Find linear transformation isomorphism I need to figure out if $\ T: \mathbb R_3[x] \rightarrow \mathbb R^{3}$ in which
$$\ T(x^2+2x) = (1,2,1) , \ T(x+1)=(0,1,1) , \ T(x^2-2)=(1,0,-1)$$ is an isomorphism.
I understand that in order for $T$ to be isomorphism $Ker(T) = \{0\} $
The set $\{x^2+2x;x+1;x^2-2\}$ is linearly dependent and I can't just determine $T$ of a basis of vectors of $\ \mathbb R_3[x] $. I'm not sure how to determine if $T(v) = 0 $ only if $v=0 $
 A: Yes as you noted we don't have sufficient information to define $T$ indeed the vectors $x^2+2x,x+1, x^2-2$ are linearly dependent since
$$(x^2-2)+2(x+1)=x^2+2x$$
Of course we can find infinitely many isomorphism, extending the given basis, such that the condition given are fulfilled but $T$ can't be determined uniquely form the given information.
For example we can set
$$\ T(x^2+2x) = (1,2,1) , \ T(x+1)=(0,1,1) , \ T(x^2)=(0,0,1)$$
and in this case $T$ is an isomorphism which satisfies the given conditions.
A: Calling 
$$
A = \left(\begin{array}{ccc}
1 & 2 & 0\\
0 & 1 & 1\\
1 & 0 & -2
\end{array}\right), X_2 = (x^2, x, 1)^{\top}, B = \left(\begin{array}{ccc}
1 & 2 & 1\\
0 & 1 & 1\\
1 & 0 & -1
\end{array}\right)
$$
we have
$$
T\left(A X\right) = B X
$$
but $A$ has not maximum rank because $\det(A) = 0$ as the second row times $-2$ added to the first row gives the third row.
hence $T$ is not an isomorphism, because $T:\mathbb{R^2}\to \mathbb{R}^3$ because $B$ has maximum rank, and then $T$ is not bijective.
