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From what I understand, you would need 3 vectors to form a basis of three dimensional space, but does this same restriction apply to a polynomial of let's say P5?

In other words, if I'm given W={x^5, x^4, x^3, x^2}, regardless of the degrees of those x's contained in the vectors, I can only have at most 4 pivots. But for P5, I would need 6 pivots, so can't I therefore automatically determine that W cannot be a basis for P5 on this alone?

Sorry for the basic question, but my book does not make this very clear at all.

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  • $\begingroup$ The definition of dimension is the number of elements in the set of basis. Do you know what is the dimension of $P_5$? $\endgroup$ Commented Jul 18, 2016 at 23:02

2 Answers 2

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You are correct.

If a set spans an n-dimensional space, it must have at least n vectors.

You are given a set with only 4 vectors. If it were a basis for the vector space of all polynomials up to degree 5, it must have at least 6 vectors, because that space is 6-dimensional. Since your set has strictly less than 6 vectors, it cannot possibly span the space, and therefore cannot be a basis for the space.

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Here's an important theorem:

Theorem. Let $V$ be a finite dimensional vector space with $\dim V=n$. Then every basis of $V$ consists of $n$ elements of $V$.

In your case, note that $\beta=\{1,t,t^2,t^3,t^4,t^5\}$ is a basis of $P_5$ which means that $\dim P_5=\#\beta=6$.

Now, note that your $W$ has $$ \#W=\#\{t^5,t^4,t^3,t^2\}=4<6=\dim P_5 $$ The theorem above implies that $W$ cannot be a basis of $P_5$.

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