# Example 2.41 in Linear Algebra Done Right 3rd edition

I'm a little confused by the solution of Example 2.41 in Linear Algebra Done Right 3th edition

Example 2.41 Show that $$1,(x-5)^2,(x-5)^3$$ is a basis of the subspace $$U$$ of $$\mathcal P_3(\mathbb R)$$ deﬁned by $$U=\{p\in \mathcal P_3(\mathbb R): p'(5)=0\}$$. $$\mathcal P_n$$ means the vector space of polynomial functions of degree at most $$n$$, $$\mathbb R$$ means the set of real numbers.

The last paragraph of the proof is:

Thus $$\dim U\ge3$$. Because $$U$$ is a subspace of $$\mathcal P_3(\mathbb R)$$, we know that $$\dim U\le\dim\mathcal P_3(\mathbb R)=4$$ (by 2.38). However, $$\dim U$$ cannot equal 4, because otherwise when we extend a basis of $$U$$ to a basis of $$\mathcal P_3(\mathbb R)$$ we would get a list with length greater than 4. Hence $$\dim U=3$$. Thus 2.39 implies that the linearly independent list $$1,(x-5)^2,(x-5)^3$$ is a basis of $$U$$.

I can understand the list $$1,(x-5)^2,(x-5)^3$$ is linearly independent. But I don't understand the sentence starting from "However"...

Theorem 2.38 If $$V$$ is finite-dimensional and $$U$$ is a subspace of $$V$$, then $$\dim U \le \dim V$$.

Theorem 2.39 Suppose $$V$$ is ﬁnite-dimensional. Then every linearly independent list of vectors in $$V$$ with length $$\dim V$$ is a basis of $$V$$.

• I agree that the writing of that part is bad. One could have a clearer argument by observing that $U$ is a proper subset of $P_3({\bf R})$.
– user9464
Commented Jul 10, 2017 at 13:38

If the list was not a basis, then a basis of $U$ would have at least $4$ vectors. Actually, it would have to have exactly four vectors, since the whole space has dimension $4$. But $U\neq\mathcal{P}_3(\mathbb{R})$ and therefore no basis of $U$ can have $4$ vectors. Therefore, the list is a basis.
• @JohnnyJi Since $U\neq\mathcal{P}_3(\mathbb{R})$, if $\dim U=4$ then, if we extended a basis of $U$ to a basis of $\mathcal{P}_3(\mathbb{R})$, that basis of $\mathcal{P}_3(\mathbb{R})$ would have more than $4$ elements. Commented Jul 10, 2017 at 13:33
• I think I got it. Because we can always extend a basis of $U$ to a basis of $\mathcal P_3(\mathbb R)$, and the $\mathrm{dim} \mathcal P_3(\mathbb R)$ is 4, and $U\neq\mathcal{P}_3(\mathbb{R})$, so $\mathrm{dim}U=3$ Commented Jul 10, 2017 at 13:34