[Linear Alg] Determining which sets are subspaces of R[x]

[iJake]

Homework Statement

[/B]
Which of the following sets are subspaces of $R[x]?$

$W_1 = {f \in \mathbf R[x] : f(0) = 0}$

$W_2 = {f \in \mathbf R[x] : 2f(0) = f(1)}$

$W_3 = {f \in \mathbf R[x] : f(t) = f(1-t) \forall t \in \mathbf R}$

$W_4 = {f \in \mathbf R[x] : f = \sum_{i=0}^n a_ix^2i}$

Homework Equations

The criteria for a given set being a subspace are that it be closed under addition and scalar multiplication, and also that the 0 vector belongs to the set.

The Attempt at a Solution

[/B]
$W_1$ is a subspace:

$f(0)+g(0) = (f+g)(0) \Rightarrow 0 + 0 = 0 \in W_1$
$\lambda \cdot f(0) = (\lambda \cdot f(0)) = \lambda \cdot 0 = 0 \in W_1$
$0v = 0 \in W_1$

$W_2$ is not a subspace:

$2f(0) = f(1) \neq (2 \cdot f(0)) \Rightarrow$ The condition of closure under scalar multiplication is not met.

I also suspect the 0 vector does not belong to this set, but I am unsure of how to show this.

The notation and executing the notation correctly is my biggest problem here, however, I appreciate all conceptual insight. I have thus far not been able to correctly describe whether or not $W_3$ and $W_4$ are subspaces. My suspicion is that $W_3$ is not a subspace because $f(0) = f(1-0) = f(1) \neq f(0)$ but I am uncertain of how to improve this train of thought. Similarly, I believe $W_4$ meets all the criteria to be a subspace of $\mathbf R[x]$ but I do not know how to correctly use notation to describe how it meets those criteria.

Additionally, I would like clarification about what exactly $\mathbf R[x]$ refers to.

Thank you very much.

 

[fresh_42]

Homework Statement

[/B]
Which of the following sets are subspaces of $R[x]?$

$W_1 = {f \in \mathbf R[x] : f(0) = 0}$

$W_2 = {f \in \mathbf R[x] : 2f(0) = f(1)}$

$W_3 = {f \in \mathbf R[x] : f(t) = f(1-t) \forall t \in \mathbf R}$

$W_4 = {f \in \mathbf R[x] : f = \sum_{i=0}^n a_ix^2i}$

Homework Equations

The criteria for a given set being a subspace are that it be closed under addition and scalar multiplication, and also that the 0 vector belongs to the set.

The Attempt at a Solution

[/B]
$W_1$ is a subspace:

$f(0)+g(0) = (f+g)(0) \Rightarrow 0 + 0 = 0 \in W_1$
$\lambda \cdot f(0) = (\lambda \cdot f(0)) = \lambda \cdot 0 = 0 \in W_1$
$0v = 0 \in W_1$

That's correct.

$W_2$ is not a subspace:

$2f(0) = f(1) \neq (2 \cdot f(0)) \Rightarrow$ The condition of closure under scalar multiplication is not met.

I don't understand this: $2f(0) = 2 \cdot f(0)$, what else should it be? And if we have e.g. $f(x)=a(x+1)$ then $2f(0)=2a=f(1)$ plus $\alpha \cdot a(x+1) + \beta \cdot b(x+1) = (\alpha a +\beta b)(x+1)$ is again of this form. So why isn't $W_2$ a subspace? I don't know whether it is, but your argumentation is wrong.

I also suspect the 0 vector does not belong to this set, but I am unsure of how to show this.

You can't. $2\cdot 0(0)= 0 = 0(1)$.

The notation and executing the notation correctly is my biggest problem here, however, I appreciate all conceptual insight. I have thus far not been able to correctly describe whether or not $W_3$ and $W_4$ are subspaces. My suspicion is that $W_3$ is not a subspace because $f(0) = f(1-0) = f(1) \neq f(0)$ but I am uncertain of how to improve this train of thought. Similarly, I believe $W_4$ meets all the criteria to be a subspace of $\mathbf R[x]$ but I do not know how to correctly use notation to describe how it meets those criteria.

Additionally, I would like clarification about what exactly $\mathbf R[x]$ refers to.

Thank you very much.

You have to show: Given any $f\; , \;g \in W_i$ and any $\alpha\; , \;\beta \in R$ - whatever $R$ is - then $\alpha \cdot f + \beta \cdot g \in W_i$. If this is the case, then the zero vector is automatically in $W_i$. Do you see why?

 

[Ray Vickson]

Homework Statement

[/B]
Which of the following sets are subspaces of $R[x]?$

$W_1 = {f \in \mathbf R[x] : f(0) = 0}$

$W_2 = {f \in \mathbf R[x] : 2f(0) = f(1)}$

$W_3 = {f \in \mathbf R[x] : f(t) = f(1-t) \forall t \in \mathbf R}$

$W_4 = {f \in \mathbf R[x] : f = \sum_{i=0}^n a_ix^2i}$

Homework Equations

The criteria for a given set being a subspace are that it be closed under addition and scalar multiplication, and also that the 0 vector belongs to the set.

The Attempt at a Solution

[/B]
$W_1$ is a subspace:

$f(0)+g(0) = (f+g)(0) \Rightarrow 0 + 0 = 0 \in W_1$
$\lambda \cdot f(0) = (\lambda \cdot f(0)) = \lambda \cdot 0 = 0 \in W_1$
$0v = 0 \in W_1$

$W_2$ is not a subspace:

$2f(0) = f(1) \neq (2 \cdot f(0)) \Rightarrow$ The condition of closure under scalar multiplication is not met.

I also suspect the 0 vector does not belong to this set, but I am unsure of how to show this.

The notation and executing the notation correctly is my biggest problem here, however, I appreciate all conceptual insight. I have thus far not been able to correctly describe whether or not $W_3$ and $W_4$ are subspaces. My suspicion is that $W_3$ is not a subspace because $f(0) = f(1-0) = f(1) \neq f(0)$ but I am uncertain of how to improve this train of thought. Similarly, I believe $W_4$ meets all the criteria to be a subspace of $\mathbf R[x]$ but I do not know how to correctly use notation to describe how it meets those criteria.

Additionally, I would like clarification about what exactly $\mathbf R[x]$ refers to.

Thank you very much.

I do not believe your argument for $W_2$. Clearly, if $2f(0)=f(1)$ then, for any $c \in \mathbf{R}$ we have $2 c f(0) = c f(1)$, so $c f \in W_2$. Does the "zero" function not satisfy the condition for $W_2?$

 

[iJake]

That's correct.
I don't understand this: $2f(0) = 2 \cdot f(0)$, what else should it be? And if we have e.g. $f(x)=a(x+1)$ then $2f(0)=2a=f(1)$ plus $\alpha \cdot a(x+1) + \beta \cdot b(x+1) = (\alpha a +\beta b)(x+1)$ is again of this form. So why isn't $W_2$ a subspace? I don't know whether it is, but your argumentation is wrong.

I understand. Then in this case, I don't see why $W_2$ isn't a subspace:

$f(x) + g(x) = (f+g)(x) \in W_2$
$c \cdot f(x) = (c \cdot f(x)) \in W_2$
$0(x) = 0 \in W_2$

You have to show: Given any $f\; , \;g \in W_i$ and any $\alpha\; , \;\beta \in R$ - whatever $R$ is - then $\alpha \cdot f + \beta \cdot g \in W_i$. If this is the case, then the zero vector is automatically in $W_i$. Do you see why?

Yes, if your statement holds, it combines the conditions for closure under addition and scalar multiplication, and if those conditions are met then unless 0 is specifically excluded from the set, you can produce the 0 vector.

Thinking more about $W_3$ :

$f(t) + g(t) = f(1-t) + g(1-t) = (f+g)(1-t) \in W_3$
$c \cdot f(t) = (c \cdot f)(t) = (c \cdot f)(1-t) \in W_3$
$0(t) = 0(1-t) = 0 \in W_3$

I am nonetheless suspicious about scalar multiplication here...

Thanks for your help, fresh_42 and Ray Vickson!

 

[fresh_42]

I understand. Then in this case, I don't see why $W_2$ isn't a subspace:

Me neither.

Thinking more about $W_3$ :

$f(t) + g(t) = f(1-t) + g(1-t) = (f+g)(1-t) \in W_3$
$c \cdot f(t) = (c \cdot f)(t) = (c \cdot f)(1-t) \in W_3$
$0(t) = 0(1-t) = 0 \in W_3$

A bit dirty, but correct. The order is as follows:
$(f+g)(t)= f(t) + g(t) = \ldots \in W_3$ which you have done right, except that you left out the first term. But now it should be the same
$(c \cdot f)(t)= c\cdot f(t) = c \cdot f(1-t) = (c \cdot f)(1-t) \in W_3$

I am nonetheless suspicious about scalar multiplication here...

Why? Do you mean the definition of $c\cdot f$ or the order which you mixed a little bit or the definition of $c\cdot 0\,$?

 

[iJake]

Me neither.
A bit dirty, but correct. The order is as follows:
$(f+g)(t)= f(t) + g(t) = \ldots \in W_3$ which you have done right, except that you left out the first term. But now it should be the same
$(c \cdot f)(t)= c\cdot f(t) = c \cdot f(1-t) = (c \cdot f)(1-t) \in W_3$

Why? Do you mean the definition of $c\cdot f$ or the order which you mixed a little bit or the definition of $c\cdot 0\,$?

Haha, I think it was just my sloppy ordering that confused me.

So I have concluded that $W_1, W_2, W_3$ are all subspaces.

To think a little more about $W_4$ then:

(note, the exponent over x is supposed to be $2i$)

$W_4 = \{f \in \mathbf R[x] : f = \sum_{i=0}^n a_ix^{2i}\}$

$a_ix^{2i} + b_ix^{2i} = (a_i+b_i)x^{2i} \in W_4$

$c \cdot a_ix^{2i} = (c \cdot a_i)x^{2i} \in W_4$

I'm not sure about the notation for a zero vector here, would it simply be $0 \cdot a_ix^{2i}$?

EDIT: Additionally I would like to know if there is anything to be said about the notation $\mathbf R[x]$. I'm not really sure what that notation tells me

 

[fresh_42]

$W_4 = \{f \in \mathbf R[x] : f = \sum_{i=0}^n a_ix^{2i}\}$

I edited the LaTeX code a bit. Exponents of more than one sign have to be put in brackets: x^{2i} and brackets as brackets $\{\,\,\}$ are actually \{ \}.

$a_ix^{2i} + b_ix^{2i} = (a_i+b_i)x^{2i} \in W_4$

$c \cdot a_ix^{2i} = (c \cdot a_i)x^{2i} \in W_4$

What has happened to the sum? We have $f(x)=\sum_{i=0}^n a_ix^{2i}\; , \;g(x)=\sum_{j=0}^m b_jx^{2j}$ and not only one summand. This doesn't change the argument, but it has to be mentioned as e.g. $n$ and $m$ doesn't have to be the same. However, we can always sum to $N=\max\{\,n,m\,\}$ and fill up the missing terms with zero coefficients $a_i$ or $b_j$. So it is basically correct, just sloppy.

I'm not sure about the notation for a zero vector here, would it simply be $0 \cdot a_ix^{2i}$?

No. The zero vector is just $0$. This is the polynomial $0=0+0\cdot x+0 \cdot x^2 + 0\cdot x^3 +\ldots$ with all $a_i=0$. What you have written is the zero scalar: $0\cdot f(x) = 0$. This is formally not the same as $c\cdot 0 = 0$ although the result is of course the same.

EDIT: Additionally I would like to know if there is anything to be said about the notation $\mathbf R[x]$. I'm not really sure what that notation tells me

Firstly I assume your $\mathbf{R}$ should be the real numbers, as you speak of vector spaces. One could also take a ring as scalar domain, which is usually noted as $\mathbf{R}$, but this wouldn't lead to a vector space. Nevertheless, the real numbers should be written as \mathbb{R} which results in $\mathbb{R}$.

$\mathbb{R}[x]$ is the vector space of all polynomials $p(x)= a_0 + a_1x+a_2x^2+a_3x^3+\ldots +a_{n-1}x^{n-1}+a_nx^n$ with real numbers $a_0,a_1,\ldots , a_n$ and some natural number $n \in \mathbb{N}$ which is the degree of the polynomial, i.e. we assume $a_n \neq 0$ for otherwise we would have stopped at a lower $n$. Another possible description is to identify $p(x)$ by the sequence $(a_0,a_1,\ldots , a_n)$, in which case we would write $p=(a_i)_i$ and almost all $a_i=0$, which means, all are zero except finitely many. But this notation is barely used. Normally one writes polynomials with a variable, $x$ in our case.

 

[iJake]

Thank you very much, fresh_42! I asked about that notation for $\mathbb{R}[x]$ because it appears in a subsequent exercise (which I might post for help shortly as it's rather involved, haha) and I wasn't entirely clear on its significance.

I appreciate your corrections regarding my proof that $W_4$ is also a subspace.

That just about concludes this thread: the answer to my exercise, with the accompanying proofs, is that all four sets $W_1, W_2, W_3, W_4$ are subspaces of $\mathbb{R}[x]$