Invertibility of a Function

**joe909** · February 16th 2010, 05:50 PM

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

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

**tonio** · February 16th 2010, 06:28 PM

Originally Posted by joe909

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

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

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

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

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

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

Tonio

**joe909** · February 16th 2010, 06:46 PM

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

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