# Show that the map is an injective linear transformation and determine a basis of the image

Hi im having trouble with this homework question:

"For any polynomial function $f : R → R$ of degree at most $n$, the function $I(f) : R → R$,$s → \int_0^s f(u) du$, is a polynomial function of degree at most $n + 1$. Show that the map $I : P_n → P_{n+1}, f→ I(f)$, is an injective linear transformation and determine a basis of the image of $I$."

I've managed to show that it is a linear transformation by checking the requirements for a linear transformation but I don't know how to show that the transformation would be injective. Usually what I would do to show a transformation is injective is I would be given a matrix and then show that the null space just consists of the zero vector by using row operations, then to find a basis for the image I would perform column operations on the same matrix.

Any help with this would be very much appreciated.

We can try this representation $$p \in P_n \\ p(x) = \sum_{k=0}^n a_k x^k \leftrightarrow (a_0, \dotsc, a_n)^T \in \mathbb{R}^{n+1}$$ Then \begin{align} I(p) &= s\to \int\limits_0^s \sum_{k=0}^n a_k x^k dx \\ &= s\to \sum_{k=0}^n a_k \left[\frac{1}{k+1} x^{k+1}\right]_{x=0}^{x=s} \\ &= s\to \sum_{k=0}^n \frac{a_k}{k+1} s^{k+1} \in P_{n+1} \\ &\leftrightarrow \left( 0, a_0, \frac{a_1}{2}, \dotsc, \frac{a_n}{n+1} \right)^T \in \mathbb{R}^{n+2} \end{align} So we can write $I$ as matrix $A$ acting on the coefficient vectors $a$: $$A = \begin{pmatrix} 0 & 0 & 0 & \dotsb & 0 \\ 1 & 0 & 0 & & 0 & \\ 0 & 1/2 & 0 & & 0 & \\ 0 & 0 & 1/3 & & 0 \\ \vdots & & & \ddots & \vdots \\ 0 & 0 & 0 & \dotsb & 1/(n+1) \end{pmatrix} \in \mathbb{R}^{(n+2)\times (n+1)}$$
A more direct approach that doesn’t require computing a matrix for $f$: The kernel of $I$ consists of polynomials $f$ such that $\int_0^sf(u)\,du=0$. The value of this integral is $F(s)-F(0)$ for some polynomial function $F$, and for this to vanish for all $s\in\mathbb R$, $F$ must be constant. The derivative of a constant polynomial is $0$, therefore the kernel of $I$ is trivial.