2
votes
1answer
41 views

Show a linear transform is self adjoint - check my answer

We are given $T:V \to V$ a normal linear transform (meaning $TT^*=T^*T$) We are also given $T^2=T$. Show that $T$ is self adjoint (meaning $T^*=T$). What I did I think I may have done something ...
2
votes
1answer
163 views

prove that if $\lambda$ is an eigenvalue of T then $\bar\lambda$ is eigenvalue of $T^*$

I have to prove that if $\lambda$ is an eigenvalue of T then $\bar\lambda$ is eigenvalue of $T^*$ (adjoint) I know that $<Tv,u> = <\lambda v,u> = ...
0
votes
0answers
40 views

Regular Sturm-Liouville Boundary Value Problem

Let $L[y]:=y''''$. Let the domain of $L$ be the set of functions that have four continuous derivatives on $[0,π]$ and satisfy $y(0)=y'(0)=0$ and $y(π)=y'(π)=0$ a) Show that $L$ is self adjoint b) ...
1
vote
0answers
27 views

Eigenvector of adjoint

I am solving a problem of the form $$ A\frac{d^2\Phi}{dy^2} + B\frac{d\Phi}{dy} + C\Phi = 0 $$ where phi is the eigenvector solution and A, B, and C are n x n square matrices. If I know matrices A, ...
-3
votes
2answers
67 views

4 Small questions about matrices, eigenvalue, consistence, lineare independant, eigenvector

i've some questions A) "A" is a real 5 x 3 matrix y € R³ and z € R^5 and for this Ay = z Why can you consider that Ax = 4z, is consistent? B) ...
2
votes
2answers
96 views

Onto and one-to-one

Let $T$ be a linear operator on a finite dimensional inner product space $V$. If $T$ has an eigenvector, then so does $T^*$. Proof. Suppose that $v$ is an eigenvector of $T$ with corresponding ...
1
vote
0answers
76 views

Can eigenvectors and eigenvalues be seen as limits is the same sense as fixed points?

Can eigenvectors and eigenvalues be seen as limits in the same sense as fixed points? What I mean by the same sense as fixed points is that if $G$ is group we can consider limits to ${\bf Set}^G$, or ...
3
votes
2answers
205 views

eigenfunctions of the adjoint of an operator

If the eigenfunctions of a linear operator are known, is there a way to calculate the eigenfunctions of the corresponding adjoint operator based on the known eigenfunctions? In other words, what's the ...
2
votes
1answer
331 views

Eigenvalues of compact operators and his adjoint.

Let $T: H \to H$ be a compact operator with $H$ a Hilbert space. Let then $\lambda \neq 0$ be an eigenvalue of $T$ with eigenfunction $v$. Is then $\lambda$ an eigenvalue for the adjoint $T^*$ ...