Is the Set {x^2 - x, 3 - x^2, 1 + x} a Vector Subspace of P^2(x)?

[kent davidge]

Homework Statement

Determine the vector subspace generated by $A = \{x^2 -x, 3 - x^2, 1+x \} \subset P^2(x)$

Homework Equations

The Attempt at a Solution

I tried the usual check of vector addition and scalar multiplication to get the conditions that $x$ and $y$ should satisfy, but according to the Mathematica software the solution for $x$ and $y$ is the empty set!

OBS: by $x$ and $y$ I mean two elements in $A$, one given by $\{x^2 - x, 3 - x^2, 1+x \}$ and the other by $\{y^2 - y, 3 - y^2, 1 + y \}$.

 

[LCKurtz]

Homework Statement

Determine the vector subspace generated by $A = \{x^2 -x, 3 - x^2, 1+x \} \subset P^2(x)$

Homework Equations

The Attempt at a Solution

I tried the usual check of vector addition and scalar multiplication to get the conditions that $x$ and $y$ should satisfy, but according to the Mathematica software the solution for $x$ and $y$ is the empty set!

OBS: by $x$ and $y$ I mean two elements in $A$, one given by $\{x^2 - x, 3 - x^2, 1+x \}$ and the other by $\{y^2 - y, 3 - y^2, 1 + y \}$.

You aren't looking for equations x and y satisfy. And you don't have to check the scalar multiplication and addition because they are automatic for a subspace. The question is asking what subspace you would get with all linear combinations of those three functions. Is is a proper subspace or all of $P^2(x)$?

 

[kent davidge]

Ah, I see.

So, $a(x^2 - x) + b(3 - x^2) + c(1+x) = (a - b)x^2 + (c - a)x + 3b +c$. Seems like this covers all of $P^2(x)$. Correct?

 

[kent davidge]

$a(x^2 - x) + b(3 - x^2) + c(1+x) = (a - b)x^2 + (c - a)x + 3b +c$. Seems like this covers all of $P^2(x)$. Correct?

Ops, I realized that this is as long as $a-b, c-a \neq 0$.

 

[Mark44]

Ah, I see.

So, $a(x^2 - x) + b(3 - x^2) + c(1+x) = (a - b)x^2 + (c - a)x + 3b +c$. Seems like this covers all of $P^2(x)$. Correct?

Not quite. You're missing an important part of the equation. You want to show that the equation $a(x^2 - x) + b(3 - x^2) + c(1+x) = 0$ either has exactly one solution or there are an infinite number of solutions, relative to the coefficients a, b, and c. If there is exactly one solution -- a = 0, b = 0, c = 0, and no others, then the three functions are linearly independent, and span $P_2(x)$. If there are multiple solutions, then one of the functions is a linear combination of the other two, and the three functions span a proper subspace of $P_2(x)$.

 

[LCKurtz]

To say it "seems like" it works isn't exactly a mathematical argument. The question you need to answer to prove more than "seems like" is whether, if I give you a polynomial $P(x)=px^2+qx+r$, you can choose $a,b,c$ to get $P(x)$.

 

[kent davidge]

In the two cases

$a(x^2 - x) + b(3 - x^2) + c(1+x) = 0$

$P(x)=px^2+qx+r$

there are three variables and one equation, so the three variables cannot be all independent. Correct?

 

[PeroK]

In the two casesthere are three variables and one equation, so the three variables cannot be all independent. Correct?

Those two hints are leading you to different approaches. You have to decide which approach to take.

Method 1) Show that any polynomial can be expressed as a linear combination of the three polynomials you have.

Method 2) Show that the three polynomials you have are linearly independent.

Which method do you prefer?

 

[Mark44]

there are three variables and one equation, so the three variables cannot be all independent. Correct?

No, and you're not thinking about this in the right way. It's not the variables that are independent. In my approach, I was setting things up to determine whether the three functions were linearly independent.

Those two hints are leading you to different approaches. You have to decide which approach to take.

Yes.
My approach was to determine whether the three functions were linearly independent. If they were, then the three polynomials were a basis for $P_2(x)$. LCKurtz's approach was to determine whether an arbitrary polynomial, $P(x) = px^2 + qx +r$ could be written as a linear combination of the three functions in the given set of functions.

 

[kent davidge]

if I give you a polynomial $P(x)=px^2+qx+r$, you can choose $a,b,c$ to get $P(x)$.

Method 1) Show that any polynomial can be expressed as a linear combination of the three polynomials you have.

I'm not getting this. Doesn't the relation I wrote in post #3

$(a - b)x^2 + (c - a)x + 3b +c$

shows that we can write any polinomial in $P^2(x)$ as a linear combination of those three in the set $A$?

My approach was to determine whether the three functions were linearly independent

Should I solve the equation you gave in post #5 for $x$ or for $a,b,c$?

 

[PeroK]

I'm not getting this. Doesn't the relation I wrote in post #3
shows that we can write any polinomial in $P^2(x)$ as a linear combination of those three in the set $A$?

What you did in post #3 is not a proof. The right-hand side of your equation does not obviously cover all polynomials. The right-hand side needs to be $ax^2 + bx + c$. Then you'd have proved it.

 

[kent davidge]

The right-hand side of your equation does not obviously cover all polynomials. The right-hand side needs to be $ax^2 + bx + c$.

That's why I said in the below post that this was as long as those conditions were met. So we could conclude that $A$ doesn't cover all of $P^2(x)$, correct? Now it remains to determine what subspace of $P^2(x)$ it covers.

 

[PeroK]

That's why I said in the below post that this was as long as those conditions were met. So we could conclude that $A$ doesn't cover all of $P^2(x)$, correct? Now it remains to determine what subspace of $P^2(x)$ it covers.

You're not getting the idea here. You could try some examples first. Are the following polynomials in the span of the ones you have?

1) $5x^2 - 2x + 8$

2) $3x^2 + 4x + 7$

Can you generalise the process to show that any polynomial:

3) $ax^2 + bx + c$

Is in the span?

 

[Mark44]

Should I solve the equation you gave in post #5 for xxx?

No, that's not it at all.
Here's what I wrote in post #5:

You want to show that the equation $a(x^2 - x) + b(3 - x^2) + c(1+x) = 0$ either has exactly one solution or there are an infinite number of solutions, relative to the coefficients a, b, and c.

The equation has to be an identity; i.e., one that is true for all values of x What you to do is to solve for the constants a, b, and c. Clearly, if a = b = c = 0, that's a solutions, but the idea is to see whether there are other solutions for these constants, as well.

This business about linear independence of vectors or functions is one of the more difficult concepts for students in linear algebra. If your set includes three vectors, $v_1, v_2$, and $v_3$, then checking for linear independence or linear dependence both involve the equation $c_1v_1 + c_2v_2 + c_3v_3 = 0$. This equation always has a solution $c_1 = c_2 = c_3 = 0$, whether the vectors are linearly independent or linearly dependent. The part that students struggle with is realizing that for lin. independent vectors, the only value for the constants is zero. For lin. dependent vectors, there will be an infinite number of possible values.

 

[LCKurtz]

To finish the problem using my hint, you need to show me what values for $a,~b$ and $c$ in your expression $(a - b)x^2 + (c - a)x + 3b +c$ will give you $px^2+qx+r$. Your answer would be $a=?,~b=?, c = ?$, each in terms of $p,~q,~r$.

 

[mattt]

First, you must know some things, I guess. Do you know that P^2(x) (polynomials with real coefficients, I guess, and of degree 2 or less) is a real vector space (relative to the typical + and * operations) of dimension 3 ?

Do you also know that the set \{ x^2, x, 1\} is a base (known as the canonical base) of this space?

Your polynomials have coordinates (1,-1,0), (-1,0,3), and (0,1,1) respectively, with respect to the canonical base, and you must prove if these three polynomials are linearly independents or not. If not, you must give a base of the space they span, or the equations of this space spanned by them. Do you understand what they want you to do?

 

[kent davidge]

Firstly, Thanks to all. Secondly, excuse me if my ignorance still prevents me of getting the solution.

So, putting together all these hints down from post #12, I came to the following:

Here your hint/notation helped a lot me in figuring things out @mattt
A vector with components $(a,b,c)$ in the basis $(x^2 - x, 3 - x^2, 1 + x)$ is $a(x^2 - x) + b(3 - x^2) + c(1 + x)$.

This same vector becomes $(a-b)x^2 + (c - a)x + (3b + c)1$ in the basis $(x^2, x, 1)$, i.e., its components transform like $(a,b,c) \longrightarrow (a-b, c-a, 3b+c)$.

So I can conclude that if a general vector of $P^2(x)$ have components that span $\mathbb{R}^3$ (in the sense that it has components $(a,b,c)$ with each component in $\mathbb{R}$) then a general vector in $A$ with respect to $P^2 (x)$ has components $(a-b, c-a, 3b +c)$, which doesn't span all of $\mathbb{R}^3$ and thus is a subspace of it. The problem is that I ended up concluding something involving $\mathbb{R}^3$ instead of $P^2(x)$ and $A$.

 

[kent davidge]

you need to show me what values for $a,~b$ and $c$ in your expression $(a - b)x^2 + (c - a)x + 3b +c$ will give you $px^2+qx+r$. Your answer would be $a=?,~b=?, c = ?$, each in terms of $p,~q,~r$.

They would be

$$a = \frac{1}{4}(r-(-3p+q)) \\ b = \frac{1}{4}(r-(p+q)) \\ c = \frac{1}{4}(r + 3(p + q))$$
and this result would show us that we can choose the above $\{a,b,c \}$ to put in $(a - b)x^2 + (c - a)x + 3b +c$ to get any $px^2+qx+r$.

 

[LCKurtz]

then a general vector in $A$ with respect to $P^2 (x)$ has components $(a-b, c-a, 3b +c)$, which doesn't span all of $\mathbb{R}^3$ and thus is a subspace of it. The problem is that I ended up concluding something involving $\mathbb{R}^3$ instead of $P^2(x)$ and $A$.

Why do you think it doesn't span $R^3$?

 

[LCKurtz]

They would be

$$a = \frac{1}{4}(r-(-3p+q)) \\ b = \frac{1}{4}(r-(p+q)) \\ c = \frac{1}{4}(r + 3(p + q))$$
and this result would show us that we can choose the above $\{a,b,c \}$ to put in $(a - b)x^2 + (c - a)x + 3b +c$ to get any $px^2+qx+r$.

Have you checked that those values work? Assuming so, do you understand how you have solved the problem and what the answer is?

 

[kent davidge]

Have you checked that those values work?

Yes, I did.

do you understand how you have solved the problem and what the answer is?

Unfortunately, no. I don't see how we can express the subspace generated by $A$ by using what we found.

 

[Mark44]

The problem is that I ended up concluding something involving $\mathbb{R}^3$ instead of $P^2(x)$ and $A$.

That's not a problem. $\mathbb R^3$ and $P_2(x)$ are isomorphic. Each vector in $\mathbb R^3$ corresponds to one function in $P_2(x)$ and vice versa. The reason for switching to the space of vectors is to allow the use of matrices in solving equations.

 

[mattt]

Because all real vector spaces of the same dimension are isomorfic, and \{(1,0,0), (0,1,0), (0,0,1)\} is a base in \mathbb{R}^3, and \{x^2, x, 1\} is a base in P^2(x), you know that the set \{x^2-x, -x^2 +3, x+1\} is linearly independent in P^2(x) if and only if the set \{(1,-1,0), (-1,0,3),(0,1,1)\} is linearly independent in \mathbb{R}^3

Do you know how to check whether a set of vectors, in a given vector space, is linearly independent?

 

[PeroK]

Yes, I did.

Unfortunately, no. I don't see how we can express the subspace generated by $A$ by using what we found.

This is the point where I think you need a lesson in linear algebra, rather than just some help with a particular homework.

What you have done is:

Taken an arbitrary polynomial.
Explicitly found the linear combination of your polynomials that equals that arbitrary polynomial.

We think this proves that the span of your polynomials is the whole space.

It's not clear why you don't see this!

 

[kent davidge]

We think this proves that the span of your polynomials is the whole space.

This statement seems to be in conflict to what is noticed by @mattt , because according to his post
$\{(1,-1,0), (-1,0,3),(0,1,1)\}$ we know that this is not linearly independent in $\mathbb{R}^3$ and so $\{x^2-x, -x^2 +3, x+1\}$ is not linearly independent in $P^2(x)$ and thus cannot form a basis for $P^2(x)$ and thus don't conver it.

It's not clear why you don't see this!

I think because the problem original statement is 'determine the subspace generated...' so I thought it must be a subspace not equal to the whole space.

 

[mattt]

This statement seems to be in conflict to what is noticed by @mattt , because according to his post
$\{(1,-1,0), (-1,0,3),(0,1,1)\}$ we know that this is not linearly independent in $\mathbb{R}^3$ and ...

Are you sure of that? :-)

 

[kent davidge]

Ohhh yea, I went through the calculation and they are linearly independent in $\mathbb{R}^3$.

 

[kent davidge]

So the answer is that the subspace generated by $A$ is $P^2(x)$. Never the less I would like to know how would we write our answer to the problem if the subspace didn't cover all of $P^2(x)$. Can someone tell me how to express the subspace in such case?

 

[PeroK]

So the answer is that the subspace generated by $A$ is $P^2(x)$. Never the less I would like to know how would we write our answer to the problem if the subspace didn't cover all of $P^2(x)$. Can someone tell me how to express the subspace in such case?

An example answer might be: All polynomials of the form $ax^2 + ax + c$.

 

[kent davidge]

Ah, ok