# Thread: Invertibility of a Function

1. ## Invertibility of a Function

This question Im really lost on, a bit of guidance would be great!

Let $\displaystyle J_n$ be the $\displaystyle R$-vector space with basis $\displaystyle B = \{1,cos(x),sin(x),...,cos(nx),sin(nx)\}$. For a fixed positive real number $\displaystyle a$, define $\displaystyle D \in Hom(J_n,J_n)$ by $\displaystyle D(f) = f'' + a^2f$. For which $\displaystyle a$ is $\displaystyle D$ invertible?

2. Originally Posted by joe909
This question Im really lost on, a bit of guidance would be great!

Let $\displaystyle J_n$ be the $\displaystyle R$-vector space with basis $\displaystyle B = \{1,cos(x),sin(x),...,cos(nx),sin(nx)\}$. For a fixed positive real number $\displaystyle a$, define $\displaystyle D \in Hom(J_n,J_n)$ by $\displaystyle D(f) = f'' + a^2f$. For which $\displaystyle a$ is $\displaystyle D$ invertible?

Check the efect of the transformation on the given basis' elements:

$\displaystyle D(1)=a^2\Longrightarrow$ so far it must be $\displaystyle a\neq 0$

$\displaystyle D(\cos x)=-\cos x +a^2\cos x=\cos x(a^2-1)\Longrightarrow$ it also has to be $\displaystyle a\neq \pm 1$

$\displaystyle D(\cos 2x)=-4\cos x+a^2\cos x =\cos 2x(a^2-4)\Longrightarrow$ it also has to be $\displaystyle a\neq \pm 2$ ....

Tonio

3. Thanks, I get it, if $\displaystyle D(x) = 0$ for $\displaystyle x \neq 0$ then it is no longer one to one. I think I just got caught up in wording of the question.