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KennethK
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<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.
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##
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##
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