# Spectral family property: $\lambda \geq M \Rightarrow E_{\lambda} = I$

I'm following Kreyszig's "Introductory Functional Analysis with Applications" and I'm trying to follow his proof about some properties of a spectral family associated with a bounded self-adjoint linear operator.

Namely, I'm trying to follow the proof of the following.

Theorem Suppose that $T: H \to H$ is a bounded self-adjoint linear operator on a complex Hilbert space $H$ and let $E_{\lambda}, \lambda \in \mathbb R$ be the projection of $H$ onto the null space $Y_\lambda := \mathscr N \left(T_\lambda ^+\right)$ of the positive part of $T_\lambda = T - \lambda I$. Then $(E_\lambda)$ is a spectral family on the interval $[m,M] := [\inf_{ \left\Vert x \right\Vert = 1} \left< Tx, x\right>, \sup_{\left\Vert x \right\Vert = 1} \left< Tx, x \right>$ and $$\lambda \geq M \Rightarrow E_\lambda = I.$$

Now the proof of this last fact begins as follows.

"Suppose that $\lambda > M$ but $E_\lambda \neq I$, so that $I - E_\lambda \neq 0$."

All fine by me so far. The next implication is that which I don't understand:

"Then $(I - E_\lambda )x = x$ for some $x$ of norm $\left\Vert x \right\Vert = 1$."

How do we justify this? Surely all we know is that for some $p \in H$, $$(I - E_\lambda) p = q \neq 0?$$

• Hint: compute $(I-E_\lambda)q$. – David Mitra May 6 '13 at 12:22

The operator $P=I-E_\lambda$ is an orthogonal projection. If it is nonzero, it means that there is some $y$ such that $Py\ne0$. So if you take $x=y/\|y\|$, you have $Px=x$ (from $P^2=P$) and $\|x\|=1$.