Finding the basis of ker(T) and im(T) Question : Let $P_2$ be the space of polynomials of degree less or equal to 2.
a.) Write down a basis $B$ for $P_2$.
b.) Define the linear transformation on $P_2$ by $T(f)(x) = f^{\prime}(x), $ the derivative of $f(x)$.
c.) Write down a basis for the $ker(T)$
d.) Write down a basis for $Image(T)$.
e.) Is $T$ an isomorphism
My Attempt:
So, $T: V \rightarrow W$. Linear transform on $V = P_2$ linear space for polynomials of degree $\le 2$. Any function $f$ in $P_2$ Would of the form $f(x)= ax^2+bx+c$
a.) Then the basis of $P_2 =  \{x^2,x,1\}$ because linear combination of $x^2,x,1$ With $a,b,c$ scalars $\in \mathbb{R} $ gives any $f$ with degree $\le 2$  in $P_2$
b.) checking if $T$ is a linear transformation-

*

*Condition 1 : $T(f+g) = (f +g )^{\prime}= f^{\prime} +g^{\prime} = T(f) + T(g) $

*condition 2 : $T(kf) = (kf)^{\prime} = k(f)^{\prime}= kT(f)$
Both conditions are true, thus $T(f) = f^{\prime}$ is a linear transformation.
c.) $Ker(T) = \{f  ~in ~P_2 : T(f) = 0 \}$

*

*so, suppose $f(x) = ax^2+bx+c$

*$T(ax^2+bx+c) = \frac{d}{dx}(ax^2+bx+c) = 2ax+b = 0 $

*$ a = 0, b = 0 $ to give a zero value in $Im(T)$

*which makes $f(x) = 0  \cdot x^2 + 0 \cdot x + 1 \cdot c = 1 \cdot c$.

*Basis of the $ ker(T) = \{1\}$

*Thus $ker(T) \neq \{0\} $, $T$ is not an isomorphism

d.) $im(T) = \{T(f) : f ~in ~P_2\}$

*

*$T(ax^2+bx+c) = 2ax+b$, suppose any function $h(x)=2ax+b$ is  in the $im(T)$.
-hence, Basis of $~im(T) = \{x,1\}$
e.) $T$ is not an isomorphism
Guys could you please check my working and point out mistake, if any. My main concerns are the bases of ker(T) and im(T).
 A: For the most part, you're correct.

*

*Part (a) is fairly straightforward; $\mathcal{B} := \{1,x,x^2\}$ is the standard basis we use for this space after all. You should probably verify that it is, indeed, a basis though -- but this is a fairly trivial exercise. (That is, show that $\mathcal{B}$ has the same cardinality as the dimension of $P_2$ over $\Bbb R$, and that $\text{span}(\mathcal{B}) = P_2$.)


*Part (b) is, again, fairly straightforward since the derivative is linear too. Can't imagine you're expected to prove that.


*Part (c), as noted, can be answered by simply noting that only constant functions have zero derivative.


*Part (d) is probably the main one where I feel some elaboration should be necessary. You claim that $\{1,x\}$ is a basis for $\text{im}(T)$. Why is this set the basis? Are all linear combinations of this in $\text{im}(T)$? Can we add elements to this basis in order to get even more? Why or why not? You're not wrong, and answering this can be easily done with an appeal to the rank-nullity theorem, but I feel elaboration is warranted.


*Part (e) is, again, straightforward, since $T$ has a nontrivial kernel.
