Finding a basis and dimension of a subspace

Question

I wish to find the basis and dimension of the subspace $U = \{p \in P_4(\mathbb{R}) \mid p(1) = p'(1) = 0\} \subseteq V = P_4(\mathbb{R})$

In my solution I find conditions on the coefficients of the fourth degree polynomials: $a_4 = -a_3 - a_2 - a_1 - a_0, \; a_3 = -2a_2 -3a_1 - 4a_0$. So any polynomials in $U$ must have the tuple of coefficients: $(a_1, a_2 , -2a_2 - 3a_1 - 4a_0, -a_3 - a_2 -a_1 - a_0)$.

But I am confused because the set of vectors I form from the above tuple has no monomials in it. So it can't span $U$ since $U$ is a subspace of polynomials.

Where have I gone wrong in my logic in my solution?

The below is my solution

Answer 1

One of the minor careless mistake is dropping $a_0$ term in the line which begins with "We have the tuple of coefficients". Also, try to express elements of $U$ in terms of polynomial.

$U$ should not contain any monomial. Suppose $x^i \in U$ where $i>0$, at $x=1$, it is evaluated to be $1$. Also, the only constant that is in $U$ would be $0$.

Answer 2

I imagine finding specific conditions on the polynomials in $\mathcal{P}_4(\mathbf{R})$ that satisfy $p(1) = p'(1) = 0$ would be quite cumbersome . My humble suggestion to you is that you observe that a possible ordered basis for $U$ is as follows $$\mathcal{K} = \{(x-1)^2,(x-1)^3,(x-1)^4\}$$

Answer 3

$2$ restrictions on a $5$-dimensional space should indeed lead to a $3$-dimensional subspace.

$\{(x-1)^2,(x-1)^3,(x-1)^4\}$ is a set of linearly independent elements of $P_4(\mathbb R)$ each of which satisfies the requirements...

It is easy to see that $\{1,x\}$ spans a $2$-dimensional subspace whose intersection with $U$ is $\{0\}$. This shows the dimension is indeed $3$...