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


4 Answers 4


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.

  • $\begingroup$ great explanation! $\endgroup$ Feb 25, 2021 at 15:52

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!

  • 3
    $\begingroup$ 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

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.


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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