Isomorphism of two vector spaces I had a problem : Prove $P = \{ax^6 +bx^3+c\mid a,b,c \in \Bbb R\}$ and $A=\operatorname{Span}\bigl((1, 0, 1, −1, 0),(0, 1, 0, 1, 1),(2, 3, 2, 1, 3),(0, −1, 0, −1, −1),(−1, 1, 0, 0, 0)\bigr)$ are isomorphic. 
Where I started is, I found the dimension of $P$, which is the number of basis vectors $\{1,0,0\}, \{0,1,0\}, \{0,0,1\}$, that is $3$, then I found the dimension of $A$, which is also $3$, with basis of the first and last 2 vectors of the Span. Thus from a theorem that states : two vector spaces are isomorphic if and only if the dimension of them is the same, the vector spaces are isomorphic.
Is that enough for a proof. Or do I need more to prove it.
 A: Given two vector spaces $V$ and $W$ such that $\dim V = \dim W$, you can prove they are isomorphic by considering any two basis, $\mathcal{B}_{V} = \{v_{1},v_{2},\ldots,v_{n}\}$ and $\mathcal{B}_{W} = \{w_{1},w_{2},\ldots,w_{n}\}$, and the linear transformation $T(v_{j}) = w_{j}$, which exists and is unique.
Thus $T:V\rightarrow W$ is injective. Indeed, if $v = \alpha_{1}v_{1} + \alpha_{2}v_{2} + \ldots + \alpha_{n}v_{n}\in V$, one has
\begin{align*}
T(v) & = T(\alpha_{1}v_{1} + \alpha_{2}v_{2} + \ldots + \alpha_{n}v_{n})\\\\
& = \alpha_{1}T(v_{1}) + \alpha_{2}T(v_{2}) + \ldots + \alpha_{n}T(v_{n})\\\\
& = \alpha_{1}w_{1} + \alpha_{2}w_{2} + \ldots + \alpha_{n}w_{n} = 0\\\\
& \Rightarrow \alpha_{1} = \alpha_{2} = \ldots = \alpha_{n} = 0\\\\
& \Rightarrow v = 0 
\end{align*}
According to the rank-nullity theorem and the given assumption, we have that
\begin{align*}
\dim W = \dim V = \dim T^{-1}(\{0\}) + \dim T(V) = \dim T(V)
\end{align*}
Since $T(V)$ is a subspace of $W$ and $\dim T(V) = \dim W$, we conclude that $T$ is surjective and, therefore, an isomorphism, as desired.
