Page 322 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes (iv) for every j, there is a polynomial e with coefficients in the scalar field such that e (T) is the
j j
orthogonal projection of V on W .
j
In the proof we use certain basic facts which we state as lemmas.
Lemma 1: Let N be a normal operator on an inner product space W. Then the null space of N is the
orthogonal complement of its range.
Proof: Suppose ( |N ) = 0 for all in W. Then (N* | ) = 0 for all ; hence N* = 0. By Theorem
10 of unit 26, this implies N = 0. Conversely, if N = 0, then N* = 0, and
(N* | ) = ( |N ) = 0
for all in W.

Lemma 2: If N is a normal operator and is a vector such that N = 0, then N = 0.
2
2
Proof: Suppose N is normal and that N = 0. Then N lies in the range of N and also lies in the
null space of N. By Lemma 1, this implies N = 0.
Lemma 3: Let T be a normal operator and f any polynomial with coefficients in the scalar field.
Then f(T) is also normal.
Proof: Suppose f = a + a x + . . . + a x . Then
n
0 1 n
f(T) = a I + a T + . . . + a T n
0 1 n
n
T
and f(T)* = a I a T *  a ( *) .
0 1 n
Since T*T = TT*, it follows that f(T) commutes with f(T)*.
Lemma 4: Let T be a normal operator and f, g relatively prime polynomials with coefficients in
the scalar field. Suppose and are vectors such that f(T) = 0 and g(T) = 0. Then ( | ) = 0.
Proof: There are polynomials a and b with coefficients in the scalar field such that af + bg = 1. Thus
a(T) f(T) + b(T) g(T) = I
and = g(T) b(T) . It follows that

( | ) = (g(T) b(T) | ) = (b(T) |g(T)* )
By assumption g(T) = 0. By Lemma 3, g(T) is normal. Therefore, by Theorem 10 of unit 26,
g(T)* = 0; hence ( | ) = 0.

Proof of Theorem 9: Recall that the minimal polynomial for T is the monic polynomial of least
degree among all polynomials f such that f(T) = 0. The existence of such polynomials follows
from the assumption that V is finite-dimensional. Suppose some prime factor p of p is repeated.
j
2
Then p = p g for some polynomial g. Since p(T) = 0, it follows that
j
2
(p (T)) g(T) = 0
j
for every in V. By Lemma 3, p (T) is normal. Thus Lemma 2 implies
j
p (T)g(T) = 0
j
for every in V. But this contradicts the assumption that p has least degree among all f such that
f(T) = 0. Therefore, p = p . . . p . If V is a complex inner product space each p is necessarily of the
1 k j
form
p = x – c
j j
with c real or complex. On the other hand, if V is a real inner product space, then p = x – c with
j j j j
c in R or
j

316 LOVELY PROFESSIONAL UNIVERSITY