Check if a transformation $L: u \rightarrow (u(2),u'(2))$ is linear.Find all solutions for this equation: $L(u)=(3,-2)$ Check if a transformation defined like this $L: u \rightarrow (u(2),u'(2))$ is an linear operator where $L:\mathbb{R[x]} \rightarrow \mathbb{R^2}$. Note that $\mathbb{R[x]}$ is a vector space of real polynomials.
After that find all solutions for the given equation: $L(u)=(3,-2)$.
The first part of this example is easy but how can I find all the solution for this equation? When I try to solve this equation I get the following system of two equations:
$a_0+2a_1+4a_2 + ... +a_n2^n+... = 3$
$0 +a_1+4a_2 + ... +na_n2^{n-1}+... = -2$
 A: The basic theory of linear equations is at play here: all you need to do is to find the kernel of $L$, and a single solution, and the rest of the solutions lie in the same coset as the solution you found.
So let's work on the kernel first. If $L(p(x))=(0,0)$, that means that $2$ is a root of both $u$ and $u'$. You've probably seen the proof (or maybe not) that when $u$ and $u'$ share a root, that means it had multiplicity greater than $1$ in $u$'s factorization. That is, $(x-2)^2|u(x)$. Clearly then, any polynomial divisible by $(x-2)^2$ is in the kernel of $L$. Let's call the set of these polynomials $V$. They are a subspace of $\mathbb R[x]$, and moreover it is a subspace of index $2$.
Now, it is easy to see that $L$ is onto $\mathbb R\times \mathbb R$. Obviously $L(b(x-2)+a)=(a,b)$ for any $a,b\in\mathbb R$. This means the kernel of $L$ has index $2$ as well, and that $V$ is precisely the kernel of $L$. And here we also scoop up our "one solution" to our original problem: $L(-2(x-2)+3)=(3,-2)$.
So the complete set of solutions is everything in the coset
$$
-2(x-2)+3+(x-2)^2\mathbb R[x]
$$
