Chapter 1

[bestsauce]

Chapter 1:

• [x] What is the criteria that when fulfilled ensures that the sum of two periodic functions is also periodic?
• [x] Can any vector space other than $\{0\}$ be written as a direct sum of at least two subspaces?
• [ ] Solve 1.C Ex 13 and explain why it fails for a field containing only two elements.

[bestsauce]

Just a note to @masamu : The first is related to Ex9 of subchapter 3 (on subspaces) while the other is related to the concept of direct sums (introduced in subchapter 3).

[bestsauce]

I got some progress with question 1: If $a,b$ are rational and they're, respectively, the periods of $f$ and $g$, then $f+g$ is periodic. Proof: Since $a,b$ are rational we can write their ratio as $a/b=c/d$ where $c,d\in\mathbf{Z}$. Thus $ad=bc$, now $$g(x+ad)+f(x+ad)=g(x+bc)+f(x)=g(x)+f(x).$$ Now and I'm not sure if that would be achievable with our current toolset. Do you think that this part is sufficient for our purposes @MaxisJaisi ?

[MaxisJaisi]

@bestsauce I wrote the proof (although I wasn't being very explicit) in the previous exercise thread already; if $a$ and $b$ are the (rational) periods of $f$ and $g$ respectively, then $f+g$ is periodic. Proof: $f+g$ is periodic with respect to $\operatorname{LCM}(a,b)$. This is the easy part.

The hard part is to show that if $a$ and $b$ aren't rational, then $f+g$ cannot be periodic. I thought this was what you had in mind when you included the question in this section.

[MaxisJaisi]

@bestsauce By the way, what made you guess the second conjecture?

[MaxisJaisi]

The second conjecture has an answer on Page 42, Proposition 2.34! (=

[bestsauce]

@MaxisJaisi Oh, right, seems like I predicted it in advance. =) (Note: it only proves it for a finite-dimensional vector space, what about an infinite-dimensional one? My intuition tells me that it wont hold.)

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5. May 2017 04:22 by notifications@github.com:

The second conjecture has an answer on Page 42, Proposition 2.34! (=

— You are receiving this because you were mentioned. Reply to this email directly, > view it on GitHub> , or > mute the thread> .

[bestsauce]

@MaxisJaisi That was more of a question than a conjecture from my part, was just a random thought.

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5. May 2017 04:09 by notifications@github.com:

@bestsauce> By the way, what made you guess the second conjecture?

— You are receiving this because you were mentioned. Reply to this email directly, > view it on GitHub> , or > mute the thread> .

[bestsauce]

@MaxisJaisi Oh, ok. Are you sure we can tackle that problem with our current toolset? It may be tempting to suggest that since they're irrational (or if one of them is) then we can't have two integers $k,m$ such that $f(x+ka)=f(x)$ and $g(x+mb)=g(x)$ and $mb=ka$. But I'm not sure if this reasoning is correct.

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5. May 2017 03:58 by notifications@github.com:

@bestsauce> I wrote the proof (although I wasn't being very explicit) in the previous exercise thread already; if $a$ and $b$ are the (rational) periods of $f$ and $g$ respectively, then $f+g$ is periodic. Proof: $f+g$ is periodic with respect to $\operatorname{LCM}(a,b)$. This is the easy part.

The hard part is to show that if $a$ and $b$ aren't rational, then $f+g$ cannot be periodic. I thought this was what you wrote when you included the question in this section.

— You are receiving this because you were mentioned. Reply to this email directly, > view it on GitHub> , or > mute the thread> .

[bestsauce]

Let us try to give an example for an infinite-dimensional vector space: Let $U$ be the subspace of $\mathcal{P}(\mathbf{R})$ made of all polynomials of the form $c+a_1x^2+a_2x^4+\ldots+a_nx^{2n}$, and let $V$ be the subspace of $\mathcal{P}(\mathbf{R})$ made of all polynomials of the form $a_1x+\ldots+a_nx^n$ where $n$ is odd (+ the $0$ polynomial). Then $U\cap V={0}$ hence $U\oplus V=\mathcal{P}(\mathbf{R})$.

Of course this is only an example, I still think that it's more likely that it may not hold for a general infinite-dimensional vector space.

[bestsauce]

@MaxisJaisi Ex 13 is nearly done, its final element may be the key to understanding why it doesn't hold when we're dealing with a field of only two elements. It uses the same idea and intuition that you used in your earlier proof of Ex 12.

The case where we start with the assumption that one of the subspaces contains the other two is trivial.

Now, suppose that none of the subspaces contains the other two. That means we can find elements $x,y$ and $z$ such that, $$x\in U_1,x\notin U_2,U_3\ y\in U_2,y\notin U_1,U_3\ z\in U_3,z\notin U_1,U_2.$$ $S=U_1\cup U_2\cup U_3$ is a subspace iff it contains $0$ (trivial) and for all scalars $a,b,c$: $ax+by+cz\in U_1\cup U_2\cup U_3$, which implies that $y+z\in U_1\cup U_2\cup U_3$", ie

1. $y+z\in U_1$ or
2. $y+z\in U_2$ or
3. $y+z\in U_3$.

The two last options are impossible since they would imply that $(y+z)+(-y)=z\in U_2$ or $(y+z)+(-z)=y\in U_3$. We only need now to show that the first option $y+z\in U_1$ will also not hold, which is the key since for vector spaces over fields such as $\mathbb F_2$ this wont hold.

[bestsauce]

I'll leave this open as we're moving towards Chapter 3, 1.C Ex 13 is worth the effort.

[MaxisJaisi]

I haven't had time to think about Ex 13, will come back to it once I have the time.

[celiopassos]

@bestsauce Suppose $y + z \in U_1$. You can repeat the same argument to show that $y - z \in U_1$. Since $U_1$ is closed under addition and scalar multiplication, we have $y + z + y - z = 2y \in U_1$, which implies $y$ is in $U_1$, a contradiction.

[bestsauce]

@cpassos That seems to be correct, the "which implies $y+y$ is in $U_1$, a contradiction" part is the important one: In fields over just two elements we can definitely have $y+y=x\in U_1$ which explains why the result doesn't hold over a field with just two elements. (writing $2y$ implicitly assumes that we're working over a field of more than two elements ;)

@MaxisJaisi What do you think?