# Calculate the spectrum of a linear operator

<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 $T$

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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 $T$
If $|1/\lambda|<1$ then this sum would converge to $\frac {2}{1- \lambda}$

Last edited: