$U=\{ p\in P_{4}\left( \mathbb{R} \right) : p''\left( 6\right) =0\}$ Find a basis for $U$

Question

Find a basis for $U$, where $$U=\{ p\in P_{4}\left( \mathbb{R} \right) : p''\left( 6\right) =0\}.$$

This question is 2.C 5 of Linear Algebra Done right.

I would like to know how this person got the example basis: $$\{1, x, x^{3}-18x^{2},x^{4}-12x^{3}\}.$$

My attempt:

Take $ f\left( x\right) = ax^{4}+bx^{3}+cx^{2}+dx + e$

$f''= 12ax^{2}+6bx+2c $

$\implies f''\left( 6\right) = 0 \implies c = -6^{3}a-18b$

$\implies f\left( x\right) = a(x^{4}-6^{3}x^{2})+b(x^{3}-18x^{2})+dx + e$

$\implies$ my basis is $\{1, x, x^{3}-18x^{2},x^{4}-6^{3}x^{2}\}$

Where have I gone wrong?

Answer 1

There is nothing wrong with both bases. Observe that your basis element $x^4-6^3x^2$ is a linear combination of the "other" basis: $$x^4-6^3x^2=(x^4-12x^3)+12(x^3-18x^2)\,.$$ And vice versa, the basis element $x^4-12x^3$ of the "other" basis is obtained via a linear combination of your basis: $$x^4-12x^3=(x^4-6^3x^2)-12(x^3-18x^2)\,.$$ As Kavi Rama Murthy said, a vector space does not have a unique basis.

Answer 2

Alternatively, write

$$f(x) = a_0 + a_1(x-6) + a_2(x-6)^2 + a_3(x-6)^3 + a_4(x-6)^4$$

so $f \in U$ if and only if $$0 = f''(6) = \Big(2a_2 + 6a_3(x-6) + 12a_4(x-6)^4\Big)\Bigg|_{x = 6} = 2a_2$$

meaning $a_2 = 0$. Therefore, another basis for $U$ is $$\Big\{1, x-6 , (x-6)^3, (x-6)^4\Big\}$$