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I have the following question and am completely lost on how to start:

Let $T: P_{2} \to P_{3}$ be the linear transformation

$$[T(p)](x) = p^{\prime}(x) + xp(x)$$

Find $\ker(T)$ and find a basis for the range of T.

Please try to explain in simple terms.

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The standard boilerplate to determine the kernel of a linear mapping $\varphi:E\longrightarrow F$ is:

Let $u\in E$. Then: $$u\in\ker\varphi\iff\varphi(u)=0_F\iff\cdots$$ and then try (if possible) to go step by step to an explicit description of the equation $\varphi(u)=0_F$ (i.e., solve it—whatever that means).

In your specific case:

Let $p\in P_2$, say $p(x)=ax^2+bx+c$ for some $a,b,c\in\mathbb{K}$. (notice that I explicitly wrote what it means to be in $P_2$, this is something students usually miss). Then: \begin{align*} p\in\ker T &\iff T(p)=0\\ &\iff p'(x)+xp(x)=0\\ &\iff 2ax+b+x(ax^2+bx+c)=0\\ &\iff ax^3+bx^2+(2a+c)x+b=0\\ &\iff\begin{cases}a=0\\b=0\\2a+c=0\\b=0\end{cases}\\ &\iff a=b=c=0\\ &\iff p=0. \end{align*} Hence $\ker T=\{0\}$.


Now for the image: it is usually easy to determine a generating family of the range of a linear mapping:

Let $T:E\longrightarrow F$ be a linear mapping and let $\mathscr{B}=(u_1,\ldots,u_n)$ be a basis of $E$. Then the family $\mathscr{C}=\bigl(f(u_1),\ldots,f(u_n)\bigr)$ is a generating family of $\operatorname{Im}(T)$. Moreover, if $T$ is injective, then the family $\mathscr{C}$ is a basis of $\operatorname{Im}(T)$.

In your particular case, we know a basis of $P_2$, namely the standard basis $\mathscr{B}=(1,x,x^2)$ and we know that $T$ is injective (as its kernel is the nil space). Hence a basis of $\operatorname{Im}(T)$ is given by $\mathscr{C}=\bigl(T(1),T(x),T(x^2)\bigr)$. Now: $$T(1)=x,\quad T(x)=1+x^2,\quad T(x^2)=2x+x^3,$$ hence $\mathscr{C}=\bigl(x,1+x^2,2x+x^3\bigr)$ is a basis of $\operatorname{Im}(T)$. We can find a slightly simpler basis of $\operatorname{Im}(T)$ by observing that $(2x+x^3)-2x=x^3$, hence the following family: $$(x,1+x^2,x^3)$$ is a basis of $\operatorname{Im}(T)$

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