Page 292 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes There are several comments we should make about the proof.
1. The proof of the existence of a non-zero X such that AX = cX had nothing to do with the fact
that A was Hermitian (self-adjoint). It shows that any linear operator on a finite-dimensional
complex vector space has a characteristic vector. In the case of a real inner product space,
the self-adjointness of A is used very heavily, to tell us that each characteristic value of A
is real and hence that we can find a suitable X with real entries.
2. The argument shows that the characteristic polynomial of a self-adjoint matrix has real
coefficients, in spite of the fact that the matrix may not have real entries.
3. The assumption that V is finite-dimensional is necessary for the theorem; a self-adjoint
operator on an infinite-dimensional inner product space need not have a characteristic
value.

Example 8: Let V be the vector space of continuous complex-valued (or real-valued)
continuous functions on the unit interval, 0  t  1, with the inner product

1
g
t
t
(f|g) = f ( ) ( )dt .
0
The operator ‘multiplication by t,’ (Tf)(t), is self-adjoint. Let us suppose that Tf = cf. Then
(t – c) f(t) = 0, 0  t  1
and so f(t) = 0 for t  c. Since f is continuous, f = 0. Hence T has no characteristic values (vectors).
Theorem 8: Let V be a finite-dimensional inner product space, and let T be any linear operator on
V. Suppose W is a subspace of V which is invariant under T. Then the orthogonal complement of
W is invariant under T*.

Proof: We recall that the fact that W is invariant under T does not mean that each vector in W is
left fixed by T; it means that if  is in W then T is in W. Let be in W . We must show that T*
is in W , that is, that ( |(T* ) = 0 for every in W. If is in W, then T is in W, so (T | ) = 0. But
(T | ) = ( |T* ).
Theorem 9: Let V be a finite-dimensional inner product space, and let T be a self-adjoint linear
operator on V. Then there is an orthonormal basis for V, each vector of which is a characteristic
vector for T.

Proof: We are assuming dim V > 0. By Theorem 7, T has a characteristic vector . Let = /
1
so that is also a characteristic vector for T and = 1. If dim V = 1, we are done. Now we
1 1
proceed by induction on the dimension of V. Suppose the theorem is true for inner product
spaces of dimension less than dim V. Let W be the one-dimensional subspace spanned by the
vector . The statement that is a characteristic vector for T simply means that W is invariant
1 1
under T. By Theorem 8, the orthogonal complement W is invariant under T* = T. Now W , with
the inner product from V, is an inner product space of dimension one less than the dimension of
V. Let U be the linear operator induced on W by T, that is the restriction of T to W . Then U is
self-adjoint and by induction hypothesis, W has an orthonormal basis { , . . ., } consisting of
2 n
characteristic vectors for U. Now each of these vectors is also a characteristic vector for T, and
since V = W W , we conclude that { , . . ., } is the desired basis for V.
1 n
Corollary: Let A be an n n Hermitian (self-adjoint) matrix. Then there is a unitary matrix P such
that P AP is diagonal (A is unitary equivalent to a diagonal matrix). If A is real symmetric
–1
–1
matrix, there is a real orthogonal matrix P such that P AP is diagonal.
n 1
Proof: Let V be C , with the standard inner product, and let T be the linear operator on V which
is represented by A in the standard ordered basis. Since A = A*, we have T = T*. Let  = { , ..., }
1 n

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