Spectra of restrictions of bounded operators 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.
 A: 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.
A: 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.
