Invertible Matrix

**ejgmath** · May 4th 2010, 04:49 PM

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

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

**Opalg** · May 5th 2010, 12:50 AM

Originally Posted by ejgmath

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

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

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

Therefore the spectrum of T consists of the eigenvalues $2^{-j}\ (j\in\mathbb{N})$ and the approximate eigenvalue 0.

**ejgmath** · May 5th 2010, 09:25 AM

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

Thanks again

**Opalg** · May 5th 2010, 11:56 AM

Originally Posted by ejgmath

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

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

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