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$$\neq$$P_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.
