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Title: Spaces of polynomials
Description: Linear algebra course

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Spaces of Polynomials

4
...
and sets of matrices
...
q(x) =bo + b1x+···+ bnr, and t E JR?
...
x'1
1
Moreover, two polynomials p and q are equal if and only if ai =bi for 0 :: i :: n
...

(a) (2 + 3x + 4x2 + x3) + (5 + x - 2x2 + 7x3)
Solution:

(2+3x+4x2+x3)+(5+x-2x2+7x3) =2+5+(3+l)x+(4-2)x2+(1+7)x3
=7 + 4x + 2x2 + 8x3

(b) (3 + x2 - 5x3) - (1 - x - 2x2)
Solution:

(3+x2-5x3)-(l -x-2x2)

=

=

3-1 + [0-(-l)]x+ [1-(-2)]x2+ [-5-0]x3
2 + x + 3x2 - 5x3

EXAMPLE 1

(c)

5(2 + 3x

+

4x2 + x3)

(continued)

Solution: 5(2+3x+4x2+
...
x3=10+15x+20x2+5
...
x3)

Solution: 2(1 + 3x - x3) + 3(4 + x2 + 2
...
x3

+ 3(4) + 3(0)x + 3(1)x2 + 3(2)
...
x3

=14 + 6x + 3x2 + 4
...
These properties follow easily from the definitions of addition and scalar mul­
tiplication and are very similar to those for vectors in !Rn
...


2
...
11 (Theorem 1
...
1) and of matrices
(Theorem 3
...
1)
...
When we look at polynomials in this way, it is the coefficients of the polynomi­
als that are important
...
11 and matrices, we can also consider linear combinations of
polynomials
...


Definition

Let 13= (p1 (x),
...
Then the span

Span

of 13 is defined as

{p1 (x),
...


{1+x,1 +x3,x+x2,x+x3}
...

In particular, we have t1

EXAMPLE3

Determine if the set 13

-2, t2

=

=

=

3, t

3

=

3, and t4

=

1
...


Solution: Consider
0

=

=

t1 (1 +2x+2x2 - x3)+t2(3+2x+x2+x3)+t (2x2+2x3)
3
(t1 +3t2)+(2t1 +2t2)X+(2t1 +t2+2t )x2+(-t1 +t2+2t )X3
3
3

Comparing coefficients of the powers of x, we get a homogeneous system of linear
equations
...
Hence 13 is linearly independent
...
Is p(x)
1+5x - 5x2+x3 in the span

EXERCISE 1

Determine if :B

==

==

of :B?

EXERCISE 2

{1,x,x2,x3)
...


Consider :B

==

PROBLEMS 4
...


(2 - 2x+3x2+4x3) +(-3 - 4x+x2+2x3)
(b) (-3)(1- 2x+2x2+x3 + 4x4)
(c) (2 +3x+x2 - 2x3) - 3(1 - 2x+4x2 + 5x3)
(d) (2 +3x + 4x2) - (5 +x - 2x2)
(e) -2(-5 +x+x2) +3(-1 - x2)
( f) 2 ( � - tx+2x2) + H3 - 2x+x2)
(g) V2o +x+x2) + 7r(-1+x2)
Let :B
{1+x2+x3,2 +x+x3, -1+x+2x2+x3)
...
If a set is linearly dependent, find all
linear combinations of the polynomials that equal
the zero polynomial
...

(a) 0

2 +4x + 3x2+4x3
( c) -x+2x2+x3
(d) -4 - x+3x2

{1+2x+x2- x3,5x+x2,l- 3x+2x2+x3}
(b) {1+x+x2,x,x2+x3,3 +2x+2x2 - x3}
(c) {3 +x+x2,4 +x - x2,1+2x+x2+2x3,
-1+5x2+x3}
(d) {1+x+x3+x4,2 +x - x2+x3+x4,
x+x2+x3+x4}
Prove that the set :B
{1,x - 1,(x - 1)2} is linearly
(a)

==

(b)

-1+7x+Sx2+4x3
2 +x+Sx3

A4

==

independent and show that Span :B is the set of all
polynomials of degree less than or equal to

2
...


(3 +4x - 2x2+5x3) - (1 - 2x+Sx3)
(-2)(2 +x+x2+3x3 - x4)
(c) (-1)(2 +x+4x2+2x3) - 2(-1 - 2x - 2x2 - x3)
(d) 3(1+x+x3) +2(x - x2+x3)
(e) 0(1 +3x3 - 4x4)
(f) H3 - �x+x2) + !(2 +4x+x2)
(g) (1 + -Y2) (1 - -Y2+ < -Y2 - l ) x2 ) - H-2 + 2x2)
Let :B
{l+ x,x + x2,1 - x3}
...
If a set is linearly dependent, find all
linear combinations of the polynomials that equal
the zero polynomial
...

(a)
(b)
(c)
(d)

p(x)
p(x)
q(x)
q(x)

==

==

==

==

1
Sx+2x2+3x3
3 +x2 - 4x3
1+x3

{x2,x3,x2+x3+x4}

(b) {1+2•1
- �2• x+ �
6•x - �}
6
(c) {1 + x+x3,x+x3+ x5, 1 - x5 }
(d) (1 - 2x+x4,x - 2x2+ x5, 1 - 3x+x3}
(e) {l+2x+x2-x3, 2+3x-x2+x3+x4,1+x-2x2
+2x3+x4,1+2x+x2+x3 - 3x4,
4 + 6x - 2x2+Sx4)
Prove that the set 13
{1,x - 2,(x - 2 ) 2,( x - 2 ) 3}
(a)

B4

=

is linearly independent and show that Span :B is

the set of all polynomials of degree less than or
equal to

3
...
, Pk(x)} be a set of polynomials
of degree at most n
...
Span 13
...



Title: Spaces of polynomials
Description: Linear algebra course