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Complex Analysis and Differential Geometry
Notes The fact is that in some books the linear operator is determined first, then its matrix is introduced
by formula (13). Explain why if we know three vectors F(e ), F(e ), and F(e ), then we can
1
2
3
reconstruct the whole matrix of operator F by means of formula (13).
Suppose we have two linear operators F and H. We can apply H to vector x and then we can
apply F to vector H(x). As a result we get
F H(x) = F(H(x)). ...(14)
Here F H is new linear operator introduced by formula (14). It is called a composite operator,
and the small circle sign denotes composition.
Exercise 15.12: Find the matrix of composite operator F H if the matrices for F and H in the
basis e , e , e are known.
1
3
2
Exercise 15.13: Remember the definition of the identity map in mathematics (see on-line Math.
Encyclopedia) and define the identity operator id. Find the matrix of this operator.
Exercise 15.14: Remember the definition of the inverse map in mathematics and define inverse
operator F for linear operator F. Find the matrix of this operator if the matrix of F is known.
1
15.4 Bilinear and Quadratic Forms
Vectors, covectors, and linear operators are all examples of tensors (though we have no definition
of tensors yet). Now we consider another one class of tensorial objects. For the sake of clarity,
lets denote by a one of such objects. In each basis e , e , e this object is represented by some
3
1
2
square 3 × 3 matrix a of real numbers. Under a change of basis these numbers are transformed
ij
as follows:
3 3
q
p
ij
a S S a , ...(1)
i
pq
j
p 1 q 1
3 3
a = T T a . ...(2)
p
ij
q
j
i
pq
p 1 q 1
Transformation rules (1) and (2) can be written in matrix form:
T
T
a S aS, a T aT. ...(3)
Here by S and T we denote the transposed matrices for S and T respectively.
T
T
Exercise 15.15: Derive (2) from (1), then (3) from (1) and (2).
Definition: A geometric object a in each basis represented by some square matrix aij and such
that components of its matrix a obey transformation rules (1) and (2) under a change of basis is
ij
called a bilinear form.
Lets consider two arbitrary vectors x and y. We use their coordinates and the components of
bilinear form a in order to write the following sum:
3 3
a(x, y) = a x y . j ...(4)
i
ij
i 1 j 1
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