Page 317 - DMTH502_LINEAR_ALGEBRA

Unit 29: Spectral Theory and Properties of Normal Operators

Theorem 5: Let V be a finite-dimensional inner product space and T a non-negative operator on Notes
V. Then T has a unique non-negative square root, that is, there is one and only one non-negative
2
operator N on V such that N = T.
Proof: Let T = c E + ... + c E be the spectral resolution of T. By Theorem 4, each c 0. If c is any
1 1 k k j
non-negative real number, let c denote the non-negative square root of c. Then according to
Theorem 3 and (2) N = T is a well-defined diagonalizable normal operator on V. It is non-
negative by Theorem 4, and, by an obvious computation, N = T.
2
2
Now let P be a non-negative operator on V such that P = T. We shall prove that P = N. Let
P = d F + ... + d F
1 1 r r
be the spectral resolution of P. Then d 0 for each j, since P is non-negative. From P = T we have
2
j
2
2
T = d F + ... + d F .
1 1 r r
Now F , ..., F satisfy the conditions I = F + ... + F , F F = 0 for i j, and no F is 0. The numbers
1 r 1 r i j j
2
d ..., d are distinct, because distinct non-negative numbers have distinct squares. By the
2
1 r
uniqueness of the spectral resolution of T, we must have r = k, and (perhaps reordering) F , = E,
j j
2
d = c . Thus P = N.
j j
Theorem 6: Let V be a finite-dimensional inner product space and let T be any linear operator on
V. Then there exist a unitary operator U on V and a non-negative operator N on V such that
T = UN. The non-negative operator N is unique. If T is invertible, the operator U is also unique.
Proof: Suppose we have T = UN, where U is unitary and N is non-negative. Then T* = (UN)* =
N*U* = NU*. Thus T*T = NU*UN = N . This shows that N is uniquely determined as the non-
2
negative square root of the non-negative operator T*T.
So, to begin the proof of the existence of U and N, we use Theorem 5 to define N as the unique
non-negative square root of T*T. If T is invertible, then so is N because
(N N ) = (N 2 ) = (T*T ) = (T T ).
–1
In this case, we define U = TN and prove that U is unitary. Now U* = (TN )* = (N )*T* =
–l
–1
(N*) T* = N T*. Thus
–1
–1
–1
UU* = TN N T*
–1
–1 2
= T(N ) T*
2 –1
= T(N ) T*
= T(T*T) T*
–1
–1
= TT (T*) T*
–1
= I
and C is unitary.
If T is not invertible, we shall have to do a bit more work to define U. We first define U on the
range of N. Let be a vector in the range of N say = N . We define U = T , motivated by the
fact that we want UB = T . We must verify that U is well-defined on the range of N in other
words, if N = N then T = T . We verified above that N 2 = T for every in V. Thus, with
2
= – , we see that N ( – ) = 0 if and only if T( – ) = 0. So U is well-defined on the range
of N and is clearly linear where defined. Now if W is the range of N, we are going to define U on
W . To do this, we need the following observation. Since T and N have the same null space, their
ranges have the same dimension. Thus W has the same dimension as the orthogonal complement
of the range of T. Therefore, there exists an (inner product space) isomorphism U of W onto
0
T(V) . Now we have defined U on W, and we define U on W to be U .
0

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