Page 298 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes for all , in V. If f and g are forms and c a scalar, then
(cf + g) ( , ) = (T | )
cf + g
= cf( , ) + g( , )
= c(T | ) + (T | )
f g
= (cT + T | | )
f g
for all and in V. Therefore,
T = cT + T
cf + g 1 g
so f T is a linear map. For each T in L(V, V) the equation
f
f( , ) = (T | )
defines a form such that T = T, and T = 0 if and only if f = 0. Thus f T is an isomorphism.
f f f
Corollary: The equation
(f|g) = tr(T T* )
f g
defines an inner product on the space of forms with the property that
g
(f|g) = ( f k , j ) ( k , j )
, j k
for every orthonormal basis { , ..., } of V.
1 n
Proof: It follows easily from Example 3 of unit 24 that (T, U) tr (TU*) is an inner product on
L(V, V). Since f T is an isomorphism, Example 6 of unit 24 shows that
f
(f|g) = tr (T T* )
f g
is an inner product. Now suppose that A and B are the matrices of T and T in the orthonormal
f g
basis = { , ..., }. Then
1 n
A = (T | ) = f( , )
jk f k j k j
and B = (T | ) = g( , ). Since AB* is the matrix of T T* in the basis , it follows that
jk g k j k j f g
(f|g) = tr (AB*) = A B
jk jk
, j k
Definition: If f is a form and = { , ..., } an arbitrary ordered basis of V, the matrix A with
1 n
entries
A = f( , )
jk k j
is called the matrix of f in the ordered basis .
When is an orthonormal basis, the matrix of f in is also the matrix of the linear transformation
T , but in general this is not the case.
f
If A is the matrix of f in the ordered basis = ( , ... ), if follows that
1 n
y A x
f x s s y r r = r rs s ...(1)
s r , r s

for all scalars x, and y (1 r, s n). In other words, the matrix A has the property that
f( , ) = Y*AX
where X and Y are the respective coordinate matrices of and in the ordered basis .

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