Show that $1, (x-5)^2, (x-5)^3$ is a basis of the subspace $U$ of $\mathcal P_3(\Bbb R)$ In Axler's LADR book example 2.41:
Show that $1, (x-5)^2, (x-5)^3$ is a basis of the subspace $U$ of $\mathcal P_3(\Bbb R)$ defined by: $$U = \{p \in P_3(\Bbb R):p'(5) = 0\}$$
Book shows that $1, (x-5)^2, (x-5)^3$ is linear independent, hence $\dim U \ge 3$.
It also must be $\dim U \le 4$.
But I dont understand this part: 

However, $\dim U$ cannot equal $4$, because otherwise when we extend a basis of $U$ to a basis of $P_3(\Bbb R)$ we would get a list with length greater than 4.

Why we will get basis with length greater than $4$?
 A: I think what Axler is saying is that $\dim U$ cannot be equal to 4 because $U$ is not all of $\mathcal P_3(\mathbb R)$. It is easy to see that they are not equal by just finding a polynomial whose derivative at $x=5$ is not 0.  (For example, try $p(x)=x$.)  If you take this polynomial and add it to a basis for $U$, you should still have a linear independent set because it is not a linear combination of the basis elements of $U$.  But if there are already 4 elements in a basis for $U$ then we would end up with 5 independent vectors in $\mathcal P_3(\mathbb R)$, which is impossible as the dimension of this space is 4.
So I guess Axler is saying, whatever the dimension of $U$ is, you can add vectors to a basis for $U$ until you get a basis for $\mathcal P_3(\mathbb R)$, and you would have to add at least one vector.
A: $\dim P_3(\Bbb R) = 4$ because $\{1, x, x^2, x^3\}$ is a basis of $P_3(\Bbb R)$.
Clearly $p'(5) = 0$ does not hold for all $p \in P_3(\Bbb R)$.
So your $U$ is a subspace of $P_3(\Bbb R)$ which does not contain all the vectors of $P_3(\Bbb R)$.
Therefore, its dimension must be lower.
A: Abstract-Algebra Argument
The linear map $\phi:\mathcal{P}_3(\mathbb{R})\to\mathbb{R}$ sending $f\mapsto f'(5)$ is surjective with kernel $U$.  Therefore, the Rank-Nullity Theorem gives $$\dim_\mathbb{R}(U)+1=\dim_\mathbb{R}(U)+\dim_\mathbb{R}(\mathbb{R})=\dim_\mathbb{R}\big(\mathcal{P}_3(\mathbb{R})\big)=4\,.$$
A: Since $U$ is a subspace of $\mathbb{R}[X]_{\leqslant 3}$, if $\dim(U)=4=\dim(\mathbb{R}[X]_{\leqslant 3})$, then $U=\mathbb{R}[X]_{\leqslant 3}$, which is not the case. Indeed, $X$ is in $\mathbb{R}[X]_{\leqslant 3}$ but not in $U$.
A: 
Let $p\in\mathcal{P}_3(\mathbb{R})$ be a polynomial such that $p(x)=x$.
Then, $p^{'}(x) = 1$.
$p^{'}(5) = 1\neq 0$.
So, $p\notin U$.
So, $U\nsubseteq\mathcal{P}_3(\mathbb{R})$.
Assume that $\dim U=4$.
Then, $U$ has a basis $u_1,u_2,u_3,u_4$.
$p\notin U=\text{Span}(u_1,u_2,u_3,u_4)$.
So, by Linear Dependence Lemma (2.21 on p.34), $u_1,u_2,u_3,u_4,p$ is linearly independent.
$\mathcal{P}_3(\mathbb{R})=\text{Span}(1,x,x^2,x^3)$.
This contradicts 2.23 on p.35.
So, $\dim U\neq 4$.

But I don't like the author's proof.
My proof is here:

Let $p\in\mathcal{P}_3(\mathbb{R})$ be a polynomial such that $p(x)=x$.
Then, $p^{'}(x) = 1$.
$p^{'}(5) = 1\neq 0$.
So, $p\notin U$.
So, $U\nsubseteq\mathcal{P}_3(\mathbb{R})$.
Assume that $\dim U=4$.
Then, $U$ has a basis $u_1,u_2,u_3,u_4$.
$u_1,u_2,u_3,u_4$ is a linearly independent list of length $4=\dim \mathcal{P}_3(\mathbb{R})$.
So, by 2.39 on p.45, $u_1,u_2,u_3,u_4$ is also a basis of $\mathcal{P}_3(\mathbb{R})$.
So, $U=\mathcal{P}_3(\mathbb{R})$.
This is a contradiction.

