Page 277 - DMTH502_LINEAR_ALGEBRA

Unit 25: Linear Functional and Adjoints of Inner Product Space

= (T k j ) Notes

= A jk .

Example 2: Let V be a finite-dimensional inner product space and E the orthogonal
projection of V on a subspace W. The for any vectors and in V.
(E ) = (E E + (1 – E) )

= (E E )
= (E + (1 – E) E )
= ( E )

From the uniqueness of the operator E* it follows that E* = E. Now consider the projection E
described in Example 14 of unit 24. Then

9 36 3
1
A = 36 144 12
154
3 12 1
is the matrix of E in the standard orthonormal basis. Since E = E*, A is also the matrix of E*, and
because A = A*, this does not contradict the preceding corollary. On the other hand, suppose
= (154, 0, 0)
1
= (145, –36, 3)
2
= (–36, 10, 12)
3
Then { , , } is a basis, and
1 2 3
E = (9, 36, –3)
1
E = (0, 0, 0)
2
E = (0, 0, 0)
3
Since (9, 36, –3) = –(154, 0, 0) – (145, –36, 3), the matrix B of E in the basis { , , } is defined by
1 2 3
the equation
1 0 0
B = 1 0 0
0 0 0

In this case B B*, and B* is not the matrix of E* = E in the basis { , , }. Applying the corollary,
1 2 3
we conclude that { , , } is not an orthonormal basis. Of course this is quite obvious anyway.
1 2 3
Definition: Let T be a linear operator on an inner product space V. Then we say that T has an
adjoint on V if there exists a linear operator T* on V such that (T ) = ( T* ) for all and in V.
By Theorem 2 every linear operator on a finite-dimensional inner product space V has an adjoint
on V. In the finite-dimensional case this is not always true. But in any case there is at most one
such operator T*; when it exists, we call it the adjoint of T.

Two comments should be made about the finite-dimensional case.
1. The adjoint of T depends not only on T but on the inner product as well.

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