# Prove mapping is invertible

1. Oct 27, 2015

### PsychonautQQ

1. The problem statement, all variables and given/known data
Let P be the vector space of one variable polynomials with complex coefficients. if D: P-->P is the derivative mapping, show that the linear mapping D^2+2D+I is invertible.

2. Relevant equations
show that D^2+2D+I is both injective and surjective

3. The attempt at a solution
Showing injectivity: assume D^2+2D+I(u)=D^2+2D+I(v), we want to show that this implies that u=v.

D^2+2D+I(u)=D^2+2D+I(v)
(D+I)^2(u) = (D+I)^2(v)
(D+I)(D+I)(u) = (D+I)(D+I)(v)

Since D and I are both given as linear mappings, we know D+I is linear, so (D+I)(D+I)(u) = (D+I)(u)(D+I)(u) = (D+I)(v)(D+I)(v)

And now i'm kind of lost.... Am I on the right track with showing injectivity?

edit: Maybe I would have better luck with showing injectivity by showing that the Ker(D^2+2D+I) = {0}? Would this involve solving an ODE? Ahh somebody help me i'm so confused ;-(.

Surjective: We want to show that if u is an element of P, then there exists a v in P such that (D^2+2D+I)(v) = u. I have no idea how to show this one. Anyone got any tips to offer?

Last edited: Oct 27, 2015
2. Oct 27, 2015

### Samy_A

That looks like a good idea indeed.

If you don't yet know the general solutions for this kind of ODE, just check what it means for a non-zero polynomial u to satisfy (D^2+2D+I)u=0.

3. Oct 27, 2015

### Staff: Mentor

Another way to say this is:
Assume u'' + 2u' + u = v'' + 2v' + v
Or equivalently, (u'' - v'') + 2(u' - v') + (u - v) = 0

A simple substitution yields w'' + 2w' + w = 0, which is pretty easy to solve.

4. Oct 27, 2015

### geoffrey159

If you show that a map $f$ is bijective, then $f\circ f$ is also bijective, and $(f\circ f)^{-1} = f^{-1} \circ f^{-1}$.
Also, remember that linear maps between two vector spaces of same dimension (finite) are bijective iff they are injective iff they are surjective. Therefore it is enough to show injectivity or surjectivity, no need to bother showing both.

5. Oct 27, 2015

### pasmith

Yes: the kernel of the operator consists of those *polynomials* which solve $y'' + 2y' + y = 0$.

Define $L(f) = (D^2 + 2D + I)(f)$ to save writing.

Let $P_n$ be the finite-dimensional subspace of polynomials of order exactly $n$. Observe that $L(P_n) \subset P_n$ and $L$ is injective. Apply the rank-nullity theorem. Explain why $L(P) = P$ follows.

Last edited: Oct 27, 2015