- #1

- 2

- 0

<mod note: moved to homework>

Calculate the spectrum of the linear operator ##T## on ##B(\ell^1)##.

$$T(x_1,x_2,x_3,\dots)=(\sum_{n=2}^\infty x_n, x_1, x_2, x_3, \dots)$$

I think the way to do it is to find the point spectrums of ##T## and its adjoint ##T^*##. But I don't know how to calculate them.

Let ##T(x_n)=\lambda (x_n)##. Then we have ##x_n=\lambda^{-n+1}x_1## for ##n \geq 2## and ##x_1=\frac{\sum_{n=2}^\infty x_n}{\lambda}##. Then

$$x_1=\frac{\sum_{n=2}^\infty x_n}{\lambda}=\frac{\sum_{n=2}^\infty \lambda^{-n+1}x_1}{\lambda}=\sum_{n=2}^\infty \lambda^{-n}x_1$$

Therefore, ##(1-\sum_{n=2}^\infty \lambda^{-n})x_1=0##.

Calculate the spectrum of the linear operator ##T## on ##B(\ell^1)##.

$$T(x_1,x_2,x_3,\dots)=(\sum_{n=2}^\infty x_n, x_1, x_2, x_3, \dots)$$

I think the way to do it is to find the point spectrums of ##T## and its adjoint ##T^*##. But I don't know how to calculate them.

**Some progress:**Let ##T(x_n)=\lambda (x_n)##. Then we have ##x_n=\lambda^{-n+1}x_1## for ##n \geq 2## and ##x_1=\frac{\sum_{n=2}^\infty x_n}{\lambda}##. Then

$$x_1=\frac{\sum_{n=2}^\infty x_n}{\lambda}=\frac{\sum_{n=2}^\infty \lambda^{-n+1}x_1}{\lambda}=\sum_{n=2}^\infty \lambda^{-n}x_1$$

Therefore, ##(1-\sum_{n=2}^\infty \lambda^{-n})x_1=0##.

**Hence, ##\sum_{n=2}^\infty \lambda^{-n}=1##.**Then we can solve for ##\lambda## to get the point spectrum of ##T##
Last edited: