λ appears on
the diagonal of the matrix of T precisely dim null(T − λI) dim V times. 169
Proof: We will assume, without loss of generality, that λ = 0 (once
the theorem is proved in this case, the general case is obtained by re-
placing T with T − λI).
For convenience let n = dim V. We will prove this theorem by induc-
tion on n. Clearly the desired result holds if n = 1. Thus we can assume
that n> 1 and that the desired result holds on spaces of dimension
n − 1.
Suppose (v 1 ,...,v n ) is a basis of V with respect to which T has an
upper-triangular matrix Recall that an asterisk
is often used in
 
λ 1 ∗ matrices to denote
 . 
 . .  entries that we do not

8.11   .

 λ n−1  know or care about.
 
0 λ n
Let U = span(v 1 ,...,v n−1 ). Clearly U is invariant under T (see 5.12),
and the matrix of T| U with respect to the basis (v 1 ,...,v n−1 ) is
 
λ 1 ∗
 . 
8.12  . .  .
 
0 λ n−1
Thus, by our induction hypothesis, 0 appears on the diagonal of 8.12
n
dim null(T| U ) n−1 times. We know that null(T| U ) n−1 = null(T| U ) (be-
cause U has dimension n − 1; see 8.6). Hence
n
8.13 0 appears on the diagonal of 8.12 dim null(T| U ) times.
The proof breaks into two cases, depending on whether λ n = 0. First
consider the case where λ n = 0. We will show that in this case

8.14 null T n ⊂ U.

n
Once this has been verified, we will know that null T n = null(T| U ) , and
hence 8.13 will tell us that 0 appears on the diagonal of 8.11 exactly
dim null T n times, completing the proof in the case where λ n = 0.
Because M(T) is given by 8.11, we have