Spanning sets and subspaces

Question

Note: the set of natural numbers is defined to include 0. Both the questions are separate from each other.

My work: for the first question I am not sure exactly what the space is. Here is what I understand; for all even n, the function is zero. There is a limit k for which if n is greater than k, then f(n) is zero. How do I go about finding the spanning set for this space? I am aware that the spanning set is a set of vectors which span V.

For the second question I want to show that V contains the zero vector, is closed under addition and scalar multiplication. I can show that it's closed under scalar multiplication and addition but how do I write down the proof for the existence of the zero vector in V?

For the second part of the second question, I can determine that for all $n\ge0$ f(n+1)=f(1), so basically for all f(n) where n $\ge$1, f(n)=f(1). That leaves us only with f(0). So every element in the space is either f(1) or f(0). Am I correct in my understanding? If so, how do I use this to find a basis for V?

Answer 1

I am working over base field $\mathbb{R}$

First question:

First of all note that if $X$ is any set, then a sequence in $X$ is nothing but a function from $\mathbb{N}$ to $X$ where we denote $a_n:=f(n)$. So our $V$ is subspace of set of sequences.

Let's look at some elements of $V$ to get an intuitive idea.

Consider the sequence $$(0,1,0,1,0,2,0,5,0,6,0,8,0,0,0,0\dots)$$ See in the above example that all even entries(aka $f(2n)=0$) and for $k=12$, $f(n)=0$ for $k\geq 12$

Note: If $f\in F(\mathbb{N})$, image of $f$ can be think of a tuple $(f(0),f(1),f(2),...)$

Think of elements of the form $e_{2i+1}=(0,,0,0,\dots,0,1,0,\dots)$, that is $1$ in $2i+1$ position and zero everywhere else.

For example,

$$e_1=(0,1,0,0,...)$$ $$e_3=(0,0,0,1,0,...)$$

Observation: $e_{2i+1}\in V$ as $f(2n)=0$ and if you take $k=2i+2$, $f(n)=0$ for all $n\geq 2i+2$

Now this collection of $\{e_{2i+1}: i=0,1,2,..\}$ spans $V$.

Second Question

Let $f,g\in V$ and $\lambda \in \mathbb{R}$. Since $f$ belongs to $V$ we have $f(n+2)=f(n+1)$ and similarly $g(n+2)=g(n+1)$

Now consider $f+\lambda g$, $$(f+\lambda g)(n+2)=f(n+2)+\lambda g(n+2)=f(n+1)+\lambda g(n+1)=(f+\lambda g)(n+1)$$, so $V$ is indeed a vector space.

The space is two dimensional. To see why, we impose the condition that $f(n+2)=f(n+1)$ for $n\geq 0$

Put $n=0$, we get $f(1)=f(2)$

Put $n=1$, we get $f(2)=f(3)$ and so on..

In short, if you know the value of $f(0) $ and $f(1)$ you have every information about the elements as $f(1)=f(2)=f(3)=\dots$

Answer 2

On part $1$: all sequences in $\mathbb R$ with even terms $0$ and eventually $0$. Apparently infinite dimensional. For a spanning set, take the set of all sequences with $2n+1$-st term equal to $1$; $0$ otherwise...

On part $2$: It must be $2$ dimensional. So, for any $f\in V$, $f=af_0+bf_1$, where $f_0(0)=1, f_1(0)=0, f_1(1)=1$, for instance. Also, the zero function easily satisfies the requirements to be in $V$.