Why if $0 <|\lambda| < 1,$ then $(U^{-1} - \frac{1}{\lambda}I)$ invertible?

Question

Let $U$ a unitary operator. Prove $\sigma (U)\subseteq \Bbb{S}^1$.

I'm trying to understand the proof made at this link

https://math.stackexchange.com/a/1638489/491866

but I do not understand because if $0 <|\lambda| < 1,$ then $(U^{-1} - \frac{1}{\lambda}I)$ invertible. Can anyone explain?

Answer 1

It is stated in the link that the spectrum of $U^{-1}$ is contained in the unit ball. As $1/\lambda$ is outside the unit ball, we get that $U^{-1} - 1/\lambda I$ is invertible (by the definition of spectrum).

Answer 2

If $U$ is a unitary on some Hilbert space $H$,

$U:H \to H, \; UU^\dagger = U^\dagger U = I, \tag 1$

then for any $x \in H$,

$\langle x, x \rangle = \langle x, Ix \rangle = \langle x, U^\dagger U x \rangle = \langle U x, U x \rangle; \tag 2$

if in addition $\mu$ is an eigenvalue of $U$, so that

$Ux = \mu x, \; x \ne 0, \tag 3$

(2) may be continued as

$\langle x, x \rangle = \langle U x, U x \rangle = \langle \mu x, \mu x \rangle = \bar \mu \mu \langle x, x \rangle, \tag 4$

from which, since $x \ne 0$ and hence $\langle x, x \rangle \ne 0$,

$\bar \mu \mu = 1; \tag 5$

that is,

$\mu, \bar \mu \in \Bbb S^1. \tag 6$

We note that, from (1),

$UU^\dagger = U^\dagger U = I \Longrightarrow (U^\dagger)^\dagger U^\dagger = U^\dagger (U^\dagger)^\dagger = I; \tag 7$

that is, $U^\dagger$ is also unitary; thus we see that (6) binds for the eigenvalues of $U^\dagger$ as well as for those of $U$; furthermore, we see via (3) and (5) that

$U^\dagger x = \bar \mu \mu U^\dagger x = \bar \mu U^\dagger \mu x = \bar \mu U^\dagger U x = \bar \mu (U^\dagger U)x = \bar \mu I x; = \bar \mu x; \tag 8$

that is, $\bar \mu$ is an eigenvalue of $U^\dagger$ whenever $\mu$ is of $U$, and with the same eigenvector $x$; and, in addition, (7) yields

$U^\dagger U = U^\dagger (U^\dagger)^\dagger \Longrightarrow U (U^\dagger U) = U(U^\dagger (U^\dagger)^\dagger)$ $\Longrightarrow (UU^\dagger)U = (UU^\dagger)(U^\dagger)^\dagger$ $\Longrightarrow IU = I(U^\dagger)^\dagger \Longrightarrow U = (U^\dagger)^\dagger. \tag 9$

These properties (8)-(9) of unitaries $U$ are not directly necessary here, though they may shed some useful light on unitary operators in general.

Returning to (2) we see that it implies

$\Vert x \Vert ^2 = \langle x, x \rangle= \langle U x, U x \rangle = \Vert Ux \Vert^2, \tag{10}$

or

$\Vert x \Vert = \Vert Ux \Vert, \tag{11}$

from which

$\Vert x \Vert = \Vert Ux \Vert \le \Vert U \Vert \Vert x \Vert, \tag{12}$

and hence

$1 \le \Vert U \Vert; \tag{13}$

furthermore, we may rule out the case

$\Vert U \Vert > 1, \tag{14}$

since this entails the existence of some $x$ with

$\Vert x \Vert < \Vert Ux \Vert \le \Vert U \Vert \Vert x \Vert \tag{15}$

by virtue of the definition of $\Vert U \Vert$ as the infimum of all $0 < C \in \Bbb R$ such that

$\Vert Ux \Vert \le C\Vert x \Vert, \; \forall x \in H; \tag{16}$

in addition, since

$U^\dagger = U^{-1} \tag{17}$

is also unitary, we have

$\Vert U^\dagger \Vert = \Vert U^{-1} \Vert = 1 \tag{18}$

as well.

Granted, then, that $\Vert U \Vert = 1$, we argue as follows: suppose

$0 < \vert \lambda \vert < 1; \tag{19}$

then (18) yields

$\Vert \lambda U^{-1} \Vert = \vert \lambda \vert \Vert U^{-1} \Vert < 1; \tag{20}$

but it is well-known that $I - A$ is invertible for any operator $A$ with

$\Vert A \Vert < 1; \tag{21}$

this is a standard result which is a consequence of the convergence of the series

$(I - A)^{-1} = \displaystyle \sum_0^\infty A^j = I + A + A^2 + \ldots, \tag{22}$

which is majorized term-by-term by

$(I - \Vert A \Vert)^{-1} = \displaystyle \sum_0^\infty \Vert A \Vert^j = I + \Vert A \Vert + \Vert A \Vert^2 + \ldots; \tag{23}$

we thus see that

$\exists (I - \lambda U^{-1})^{-1} \Longrightarrow \exists \lambda (I - \lambda U^{-1})^{-1} = (\lambda^{-1}I - U^{-1})^{-1}. \tag{24}$

If, on the other hand,

$\vert \lambda \vert > 1, \tag{25}$

then

$\Vert \lambda^{-1} U \Vert = \vert \lambda^{-1} \vert \Vert U \Vert < 1, \tag{26}$

whence

$\exists (I - \lambda^{-1}U)^{-1} \Longrightarrow \exists U(I - \lambda^{-1}U)^{-1} = (U^{-1} - \lambda^{-1} I)^{-1} \tag{27}$

as well.

We thus see that for all cases

$0 < \vert \lambda \vert \ne 1 \tag{28}$

the inverse of the operator

$U^{-1} - \lambda^{-1} I \tag{29}$

exists on the Hilbert space $H$.