# Invertible Matrix

• May 4th 2010, 04:49 PM
ejgmath
Invertible Matrix
I need to show that if $\displaystyle T:\ell^2(\mathbb{N},\mathbb{C})\rightarrow\ell^2(\ mathbb{N},\mathbb{C})$ is an operator defined by $\displaystyle (Tx)(j)=\frac{1}{2^j}x(j)$ and $\displaystyle \lambda\neq0$ is NOT in the set of eigenvalues of $\displaystyle T$ then $\displaystyle T-\lambda I$ is invertible. I also need to find the set of approximate eigenvalues of $\displaystyle T$ and the spectrum of $\displaystyle T$.

So far I have argued that since $\displaystyle \lambda$ is not a eigenvalue of $\displaystyle T$ and is not zero, there is no vector $\displaystyle x$ such that $\displaystyle Tx=\lambda x$ which implies that $\displaystyle Tx-\lambda x\neq0$ and therefore $\displaystyle Tx-\lambda Ix=(T-\lambda I)x\neq0$ and therefore $\displaystyle T-\lambda I\neq0$ hence $\displaystyle T-\lambda I$ is non singular and therefore invertible. Is this correct?
• May 5th 2010, 12:50 AM
Opalg
Quote:

Originally Posted by ejgmath
I need to show that if $\displaystyle T:\ell^2(\mathbb{N},\mathbb{C})\rightarrow\ell^2(\ mathbb{N},\mathbb{C})$ is an operator defined by $\displaystyle (Tx)(j)=\frac{1}{2^j}x(j)$ and $\displaystyle \lambda\neq0$ is NOT in the set of eigenvalues of $\displaystyle T$ then $\displaystyle T-\lambda I$ is invertible. I also need to find the set of approximate eigenvalues of $\displaystyle T$ and the spectrum of $\displaystyle T$.

So far I have argued that since $\displaystyle \lambda$ is not a eigenvalue of $\displaystyle T$ and is not zero, there is no vector $\displaystyle x$ such that $\displaystyle Tx=\lambda x$ which implies that $\displaystyle Tx-\lambda x\neq0$ and therefore $\displaystyle Tx-\lambda Ix=(T-\lambda I)x\neq0$ and therefore $\displaystyle T-\lambda I\neq0$ hence $\displaystyle T-\lambda I$ is non singular and therefore invertible. Is this correct?

The set $\displaystyle S = \{2^{-j}: j\in\mathbb{N}\}\cup{0}$ is closed. So if $\displaystyle \lambda\notin S$ then d, the distance from $\displaystyle \lambda$ to S, is greater than 0. The inverse of $\displaystyle T-\lambda I$ is given by $\displaystyle ((T-\lambda I)^{-1}(x))(j) = (2^{-j}-\lambda)^{-1}x(j)$. This is a bounded operator, with $\displaystyle \|(T-\lambda I)^{-1}\|\leqslant d$. So $\displaystyle \lambda$ is not in the spectrum of T.

Therefore the spectrum of T consists of the eigenvalues $\displaystyle 2^{-j}\ (j\in\mathbb{N})$ and the approximate eigenvalue 0.
• May 5th 2010, 09:25 AM
ejgmath
Thats great, thanks for the help. One question though, is my original arguement for the invertability of $\displaystyle T-\lambda I$ correct?

Thanks again
• May 5th 2010, 11:56 AM
Opalg
Quote:

Originally Posted by ejgmath
Thats great, thanks for the help. One question though, is my original arguement for the invertability of $\displaystyle T-\lambda I$ correct?

No, it's not. There are two things wrong with it. First, on an infinite-dimensional space like $\displaystyle \ell^2(\mathbb{N},\mathbb{C})$, an operator can be injective without being surjective. So the fact that $\displaystyle (T-\lambda I)x\neq0$ for all $\displaystyle x\ne0$ does not imply that $\displaystyle T-\lambda I$ is invertible. Second, I'm assuming that you are dealing with bounded operators here. So it's not enough to show that $\displaystyle T-\lambda I$ has an algebraic inverse, you need to show that the inverse is a bounded operator.