Determination of injectivity and sujectivity of linear transformation $T$ is a linear map $$T : P_2(\Bbb R)\to P_3(\Bbb R)$$ defined by $$T(f(x)) = xf(x) + f′(x).$$  $P_i$ denotes polynomials with degree less than or equal to $i$.
I tried to determine whether $T$ is surjective or injective. I first tried to find the kernel of $T$, and then come up with the following. 
$f'(x)=-xf(x)$ , this equation has solution $f(x)=\frac{1}{2}k e^{-x^2}$ for any $k \in \mathbb{R}$. But $f$ is not a polynomial and then I am stuck.Can anyone help?
 A: You showed that there are no polynomials satisfying $xf+f'=0$, hence $\ker T =\{0\}$. This guarantees that $T$ is injective.
Now, for surjectivity. What is is the dimension of $P_2$  and of $\ker T$? Is there a formula that you can apply to obtain the dimension of $im (T)$? When you find this dimension and compare with the dimension of $P_3$, wha can you conclude?
A: You can do surjectivity also like that. Take any polynomial $p(x)=ax^3+bx^2+cx+d$, we have to find $f(x) = ex^2+fx+g$ such that:
$$ xf(x)+f'(x) = p(x)$$
so 
$$ex^3+fx^2+(g +2e)x+f = ax^3+bx^2+cx+d$$
so $e=a$, $f=b$, $g+2e =c$ and $f=d$. 
So if $b\ne d$ then $p(x)$ is not in the range. 
Also we can see injectivity:
$$T(ex^2+fx+g) = ex^3+fx^2+(g +2e)x+f$$ Thus if $$T(ex^2+fx+g)= T(e'x^2+f'x+g')$$
we get, by comparing the coefficients $e=e'$, $f=f'$ and $g=g'$ and thus it is injective.
A: This is why you learn about matrices and bases and the like. There is a straightforward, simple solution method:


*

*Choose a basis for both vector spaces

*Write down the matrix that expresses the linear transformation

*Phrase the given question in terms of properties of this matrix

*Use your tools for doing computations with matrices

