Linear algebra- Basis of Range(T) 
Let $T: P_2\to P_3$ be a linear transformation defined by $$T(at^{2} +bt+c) = (a-b+c)t^{3} + (-a+3b-2c)t^{2} + (-a-b)t + (2b-c).$$
Find a basis of $\operatorname{range}(T)$.

Would the basis of $\operatorname{range}(T)$ be {$t^{3}$, $t^{2}$, $t$, $1$}? Since they span $T$ ?
 A: Choose a basis $1,t,t^2$ for $P_2$, and $1,t,t^2,t^3$ for $P_3$. The the matrix for $T$ is given by:
$$ \tau = \left[\begin{array}{rrr}
-1 & 2 & 0 \\
0 & -1 & -1 \\
-2 & 3 & -1 \\
1 & -1 & 1 
\end{array}\right].$$
It is fairly easy to see that $\tau (2,1,-1)^T= 0$ (ie, the second column is equal to the third column minus twice the first column), and that the first and last columns are linearly independent. Thus these two columns span the range space of $T$. Thus  $T(1), T(t^2)$ form a basis for the range.
A: You cannot span a linear transformation, so saying something "spans $T$" is, ipso facto, wrong.
The set you give spans the codomain of $T$; but the codomain is not the same thing as the range. The codomain is a set that contains the range; the range is just the elements that are actually images under $T$.
By dimensional considerations, the image of $P_2$ has dimension at most 3, and so cannot have a basis with four elements. So your answer is definitely wrong.
There are at least two options here:


*

*Hard way. Try to figure out exactly what vectors of $P_3$ can be images of $T$ by looking at the form of $T$; then find a basis for that subspace.

*Easy way. The image of a basis for $P_2$ will span the range of $T$. And every spanning set contains a basis. So look at the image under $T$ of your favorite basis for $P_2$, and extract a basis for its span from that set.
A: HINT: Transform every single element of a basis of the set of polynomials of degree at least two.  
