# Eigenvalues of Linear operator

#### charikaar

Hello,
Find all the eigenvalues and determine with a reason whether L is self-adjoint.
$$\displaystyle d/dt $$---------->$$\displaystyle L_2[0,1]$$ where the domain D is given by D{f belongs to $$\displaystyle L_2[0,1]$$: f is continuously differentiable and f(0)=0)}
I get $$\displaystyle f(x)=e^{\lambda x}$$ after solving first order linear differential equation. How do i find $$\displaystyle \lambda$$?
thanks

#### HallsofIvy

MHF Helper
Hello,
Find all the eigenvalues and determine with a reason whether L is self-adjoint.
$$\displaystyle d/dt $$---------->$$\displaystyle L_2[0,1]$$ where the domain D is given by D{f belongs to $$\displaystyle L_2[0,1]$$: f is continuously differentiable and f(0)=0)}
I get $$\displaystyle f(x)=e^{\lambda x}$$ after solving first order linear differential equation. How do i find $$\displaystyle \lambda$$?
thanks
our operator is just differentiation so, yes, $$\displaystyle \frac{d e^{\lambda x}}{dx}= \lambda e^{\lambda x}$$. More generally, if $$\displaystyle \frac{dy}{dx}= \lambda y$$, then $$\displaystyle y= Ce^{\lambda 0}$$$$\displaystyle e^{\lambda x}$$ is an "eigenvector" and $$\displaystyle \lambda$$ is an eigenvalue. As for finding $$\displaystyle \lambda$$, you don't! That equation is true for all $$\displaystyle \lambda$$- any real number is an eigenvalue.

As for whether or not it is "self adjoint", the definition of self-adjoint is that <Af, g>= <f, Ag> for any vectors f, g. Here, I presume you are using the "standard" inner product, $$\displaystyle <f, g>= \int_0^1 f(x)g(x)dx$$ so that we want $$\displaystyle \int_0^x \frac{df}{dx} g dt$$. Use "integration by parts" with u= g and dv= (df/dx)dx. Then du= (dg/dx)dx and v= f.

$$\displaystyle \int u dv= uv- \int vdu= f(x)g(x)- \int f (dg/dx) dx= <f, dg/dx>$$ provided f(1)g(1)- f(0)g(0)= 0.

• charikaar and Jester

#### Jester

MHF Helper
I'm curious on one thing. Don't we require that $$\displaystyle f(0) = 0$$?

#### charikaar

$$\displaystyle L_2[0,1]$$ is the Hilbert space of complex square integrable functions on [0,1] and D={f belongs to $$\displaystyle L_{2}[0,1]$$ : f is continuously differentiable and f(0)=1)}. so $$\displaystyle \lambda$$ includes complex numbers also? L is not self-adjoint because there is minus in front of the integral right?