I'm not sure how to do this, especially since the answer contradicts with mine.
Question:
Let $V$ be the vector space of all real functionson $[0,\pi]$ which are arbitrarily often differentiableon $(0,\pi)$. Let $T:V \rightarrow V$ be defined by $T(y) = y''$. Find all eigenvectors of $T$ which also satisfy the condition $y(0) = y(\pi) = 0$, and state the corresponding eigenvalues.
Answer says the eigenvalues are $\lambda = -n^2$...
This is my attempt.
Let $\lambda$ be an eigenvalue of the transformation $T$ so that $y'' = \lambda y$.
Now this is a second order homogenous differential equation. Let $y = e^{rx}$ for $r \in \mathbb{R}$.
then we have $r^2 = \lambda$ (this is exactly the opposite sign of my answer?!).
Then the solution to the second order is just
$y = c_{1}e^{\sqrt{\lambda} x} + c_{2}e^{-\sqrt{\lambda}x}$.
Hence, with our conditions we have
$c_{1} + c_{2} = 0$ and
$c_{1}e^{\sqrt{\lambda} \pi} + c_{2}e^{-\sqrt{\lambda}\pi}=0$.
I stop here because the answer says it's in the form $y = B\sin n x$ for positive intger $n$ and corresponding eigenvalues $-n^2$?
Here, $\lambda$ can be either positive, zero or negative. In your solution, you have considered only $\lambda>0$ case. Now let $\lambda=0$ then $$y''=0\cdot y=0$$ i.e $$y=c_1 x+c_2.$$ Using, $y(0)=0=y(\pi)$ we get, $y=0$
Hence $\lambda=0$ cannot be an eigen value. Now let $\lambda=-d<0$ where $d>0$ then $$y''=\lambda y$$ has solution of the form $$y = c_{1}\cos\sqrt{d} x + c_{2}\sin\sqrt{d}x.$$ Now $y(0)=0\implies c_1=0$. Again $y(\pi)=0\implies c_2 \sin\sqrt{d} \pi=0.$ To get $c_2\neq0$ we must have $\sqrt{d}=n.$ i.e $d=n^2$ and $\lambda=-n^2$