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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 $V$.

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    $\begingroup$ 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 $P_3({\bf R})$. $\endgroup$ – Jack Jul 10 '17 at 13:38
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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.

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    $\begingroup$ why "we would get a list with length greater than 4"? $\endgroup$ – Johnny Ji Jul 10 '17 at 13:28
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    $\begingroup$ @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. $\endgroup$ – José Carlos Santos Jul 10 '17 at 13:33
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    $\begingroup$ 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 $\mathrm{dim}U=3$ $\endgroup$ – Johnny Ji Jul 10 '17 at 13:34

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