# Dimensions of linear spaces (linear algebra)

## Homework Statement

Find the dimensions of the following linear spaces.
a. the real linear space C^3

n/a

## The Attempt at a Solution

So I'm not really sure what C means. I know it's the smooth functions and I think raised to the 3rd power means it the functions whose first, second, and third derivatives are equal. But if that's the case, wouldn't e^x be the only function like that? And wouldn't that mean it's a one dimensional space? I'm kind of confused about this haha. thanks

## Answers and Replies

I am almost positive it means all continuous functions that can be differentiated 3 times. Unless your book uses a different convention, I am pretty sure that is it.

so would the dimension be 4 because you could have like [x^4, x^3, x^2, x] would go to [4x^3, 3x^2, 2x, 1] because the smallest set would include those 4 parts?

Are you giving a polynomial restriction like P5? If not, it would be {x^3, x^4,...,x^n}. Any function would be in the set that can be differentiated 3 times.

no, there's no restriction. so saying that it could go up to x^n would mean that there are infinitely many dimensions? except that that can't be true because it can only be a real number answer...

Mark44
Mentor
It wouldn't go just up to xn. If so, that would imply that the space would have a basis with n + 1 functions, which would be a finite-dimensional space.

ok so, if it could be functions x^3, x^4, ..., x^n then n could be an infinite number unless an n is specified which it isn't. I have no idea how to find this then...

Mark44
Mentor
You're assuming that the space consists of power functions such as x^3, x^4, and so on. They are in the space, but it also includes functions such as e^x, sin(x), cos(x), tan(x), sqrt(3 + x^2), and on and on.

ok, that makes sense, but then how can there be a finite dimension for a space that contains infinitely many functions?

Mark44
Mentor
R3 has an infinite number of vectors, but the dimension of this vector space is 3.
P3 has an infinite number of polynomials (of degree <= 3), but the dimension of this (function) space is 4.

The upshot is that it's not how many vectors or functions are in the space that determines its degree. Can you tell me why dim(R3) is 3 or why dim(P3) is 4?

R^3 is any combination of a[1, 0, 0] + b[0, 1, 0] + c[0, 0 ,1] which means that there are three vectors in the basis that make up all of R^3 so the dimension is 3.

P3 would be made up of [1, x, x^2, x^3] which consists of 4 elements in the basis so there are 4 dimensions.
right?

Mark44
Mentor
Right. The dimension of a linear space is equal to the number of vectors in any basis for the space. BTW, you showed the standard bases for each of those spaces; there are others.

Now, how many vectors/functions do you think there might be in a basis for C3? I'm just asking for your gut feeling here.

I suspect what is actually meant is $$\mathbb{C}^3$$, the space of complex triples, considered as a vectorspace over the reals.

I was thinking 4 at first (which I know isn't right) because I thought it would be functions that can be differentiated 3 times to get a smooth function like x^3 + x^2 + x + 1 whose 3rd derivative would be 6, which would still be a smooth function. I know that's wrong, but now that the idea of having, in my mind, infinitely many functions in C^3 I don't see how you can extract a basis...

Mark44
Mentor
I don't think so, but I can't say for sure. It seems to be the space of three-times differentiable real functions.

but aren't there infinitely many three times differentiable functions?

Mark44
Mentor
Yes, but so what? Aren't there infinitely many vectors in R3? And infinitely many polynomials of degree <= 3 in P3?

ok yeah that's true haha. so then I would have to think about the basic smooth functions like sin(x), e^x, any polynomial etc?

Mark44
Mentor
Yeah, as long as C3 means thrice-differentiable functions.

Is it reasonable for me to assume that since you're in a linear algebra class you know about Taylor and Maclaurin series? E.g., the Maclaurin series for ex is 1 + x + x2/2! + x3/3! + ... + xn/n! + ... = $$\sum_{n = 0}^{\infty} \frac{x^n}{n!}$$

Care to speculate how many functions are in a basis to generate this function?

I would say n because it would depend on how many powers you're raising it to but I"m guessing that's wrong.

Mark44
Mentor
If the dimension is n, then the space is finite dimensional. So you couldn't represent, say, ex or cos(x) or tan(x), etc.