Let $V$ be the subspace of $P_3 (\Bbb R ) $ spanned by $(1+2x+x^2) , (x+x^2+x^3), (3+5x+2x^2 -x^3)$
Find $n=\dim(V) $ and write an explicit formula for an isomorphism $T: \Bbb R^n \to V $ and prove that it is an isomorphism.
I think that $P_3 (\Bbb R ) $ is spanned by 4 vectors $(1,x,x^2,x^3 )$
Looking at it $ -3(1+2x+x^2) +(x+x^2+x^3)+ (3+5x+2x^2 -x^3)=0$
So really all i need to consider is the span of the vectors $(1+2x+x^2), (3+5x+2x^2 -x^3)$ or $(1+2x+x^2), (1+x -x^3)$
This would seem to imply that $\dim(V) =2 $ as for the map from $\Bbb R^n $ i can't seem to imagine what that space looks like mapping into this one?
I meant id like to say take $T(1,0) \to(1+2x+x^2) $ and $T(0,1) \to(3+5x+2x^2 -x^3) $ where $T:\Bbb R^2 \to V $
Write down the coefficients of the polynomial in a matrix, and row or column reduce it to an upper triangular form, to find the number of free variables which the span comprises of: $$ \begin{pmatrix} 1 & 2 & 1 & 0\\ 0 & 1 & 1 & 1 \\ 3 & 5 & 2 & -1 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & -1 & -2\\ 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix} $$
(I leave you to do this yourself. If you have trouble I can explain.)
Since there are precisely two non-zero rows present in the above matrix, the rank of the subspace is $2$.
Now, it is clear what the isomorphism should be as well. Simply take $T(1,0) = (1,0,1,2) = x + x^2 + 2x^3$, and $T(0,1) = (0,1,1,1) = x+x^2+x^3$, and extend $T$ linearly to the real plane. By the fact that the above are linearly independent, this is automatically an isomorphism.