Page 116 - 35Linear Algebra

P. 116

116 Linear Transformations

2
You saw that a linear operator acting on R is completely speciﬁed by

1 0
how it acts on the pair of vectors and . In fact, any linear operator
0 1
2
acting on R is also completely speciﬁed by how it acts on the pair of vectors

1 1
and .
1 −1

Example 73 The linear operator L is a linear operator then it is completely speciﬁed
by the two equalities

1 2 1 6
L = , and L = .
1 4 −1 8

x 1 1
2
This is because any vector in R is a sum of multiples of and which
y 1 −1
can be calculated via a linear systems problem as follows:

x 1 1
= a + b
y 1 −1

1 1 a x
⇔ =
1 −1 b y
x+y
1 1 x 1 0
2
⇔ ∼ x−y
1 −1 y 0 1
2
x+y
a =
2
⇔ x−y
b = .
2
Thus
! ! !
x x + y 1 x − y 1
= + .
y 2 1 2 −1

We can then calculate how L acts on any vector by ﬁrst expressing the vector as a

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