# Shankar Exercise 1.1.3 (Vector Spaces)

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1. Apr 13, 2017

### Stephen Wright

<Mentor's note: moved from a technical forum, therefore no template.>

I'm long out of college and trying to teach myself QM out of Shankar's.
I'm trying to understand the reasoning here because I think that I am missing something...

1.1.3
1) Do functions that vanish at the endpoints x=0 and L=0 form a vector space?
2) How about periodic functions? obeying f(0)=f(L) ?
3) How about functions that obey f(0)=4 ?

If the functions do not qualify, list what go wrong.

The first 2 questions seem straight forward. X and L go to zero, and outside those boundaries you would get the null vector. If the function is periodic, you are certain to return a vector in the f(x).

#3 is tripping me up. I thought of an example function... lets say f(x)=x+4, so f(0)=4.
How is this not a vector space?

I was told that if g(x) and h(x) were in the set of f(x), g(0)+h(0)=8 which is not 4.

I do understand how this proves anything? What am I misunderstanding?

2. Apr 13, 2017

### Stephen Wright

1. The problem statement, all variables and given/known data
I'm long out of college and trying to teach myself QM out of Shankar's.
I'm trying to understand the reasoning here because I think that I am missing something...

1.1.3
1) Do functions that vanish at the endpoints x=0 and L=0 form a vector space?
2) How about periodic functions? obeying f(0)=f(L) ?
3) How about functions that obey f(0)=4 ?

If the functions do not qualify, list what go wrong.

The first 2 questions seem straight forward. X and L go to zero, and outside those boundaries you would get the null vector. If the function is periodic, you are certain to return a vector in the f(x).

#3 is tripping me up. I thought of an example function... lets say f(x)=x+4, so f(0)=4.
How is this not a vector space?

I was told that if g(x) and h(x) were in the set of f(x), g(0)+h(0)=8 which is not 4.

I do understand how this proves anything? What am I misunderstanding?

2. Relevant equations
The first 2 questions seem straight forward. X and L go to zero, and outside those boundaries you would get the null vector. If the function is periodic, you are certain to return a vector in the f(x).
3. The attempt at a solution

#3 is tripping me up. I thought of an example function... lets say f(x)=x+4, so f(0)=4.
How is this not a vector space?

I was told that if g(x) and h(x) were in the set of f(x), g(0)+h(0)=8 which is not 4.

I do understand how this proves anything? What am I misunderstanding? It said f(0)= 4, not f(x)=4. In my head 8 would be an acceptable answer.

3. Apr 13, 2017

### PeroK

Your working shows that you perhaps lack a good enough grasp of linear algebra and formal mathematics. You seem to be confused between an individual function (such as $f(x)$) and a vector space, which is an infinite set of functions. Normally, you would denote a vector space by $V$ and have something like $V$ is the set of functions that vanish on the endpoints.

Last edited: Apr 13, 2017
4. Apr 13, 2017

### MathematicalPhysicist

Well for 3) it's not closed under addition, take two functions with f(0)=g(0)=4, does h(x)=f(x)+g(x) satisfy this condition?

5. Apr 15, 2017

### nuuskur

3 is a collection of functions all satisfying $f(0)=4$. These functions can't form a vector space over $\mathbb R$ (or $\mathbb C$ for that matter), because $2f(0)=8\neq 4$. There is no homogeneity.

6. Apr 15, 2017

### vela

Staff Emeritus
The way the question is worded might be confusing you. In each case, you have a set of functions, and the question is asking you if this set along with the usual definitions of adding functions and multiplying by a scalar satisfy the requirements of being a vector space. So first, you have to know all of the conditions required of a vector space. There's probably a list in the book, or if not, you can check Wikipedia.

What you wrote here doesn't make sense. For the first question, let $V = \{f\ |\ f(0)=0\text{ and }f(L)=0\}$, and let $f, g \in V$. The fact that $f$ is an element of $V$ means you know that $f(0)=0$ and $f(L)=0$. Likewise for $g$. One of the conditions of a vector space is that of closure under addition. In other words, the function $f+g$ should also be in $V$. So you can check to see if $(f+g)(0)=0$ and if $(f+g)(L)=0$. If this is the case, then the function $(f+g) \in V$. It turns out that's true because $(f+g)(0) = f(0)+g(0) = 0+0 = 0$ and $(f+g)(L) = f(L)+g(L) = 0+0 = 0$. So $V$ is closed under addition.

You have to verify all of the other requirements before concluding $V$ is a vector space.

Given the explanation above, do you now understand why set 3 isn't closed under addition?

7. Apr 15, 2017

### eys_physics

For each of the cases 1), 2) and 3) you will have a set of functions (probably infinite number, but it doesn't effect the problem). To show that they form a vector space $V$, you have to show that for all members of the set, the conditions in Definition 1 are satisfied.