Page 276 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes :
= T* .
We have (2), but we must verify that T* is a linear operator. Let , be in V and let c be a scalar.
Then for any ,
( T* (c )) = (T c )
= (T c ) + (T )

= c (T ) (T )

= c ( T * ) ( T * )
= ( cT* ) + ( T* )
= ( c T* + T* ).

Thus T* (c ) = cT* + T* and T* is linear operator.
The uniqueness of T* is clear. For any in V, the vector T* is uniquely determined as the vector
such that (T ) = ( ) for every .

Theorem 3: Let V be a finite-dimensional inner product space and let  = { , ...., } be an
1 n
(ordered) orthonormal basis for V. Let T be a linear operator on V and let A be the matrix of T in
the ordered basis . Then A = (T ).
kj j k
Proof: Since  is an orthonormal basis, we have

n
= ( k ) k .
k 1

The matrix A is defined by
n
T = A
j j k k
k 1
and since

n
T = (T )
j j k k
k 1
we have A (T ).
j k j k
Corollary: Let V be a finite-dimensional inner product space, and let T be a linear operator on V.
In any orthonormal basis for V, the matrix of T* is the conjugate transpose of the matrix of T.
Proof: Let  = { , ...., } be an orthonormal basis for V, let A = [T] and B = [T*] . According to
1 n  
Theorem 3,
A = (T )
kj j k
B = (T* ).
kj j k
By the definition of T* we then have
B = (T* )
kj j k
= ( k T * j )

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