# Operator Eigenvalues

• Apr 30th 2010, 09:19 AM
ejgmath
Operator Eigenvalues
I need find all the eigenvalues of the operator $T:\ell^2(\mathbb{N},\mathbb{C})\rightarrow \ell^2(\mathbb{N},\mathbb{C})$ defined by $(Tx)(j)=\frac{1}{2^j}x(j)$ for all $j$ and show that $\lambda =0$ is an approximate eigenvalue.

So far I have deduced that since $(Tx)(j)=\frac{1}{2^j}x(j)$ and $\lambda$ must satisfy $Tx=\lambda x$ for $x\in\ell^2$ with $x\neq 0$ that $\lambda_j=\frac{1}{2^j}$, is this correct?

Then for the approximate value $\lambda=0$ for a sequence $x_n(j)$ such that $\|x_n\|=1$ this must satisfy $(Tx_n)(j)-\lambda x_n(j)\rightarrow0$ as $n\rightarrow\infty$ but I have no idea what $x_n(j)$ could be?
Any help would be great
• Apr 30th 2010, 11:20 AM
Opalg
Quote:

Originally Posted by ejgmath
I need find all the eigenvalues of the operator $T:\ell^2(\mathbb{N},\mathbb{C})\rightarrow \ell^2(\mathbb{N},\mathbb{C})$ defined by $(Tx)(j)=\frac{1}{2^j}x(j)$ for all $j$ and show that $\lambda =0$ is an approximate eigenvalue.

So far I have deduced that since $(Tx)(j)=\frac{1}{2^j}x(j)$ and $\lambda$ must satisfy $Tx=\lambda x$ for $x\in\ell^2$ with $x\neq 0$ that $\lambda_j=\frac{1}{2^j}$, is this correct?

That is correct as far as the eigenvalue is concerned, but you need to be a bit more precise about what the corresponding eigenvector is. let $e_n$ be the vector with a 1 in the n'th coordinate and zeros everywhere else. Then $Te_n = 2^{-n}e_n$. So $e_n$ is the eigenvector corresponding to the eigenvalue $2^{-n}$.

Quote:

Originally Posted by ejgmath
Then for the approximate value $\lambda=0$ for a sequence $x_n(j)$ such that $\|x_n\|=1$ this must satisfy $(Tx_n)(j)-\lambda x_n(j)\rightarrow0$ as $n\rightarrow\infty$ but I have no idea what $x_n(j)$ could be?

Take $x_n$ to be the same vector $e_n$ described above.