Let $\;U=\left\{p\in P_4(\mathbb{R}) : p''(6)=0\right\}.\;$ Find a basis of $U$.

My approach here was to use the fact that $$P_4(\mathbb{R})$$ is isomorphic to $$\mathbb{R}^5$$. So I wrote each polynomial in the list $$1,\;x,\;(x-6)^3,\;(x-6)^4$$ as a column vector in $$\mathbb{R}^5$$ relative to the standard basis of $$P_4(\mathbb{R})$$ and then used Gaussian elimination on the corresponding matrix to ascertain linear independence.

Because the length of any linearly independent list in $$U$$ $$\le$$ the length of any spanning list in $$U$$, we have $$\;\dim(U)\ge4$$. And because $$U$$ is a subspace of $$P_4(\mathbb{R})$$, we have $$\;\dim(U)\le5$$. Thus $$4\le\dim(U)\le5$$. Clearly $$\;\dim(U)\neq5$$. This is because $$U$$ is a subspace of $$P_4(\mathbb{R})$$, which means that if $$\;\dim(U)=5=\dim(P_4(\mathbb{R}))$$, then $$U=P_4(\mathbb{R})$$. However, $$U\neqP_4(\mathbb{R})$$ and hence $$\;\dim(U)=4$$. Thus the list of polynomials $$1,\;x,\;(x-6)^3,\;(x-6)^4$$ is a basis of $$U$$.

I've seen people use different methods here and so I'd like to know whether this is correct.

$$\frac{(x-6)^0}{0!}, \frac{(x-6)^1}{1!}, \frac{(x-6)^3}{3!}, \frac{(x-6)^4}{4!}$$ is a basis you’re looking for.
• What do you mean? Isn't the linearly independent list $1$,$x$,$(x-6)^3$,$(x-6)^4$ a basis of U? My understanding is that every linearly independent list in a vector space with length equal to the dimension of the vector space is a basis of the vector space. Commented May 9, 2021 at 9:08
• @Karam You're right. And there is no contradiction between what you wrote and my answer. I'm just highlighting the fact that considering the condition $p^{\prime\prime}(6)=0$, using the polynomials derived from the Taylor formula is very natural. Commented May 10, 2021 at 13:40