## Example of a Polynomial Basis

Dec 21 2015

This is a series of notes taken during my review of linear algebra, using Axler's excellent textbook Linear Algera Done Right, which will be heavily referenced.

p. 46, Example 2.41: Show that $$1, (x-5)^2, (x-5)^3$$ is a basis of the subpsace of $$V=\mathcal{P}_3(\mathbb{R})$$ defined by

$$U= \{p \in \mathcal{P}_3(\mathbb{R}): p^\prime (5) =0 \}$$

Solution:

Step 1. Verify the list of given polynomials are in $$U$$, and they're independent (because $$a, b, c \in \mathbb{R}, \forall x, a+b(x-5)^2+c(x-5)^3 = 0 \implies a=b=c=0$$). By the latter we know $$\operatorname{dim} U \ge 3$$

Step 2. Show that dim $$U \leq 3$$. This is because $$U$$ is a proper subspace (subset) of $$V=\mathcal{P}_3(\mathbb{R})$$, i.e. $$\exists q$$, (simply take $$q(x)=x$$), s.t. $$q \in V$$ but $$q \notin U$$. Then by p. 45, 2.39, this independent list of vectors in $$U$$ with length $$\operatorname{dim} U$$ is a basis of $$U$$.

Extension: extend the above basis of $$U$$ (call it $$B$$) to a basis of $$V$$.

Solution:

Use p. 41, 2.32, the algorithm for extending a linearly independent list to a basis. Start with a basis of $$V$$, $$A = 1, x, x^2, x^3$$. Try to add each vector in $$A$$ to $$B$$ only if it's not in the span of the vectors already in $$B$$. Clearly $$1 \in \operatorname{span}(B)$$, no need to add it. Is $$x \in \operatorname{span}(B)$$? i.e. is there $$a,b,c,d$$ not all identically zero, s.t. $$\forall{x}, a\times 1 + b(x-5)^2 + c(x-5)^3 + dx = 0$$ ? No, so we add $$x$$ to $$B$$, $$B$$ now includes $$1, (x-5)^2, (x-5)^3, x$$. We can stop here, as we already have a list of independent vectors of the right dimension. But just to verify that $$x^2 \in \operatorname{span}(B)$$, we indeed can write $$-25\times 1 + 1(x-5)^2 + 0(x-5)^3 + 10x - 1x^2 =0$$, similarly for $$x^3$$.