Polynomial Span related problem Linear Algebra

[1LastTry]

Homework Statement

Consider the vector space F(R) = {f | f : R → R}, with the standard operations. Recall that the zero of F(R) is the function that has the value 0 for all
x ∈ R:
Let U = {f ∈ F(R) | f(1) = f(−1)} be the subspace of functions which have
the same value at x = −1 and x = 1.
Define functions g; h; j and k ∈ F[R] by
g(x) = 2x^3− x − 2x^2+ 1;
h(x) = x^3+ x^2− x + 1;
k(x) = −x^3+ 5x^2+ x + 1 and
j(x) = x^3− x; ∀x ∈ R:
a) Show that g and h belong to U.
b) Show that k ∈ span{g; h}.
c) Show that j /∈ span{g; h}.
d) Show that span{g; h} /= span{g; h; j}.

Homework Equations

The Attempt at a Solution

My Attempt:

I have no idea how to do a) I tried plugging -1 and +1 in g and h however it doesn't meet the requirement of f(1)=f(-1).

I am sure I know how to do party b,c and d. However, I am doing something wrong and have no clue what it is...

So what I did for b) was:

k = a(g) + b(h)

after that you get something like (...)x^3 + (...)x^2 + (...)x + (a+b)

So I set the coefficient (...) equal to the coefficients of k (-1,5,1,1)
however I am unable to solve the linear system.
But it says it is in the span so I don't know what happened maybe mtah error, but I redid it a lot of times.

 

[HallsofIvy]

Homework Statement

Consider the vector space F(R) = {f | f : R → R}, with the standard operations. Recall that the zero of F(R) is the function that has the value 0 for all
x ∈ R:
Let U = {f ∈ F(R) | f(1) = f(−1)} be the subspace of functions which have
the same value at x = −1 and x = 1.
Define functions g; h; j and k ∈ F[R] by
g(x) = 2x^3− x − 2x^2+ 1;

I suspect you have copied this wrong. It should be g(x)= 2x^3- x^2- 2x+ 1.

h(x) = x^3+ x^2− x + 1;
k(x) = −x^3+ 5x^2+ x + 1 and
j(x) = x^3− x; ∀x ∈ R:
a) Show that g and h belong to U.
b) Show that k ∈ span{g; h}.
c) Show that j /∈ span{g; h}.
d) Show that span{g; h} /= span{g; h; j}.

Homework Equations

The Attempt at a Solution

My Attempt:

I have no idea how to do a) I tried plugging -1 and +1 in g and h however it doesn't meet the requirement of f(1)=f(-1).
I am sure I know how to do party b,c and d. However, I am doing something wrong and have no clue what it is...

So what I did for b) was:

k = a(g) + b(h)

after that you get something like (...)x^3 + (...)x^2 + (...)x + (a+b)

So I set the coefficient (...) equal to the coefficients of k (-1,5,1,1)
however I am unable to solve the linear system.
But it says it is in the span so I don't know what happened maybe mtah error, but I redid it a lot of times.

 

[1LastTry]

no I didn't copy anything wrong that was exactly how it was.

 

[1LastTry]

I suspect you have copied this wrong. It should be g(x)= 2x^3- x^2- 2x+ 1.

it is right it is -x* not -2x

 

[HallsofIvy]

Then it is an error in your text. The statements are true if g(x)= )= 2x^3- x^2- 2x+ 1, not 2x^3- x- 2x^2+ 1. (It is also peculiar that the polynomial is not in standard form. That is why I thought about 2x^3- x^2- 2x+ 1.)

 

[v1ru5]

HallsofIvy, you are right. g(x) = 2x^3 - x^2 -2x + 1. I did part a by calculating g(1) = 0 and g(-1) = 0 therefore g(1) = g(-1) so g belongs to U. Then I did h(1) = 2 and h(-1) = 2 so h(1) = h(-1) therefore h also belongs to U. I am stuck at the span part and not really sure what I am supposed to do. It would be great if someone could help me out :)

 

[v1ru5]

anyone want to help me with part b please? I understand that it is asking whether k is a linear combination of g + h. So basically k = ag + bh of a and b are all real numbers. I just don't understand what number to use for a and b because none of them get be k :( Any help would be appreciated.

Thanks!

 

[1LastTry]

It is 2x^3-x^2-2x+1 the teacher made a mistake.

To do this you do what you said set ag+bh=k

You don't have to sub in for x

so ag+bh=k would be: a(2x^3-x^2-2x+1)+b(x^3+x^2-x+1)=-x^3+5x^2+x+1

so after you expand and factor out the a,b and c you get

(2+b)x^3-(a-b)x^2-(2+b)x+a+b = -x^3+5x^2+x+1

so you set the coefficients equal to each other you get:

2a+b=-1
-a+b=5
-2a-b=1
a+b=1

Solve this linear system and you has the answer, it is solvable so it is part of the span.

 

[v1ru5]

It is 2x^3-x^2-2x+1 the teacher made a mistake.

To do this you do what you said set ag+bh=k

You don't have to sub in for x

so ag+bh=k would be: a(2x^3-x^2-2x+1)+b(x^3+x^2-x+1)=-x^3+5x^2+x+1

so after you expand and factor out the a,b and c you get

(2+b)x^3-(a-b)x^2-(2+b)x+a+b = -x^3+5x^2+x+1

so you set the coefficients equal to each other you get:

2a+b=-1
-a+b=5
-2a-b=1
a+b=1

Solve this linear system and you has the answer, it is solvable so it is part of the span.

Oh okay, thanks for your help! See you around uOttawa :D

 

[s_nirmit]

how do you do part d i don't understand it

 

[1LastTry]

after you verify c) you can just say that since j is not in {g,h} therefore {g,h} is not in {g,h,j}.

It is basically the same process if you don't do that^. :)