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I have the vector space $V$ above that belongs to $\mathbb{F}$, and $V$ is the group of all polynomials that are of degree $3$.

$W= \{ p \in V | p(1)=p(-1)=0\}$

1.) Prove that W is a subspace of $V$.

A.) Ok let's prove that the space isn't empty.

Let's take $p(x)= 0$, and this satisfies the condition of $p(1)=p(-1)=0$

B.) Now that the sum of two vectors in $W$ belong in $V$

Let's take two polynomials that satisfy $W$ and show that their sum belongs in $W$. This is pretty trivial since you can't go increase your degree from addition.

Let's take $p_1(x)=x_1^3-x_1^2-x_1+1$ and $p_2(x)=x_2^3-x_2^2-x_2+1$.

Their sum belongs in $V$.

Let's check multiplication by a scalar $\lambda \in \mathbb{R} $(also pretty trivial):

$p(\lambda x)=\lambda x^3-\lambda x^2-\lambda x+1 = \lambda p(x)$

Now onto the basis:

My thought process is that I might need to find a basis that contains at least 4 components (since $\mathbb{R}_3[x]$ contains 4 components), which I don't think is correct.

So far I have two components:

B= $\{ (x^2-1),(x(x^2-1))\} $

This is where I'm stuck and can't find anymore polynomials that satisfy the requirement.

TL;DR

1.) Is my thought process for proving that $W$ is a subspace of $V$ correct?

2.) How can I find a basis for $W$? Is there a simpler way using a system of equations rather than brute force trial and error?

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4 Answers 4

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Sorry, but your line of thought is quite wrong.

Your part B is the worst one. Apart from using $x_1$ and $x_2$, which is already bad, you can't prove the set is closed under addition by showing that the sum of two particular element is in the set. You need to show that for any choice of polynomials $p,q\in W$, the sum $p+q\in W$. Similarly for the closure with respect to scalar multiplication.

Let's try a more systematic approach.

Consider the vector space $V$ of polynomials of degree at most $3$ (not *of degree $3$) with real coefficients.

For $a\in\mathbb{R}$, consider $W_a=\{p\in V: p(a)=0\}$. Then $W_a$ is a subspace of $V$. Indeed, the zero polynomial clearly belongs to $W_a$. Moreover, if $p,q\in W_a$, we have $p(a)=0$ and $q(a)=0$; so, if $r(x)=p(x)+q(x)$, we have $$ r(a)=p(a)+q(a)=0+0=0 $$ because of standard laws of polynomials. Finally, if $\lambda\in\mathbb{R}$ and $s(x)=\lambda p(x)$, we have $$ s(a)=\lambda p(a)=0 $$ Therefore $p(x)+q(x)\in W_a$ and $\lambda p(x)\in W_a$.

Your set is $W=W_1\cap W_{-1}$ and is a subspace being the intersection of two subspaces.

In order to find a basis, you can consider the linear map $$ f\colon V\to\mathbb{R}^2,\qquad f(p)=\begin{bmatrix}p(1)\\p(-1)\end{bmatrix} $$ and notice that $W=\ker f$. (Note that linearity of this map is basically the same as the proof above.)

It's easy to find the matrix of $f$ relative to the standard bases on $V$ (that is, $\{1,x,x^2,x^3\}$ and of $\mathbb{R}^2$. It is $$ \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & -1 & 1 & -1 \end{bmatrix} $$ A standard Gaussian elimination brings to the reduced row echelon form $$ \begin{bmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \end{bmatrix} $$ and the null space of this matrix has as basis $$ \left\{ \begin{bmatrix}-1\\0\\1\\0\end{bmatrix}, \begin{bmatrix}0\\-1\\0\\1\end{bmatrix} \right\} $$ The polynomials having these coordinates with respect to the standard basis are $x^2-1$ and $x^3-x$, so a basis for $W$ is $$ \{x^2-1,x^3-x\} $$ as was your intuition.

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$P(x)=ax^3+bx^2+cx+d$ $$\begin{aligned} p(1)&=a+b+c+d \\ p(-1)&=-a+b-c+d \end{aligned} $$ so $p(1)-p(-1)=0 $ $$ p(1)-p(-1)=a+b+c+d -(a+b-c+d)=2a+2c$$ 2-d plane like x-y axis

your basis does work $\{x^2-1,x^3-x\}$ feel that grader would not like how you show it is a subspace. Do not use lambda it means to grader e-value. I seen a peer get docked a whole letter grade for not using proper notation. Also, when closed under addition you used a particlar case this how I would do it

Closed add $p,g\in W$ consider $h(x)=p(x)+g(x)$ $$ \begin{aligned} h(1)&=(p+g)(1)=p(1)+g(1)=0+0=0 \\ h(-1)&=(p+g)(-1)=p(-1)+g(-1)=0+0=0 \end{aligned}$$ so $$ h(1)=h(-1)=0$$

closed under scalar mult $ap(1)=a(0)=0$and $ap(-1)=a*0=0$ so $ap(1)=ap(-1)=0$

lastly show $0\in W$ treating 0 like a poly $0(1)=0=0(-1)$

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  • $\begingroup$ That's exactly what I was looking for ! Thanks! $\endgroup$
    – user300011
    Commented Apr 29, 2016 at 19:20
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    $\begingroup$ What's wrong with $\lambda$? $\endgroup$
    – djechlin
    Commented Apr 29, 2016 at 23:25
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Your thought process is correct, you are also right in finding a base that has 2 elements: the space $W$ is 2-dimensional.

You can show this by writing the general element $p = ax^3+bx^2+cx+d$ and using $p(1)=0=p(-1)$ on it, to get a system of linear equations that yield $a = -c, b = -d$ and thus rewrite the space as $\{ax^3+bx^2-ax-b\}$ (abusing notation a bit).

You can get a basis by substituting two linearly independent pairs for $(a, b)$ (and in fact, using $(1, 0)$ and $(0, 1)$ yields exactly your base).

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  • $\begingroup$ That's exactly what I just did just now thanks to another question. Another question is: how do we know that W is 2 dimensional? $\endgroup$
    – user300011
    Commented Apr 29, 2016 at 19:20
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    $\begingroup$ @RonaldB Intuitively, you start with a 4-dimensional space, and fix two conditions. $\endgroup$ Commented Apr 29, 2016 at 19:22
  • $\begingroup$ So then let's say we have one condition, the basis would need 3 components? $\endgroup$
    – user300011
    Commented Apr 29, 2016 at 19:22
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    $\begingroup$ Intuitively, you start with 4 dimensions and lose two because of the two conditions you set. You can also get the same result more formally by solving the equations like Hugo did, and seeing that you are left with two parameters; that the entire space is then generated by a base with two elements, and therefore is 2-dimensional. $\endgroup$ Commented Apr 29, 2016 at 19:25
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    $\begingroup$ @RonaldB Indeed: as long as it isn't too bad a condition, like an impossible one. Try doing the same thing without the $p(-1)=0$ constraint $\endgroup$ Commented Apr 29, 2016 at 19:26
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If $W= \{ p \in V | p(1)=p(-1)=0\} $, then, independent of the degree restriction of $V$, since each $p$ is a polynomial, $p(x) =(x-1)(x+1)q(x) =(x^2-1)q(x) $, where $q$ is a polynomial of degree $2$ less than $p$.

Since $\deg(p) = 3$ by the definition of $V$, then $\deg(q) = 1$, so $q(x) =ax+b $ for some $a$ and $b$.

It is now easy to prove that $W$ is a subspace, since that follows from the set of linear polynomials being closed under addition and multiplication by a scalar.

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