Page 383 - 35Linear Algebra

G.5 Linear Transformations 383

You can also model the new vector 2J obtained by scalar multiplication by
2 by thinking about Jenny hitting the puck twice (or a world with two Jenny
Potters....). Now ask yourself questions like whether the multiplicative
distributive law
2J + 2N = 2(J + N)

make sense in this context.

Hint for Review Question 5

Lets worry about the last part of the problem. The problem can be solved
by considering a non-zero simple polynomial, such as a degree 0 polynomial,
and multiplying by i ∈ C. That is to say we take a vector p ∈ P R and then
3
considering i·p. This will violate one of the vector space rules about scalars,
and you should take from this that the scalar field matters.
As a second hint, consider Q (the field of rational numbers). This is not
√ √
a vector space over R since 2 · 1 = 2 /∈ Q, so it is not closed under scalar
multiplication, but it is clearly a vector space over Q.

G.5 Linear Transformations

Hint for Review Question 5

The first thing we see in the problem is a definition of this new space P n .
Elements of P n are polynomials that look like

2
a 0 + a 1 t + a 2 t + . . . + a n t n
where the a i ’s are constants. So this means if L is a linear transformation
from P 2 → P 3 that the inputs of L are degree two polynomials which look like
a 0 + a 1 t + a 2 t 2

and the output will have degree three and look like

2
b 0 + b 1 t + b 2 t + b 3 t 3
We also know that L is a linear transformation, so what does that mean in
this case? Well, by linearity we know that we can separate out the sum, and
pull out the constants so we get
2
2
L(a 0 + a 1 t + a 2 t ) = a 0 L(1) + a 1 L(t) + a 2 L(t )
Just this should be really helpful for the first two parts of the problem. The
third part of the problem is asking us to think about this as a linear algebra
problem, so lets think about how we could write this in the vector notation we
use in the class. We could write

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