*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

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 \(x^3\).