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Explain why the space $\mathbb{P}$ of all polynomials is an infinite-dimensional space.

For any integer $n,$ the set $\left\{1, x, x^{2}, \ldots, x^{n}\right\}$ is linearly independent. Let $H_{n}$ be the span of this set, then $\operatorname{dim} H_{n}=n+1 .$ since $H_{n}$ is a subspace of $V,$ by theorem 11 on pg. $227, \operatorname{dim} H_{n}=n \leq \operatorname{dim} V .$ since we can construct $H_{n}$ for any integer $n$ , the dimension of $V$ is greater than every number. Therefore, the dimension of $V$ cannot be finite and $V$ must be infinite dimensional.

Calculus 3

Chapter 4

Vector Spaces

Section 5

The Dimension of a Vector Space

Vectors

Oregon State University

Harvey Mudd College

University of Nottingham

Idaho State University

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Hello there. Okay, So for this exercise, we need to prove that the set off polynomial. Sorry. Hey, this is our I'm meaning. This set the set off polynomial actually corresponds. Finito, dimensional vector space. So how to prove that? Okay, so it could be quite a interesting, but how is defined the set of polynomial, So basically, we can expand. Okay, So instead of polynomial is basically the span off one x x two extra square here X to the square X to the third you want until X m. Okay. So we can define, um, integral end. So in this case, and corresponds to an element off natural. Ah, on by the the the space off phenomenally defined by the span of these. So just to recall that the spun means Elinor combinations off these elements here on, it's clear that that corresponds to a paranormal off degree and right, So, for example, let's say that n equals 22 We got a X square plus b x plus c. So here we got a linear combination off this elements here until degree to because we fix it and equal to two okay on. In this case, we're responding. The set off point, however, were interested on spend the whole set off polynomial and actually we confined Objection. So let's define the following the following function. So let's say we can define a function from the natural to the set off polynomial B n. That is really easy surgeries people to the P. That actually is really easy to construct this because we're going to take for any end on the natural on. We're going to translate into X to the end. Okay. And we basically were constructing polynomial we find, ah, function that take any any value off here from the natural here. So let's say 123 on we're map into the space off polynomial we're taking one. For example. On we get X to the one is an element off. The point on this again for two we can map two x square for three is the same X three okay on this F form by ejection because for every number for every element here there exists one element on the side of Paul Anonymous. So is one by one, correspondence on for ah, on every every every element. Here is map to the side of polygamy is very subjective. So this function f is onto us objective. So that means that f responsive rejection So great. So we have defined that f is a rejection on Actually, if you notice f form a basis for the set off for a Nahmias Why? Because we have a state before we have already see that this span at the set, the space of pollen animals. So basically, we can say a basis for P will be the basis for the space. P will be the set off functions off end. Such that end is an element off naturals. What I am putting into these ways basis is that fn Well, what is f one If one corresponds to x f zero corresponds toe one and so on. So basically here in this said I'm putting one x x square to the third until extend But here we got something And is any element off the natural aunt, we know that the set off natural is an infinite set is a new infinite comfortable on we have fine Ah, by ejection f from n to the set of polynomial. So that means on also f is rejected, so f is by objective. So that means that and at least, um, this off the set off Paula normals is, uh, has Dimension has the same dimension that end and in this case and actually objectives of have so have the same dimension on and is an infinite set. Soapy should be on infinite dimensional space as well, and that completes the proof.

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