Suppose $T$ is a bounded operator on a Banach Space $X$ and $Y$ is a non-trivial closed invariant subspace for $T$. It is fairly easy to show that for the point spectrum one has $\sigma_p(T_{|Y})\subseteq\sigma_p(T)$ and this is also true for the approximate point spectrum, i.e. $\sigma_a(T_{|Y})\subseteq\sigma_a(T)$. However I think it is not true in general that $\sigma(T_{|Y})\subseteq\sigma(T)$. We also have

$$ \partial(\sigma(T_{|Y}))\subseteq\sigma_a(T_{|Y})\subseteq\sigma_a(T) $$

Hence $\sigma(T_{|Y})\cap\sigma(T)\ne\emptyset$. Moreover, if $\sigma(T)$ is discrete then $\partial(\sigma(T_{|Y}))$ is also discrete, which implies that $\partial(\sigma(T_{|Y}))=\sigma(T_{|Y})$, so at least in this case the inclusion $\sigma(T_{|Y})\subseteq\sigma(T)$ holds true. So for example holds true for compact, strictly singular and quasinilpotent operators.

Question 1: Is it true, as I suspect, that $\sigma(T_{|Y})\subseteq\sigma(T)$ doesn't hold in general? A counterexample will be appreciated. On $l_2$ will do, as I think that on some Banach spaces this holds for any operators. For example, if $X$ is hereditary indecomposable (HI), the spectrum of any operator is discrete.

Question 2 (imprecise): If the answer to Q1 is 'yes', is there some known result regarding how large the spectrum of the restriction can become?

Thank you.

  • $\begingroup$ Given any two nonempty compact subsets $K_1,K_2$ of the complex plane with $\partial K_2\subseteq K_1$, you should be able to find a bounded normal operator $T$ on $\ell^2$ with $\sigma(T)=K_1$ and $\sigma(T\vert_Y)=K_2$ for some subspace $Y$. $\endgroup$ – George Lowther May 24 '12 at 1:21
  • $\begingroup$ In fact, given any bounded operator $T$ on a Banach space with spectrum $K$ then, for any closed bounded set $K^\prime$ containing $K$ and with boundary contained in $K$, there exists a closed invariant subspace on which $T$ has spectrum $K^\prime$. $\endgroup$ – George Lowther May 24 '12 at 12:31
  • $\begingroup$ @GeorgeLowther I am a bit suspicious about your last statement. What if $T$ doesn't have any invariant subspaces? $\endgroup$ – Theo May 24 '12 at 13:20
  • $\begingroup$ @GeorgeLowther When you have time, could you write a proof of this claim, or give a reference if there is any, I think it is quite interesting. Thanks you! $\endgroup$ – Theo May 24 '12 at 15:57
  • $\begingroup$ I thought that was true, but am no longer so sure. It could be false. The idea is to consider the closed subspace $Y$ generated by $(\lambda-T)^{-1}x$ for $\lambda\in\mathbb{C}\setminus K^\prime$ and some fixed nonzero $x\in X$. This is $T$-invariant with spectrum lying between $\partial K$ and $K^{\prime}$. The idea is then to show that for "almost all" possible choices of $x$, then for $\lambda\in K^\prime\setminus K$, $(\lambda-T)^{-1}x$ is not in $Y$. I'm not convinced that this true now, although it does work in many cases. $\endgroup$ – George Lowther May 26 '12 at 11:07

For example, consider the right shift operator $R$ on $X = \ell^2({\mathbb Z})$, $Y = \{y \in X: y_j = 0 \ \text{for}\ j < 0\}$. Then $Y$ is invariant under $R$, and $\sigma(R)$ is the unit circle while $\sigma(R|_Y)$ is the closed unit disk.


E.g., looking for a case where $0$ is in the spectrum of $T|_Y$ but not in the spectrum of $T$ amounts to looking for an invertible extension of an injective nonsurjective operator. To have any hope, $T|_Y$ should also be bounded below. In the case where $T_Y$ is an isometry on Hilbert space, there is always a unitary extension, and Robert Israel's answer gives the canonical example of this.

If $A$ is a bounded, bounded below, nonsurjective operator on a Hilbert space $H$, then let $T$ be defined on the Hilbert space $H\oplus H$ by the operator matrix $\begin{bmatrix}A&I+A^*\\0&A^*\end{bmatrix}$. We recover $A$ as the restriction of $T$ to $H\oplus \{0\}$, and $A$ is not invertible.

However, $T$ is invertible. If $T\begin{bmatrix}x\\y\end{bmatrix}=0$, then $A^*y=0$ and $Ax+y=0$. Since $y\in\ker(A^*)=(AH)^\perp$, it follows that $Ax=y=0$, hence $x=0$. This shows that $T$ is injective. To see that $T$ is surjective, let $a$ and $b$ be arbitrary elements of $H$. Since $A$ is bounded below, $A^*$ is onto, so there exists $y_0\in H$ with $A^*y_0=b$. Since $A$ has closed range, $H=AH\oplus \ker(A^*)$, so there exists $x\in H$ and $z\in\ker(A^*)$ such that $Ax+z=a-b-y_0$. It follows that $T\begin{bmatrix}x\\y_0+z\end{bmatrix}=\begin{bmatrix}a\\b\end{bmatrix}$.

If $A$ is an isometry, then $\begin{bmatrix}A&I-AA^*\\0&A^*\end{bmatrix}$ is a unitary operator that extends $A$. If $A$ is not unitary, its spectrum is the closed unit disk, and the spectrum of the extension is the unit circle.


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