Find the dimension of the kernel

Question

Consider the map $T$ from the vector space of polynomials of degree at most $5$ over the reals to $\mathbb{R}\times \mathbb{R}$, given by sending a polynomial $P$ to the pair $(P(3),P′(3))$, where $P′$ is the derivative of $P$. How to find the dimension of the kernel.

Answer 1

If we call the domain vector space $V$, this has dimension $6$ (it has a basis $\{1, x, \ldots, x^5\}$).

If you show $T$ is surjective then the theorem that says $\dim(V) = \dim(\ker(T)) + \dim(\operatorname{Im}(T))$ implies that the dimension of the kernel must be $4$.

Or directly:

Any polynomial $p$ in the kernel must be of the form $p(x) = a(x)(x-3)^2$:

Write $p(x) = a(x)(x-3)^2 + b(x)$ by Euclidean division, where $a(x)$ and $b(x)$ are polynomials with $\deg(b(x)) < \deg((x-3)^2) = 2$. We must have $p(3) = 0$ so $b(3) =0$ and $p'(3) = 0$ forces $b'(3) = 0$. ($p'(x) = a'(x)(x-3)^2 + 2a(x)(x-3) + b'(x)$ and the first two terms vanish at $3$). And as $b(x)$ is linear, these two facts force $b(x) = 0$)

As the degree of $a$ can be at most $3$, a basis for the kernel of $T$ is $\{(x-3)^2, x(x-3)^2, x^2(x-3)^2, x^3(x-3)^2\}$.