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I don't understand how to find a basis for a polynomial vector space. Can someone help me with an example?

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4 Answers 4

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A basis for a polynomial vector space $P=\{ p_1,p_2,\ldots,p_n \}$ is a set of vectors (polynomials in this case) that spans the space, and is linearly independent. Take for example, $$S=\{ 1,x,x^2 \}.$$ This spans the set of all polynomials ($P_2$) of the form $$ax^2+bx+c,$$ and one vector in $S$ cannot be written as a multiple of the other two. The vector space $\{ 1,x,x^2,x^2+1 \}$ on the other hand spans the space, but the 4th vector can be written as a multiple of the first and third (not linearly independent), thus it is not a basis.

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  • $\begingroup$ great explanation! $\endgroup$ Feb 25, 2021 at 15:52
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The simplest possible basis is the monomial basis: $\{1,x,x^2,x^3,\ldots,x^n\}$.

Recall the definition of a basis. The key property is that some linear combination of basis vectors can represent any vector in the space.

If, instead of thinking of vectors as tuples such as $[1\ 2\ 4]$, you think of them as polynomials in and of themselves, then you see that you can make any real-valued polynomial of degree less than or equal to $n$ out of the monomial basis listed above.

You don't have to take the monomial basis. For example, you could have $\{1, x^2-4, x^3+x\}$ as a basis. But you cannot make every possible polynomial of degree $\le 3$ out of this basis!

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    $\begingroup$ But can you prove it is really a basis, namely they are independent $\endgroup$
    – Userkkr
    Aug 31, 2017 at 0:15
  • $\begingroup$ @Userkkr: Yes! Just create a matrix representation of Emily's suggestion. You will find the matrix is identity matrix. $\endgroup$
    – Beta
    Feb 3, 2021 at 18:18
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For an example take the vector subspace of $\mathbb{R}[x]$ of all polynomials with real coefficients of degree at most 3. The elements in there will all look like $a_0+a_1x+a_2x^2+a_3x^3$, where the coefficients $a_i \in \mathbb{R}$ can be thought of as coordinates with respect to the set of vectors $(1,x,x^2,x^3)$. You can check that these are linearly independent and span the space, so that you have a 4 dimensional vector space over $\mathbb{R}$. The distinct powers of $x$ act as independent placeholders with their coefficients being coordinates.

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Polynomial vector spaces are easier to work with if you forget that they're polynomials and just focus on the coefficients. Then it becomes mostly notational. Let me show you what I mean.

Lets write the canonical basis vectors as $x_0=(1,0,0),x_1=(0,1,0),x_2=(0,0,1) \in \mathbb{R}^3$. This means I can write any vector $(a,b,c)$ as $ax_0 + bx_1 + cx_2$. But what if we used superscripts instead of subscripts to index them? This would give us $ax^0+bx^1+cx^2$. Now all we have to do is remember that they're polynomials again to get $a+bx+cx^2$. That gives us three ways of writing exactly the same thing: $$(a,b,c)\\ax_0 + bx_1 + cx_2 \\a+bx+cx^2$$

Notice that the polynomial being backwards is immaterial. Left, right, up, down, or stacking them in a matrix they're just placeholders for coefficients. We can just read off the canonical basis in our alternate notation as $1,x,x^2$. If I had used different basis vectors other than the canonical ones I would get different polynomials.

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  • $\begingroup$ Is this concept can be applied to starting point for the construction of Regularity structure? If so how can I add a noise basis into it? $\endgroup$
    – Creator
    Dec 10, 2023 at 4:45

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