= Spectrum of a bounded operator
{title2=$\sigma(T)=\mathbb C\setminus\rho(T)$}
Let $X\ne\{0\}$ be a <Banach space> over the <complex numbers>, and let $T:X\to X$ be a <bounded linear operator> defined on all of $X$. Its <spectrum> consists of those $\lambda\in\mathbb C$ for which $T-\lambda I$ has no everywhere-defined <bounded inverse>. Its complement is the <resolvent set of an operator> $\rho(T)$. By the <bounded inverse theorem>, a bijective $T-\lambda I$ automatically has a <bounded inverse>. The <spectrum> is a nonempty <compact subset> of the <complex plane>, with
$$
\sigma(T)\subseteq\{\lambda\in\mathbb C:|\lambda|\leq\|T\|\}.
$$
The bound uses the <operator norm>, and nonemptiness uses the complex <field>.https://www.math.hkust.edu.hk/~makyli/4063/fanotes2_150615.pdf#page=38[HKUST Math 4063 notes, chapter 4, section 3: compactness and nonemptiness of the spectrum]{ref} For a real <linear operator>, complex spectral values are understood by extending it to a complex <vector space>; restricting $\lambda$ to the real <field> can give an empty set.
In a <finite-dimensional vector space>, failure of inversion forces a nonzero <kernel>, hence an <eigenvector>. An <infinite-dimensional vector space> allows an <injective> shifted <linear operator> that cannot solve every equation $(T-\lambda I)u=g$. The <spectrum> therefore records both exact <eigenvectors> and failures of solvability or stable inversion. Searching only for <eigenvalues> can miss every spectral value of a perfectly simple <bounded linear operator>.
Consider the <multiplication operator> $M$ on the complex <L2 space> $L^2([0,1],dx)$, with <Lebesgue measure>:
$$
(Mf)(x)=xf(x),\qquad \|f\|_2^2=\int_0^1|f(x)|^2\,dx.
$$
Its <operator norm> is one. The inequality $\|Mf\|_2\leq\|f\|_2$ gives the upper bound; <unit vectors> concentrated near $x=1$ approach it. The result is
$$
\sigma(M)=[0,1],\qquad \sigma_p(M)=\varnothing.
$$
This is the interval example of the general connection between <multiplication operators> and the <essential range> of their multiplier.https://hackmd.io/@knikct/BkX1s2PFr[Berkeley Math 224A, lecture 13: spectra of multiplication operators]{ref} Indeed, an <eigenfunction> would satisfy $(x-\lambda)f(x)=0$ <almost everywhere>. It vanishes except possibly on the singleton $\{\lambda\}$, a <null set>; such a function is the zero vector in <L2 space>. Concentrating at one point cannot produce a nonzero <L2 function>.
For $\lambda\notin[0,1]$, the inverse is completely explicit:
$$
((M-\lambda I)^{-1}g)(x)=\frac{g(x)}{x-\lambda},\qquad
\|(M-\lambda I)^{-1}\|=\frac1{\operatorname{dist}(\lambda,[0,1])}.
$$
The <distance to a set> is positive, so the inverse multiplier is bounded. Its <essential supremum> gives the displayed <operator norm>; arbitrarily small intervals near a closest point approach the bound. Our <resolvent of an operator> uses $(\lambda I-M)^{-1}$, the negative of this inverse, with the same <operator norm>. For $\lambda=-1/4$, $5/4$, and $1/2+i/10$, the inverse norms are respectively $4$, $4$, and $10$. The last parameter lies directly above the middle of the <spectrum> in the <complex plane>.
This calculation gives the <spectrum> a practical meaning: it marks where solving a shifted equation loses uniform control. At $\lambda=1/2+i\eta$, the largest amplification of an input error is $1/|\eta|$. Changing $\eta$ from $0.1$ to $0.01$ increases that factor from ten to one hundred. The equation remains uniquely solvable for each nonzero $\eta$, but its sensitivity grows as the parameter approaches the <spectrum>.
\Image[/operator-spectrum-examples.png]
{title=An interval without eigenvectors and a diagonal spectrum accumulating at zero}
To see directly why every $\lambda\in[0,1]$ belongs to the <spectrum>, choose $0<\varepsilon<1$ and define a clipped interval and a normalized <indicator function>:
$$
a=\max(0,\lambda-\varepsilon),\quad b=\min(1,\lambda+\varepsilon),\quad
\ell=b-a,\qquad u_{\lambda,\varepsilon}(x)=\ell^{-1/2}\mathbf1_{[a,b]}(x).
$$
Its <L2 norm> is exactly one, including at both endpoints. On its <support>, $|x-\lambda|\leq\varepsilon$, giving
$$
\|(M-\lambda I)u_{\lambda,\varepsilon}\|_2^2
=\frac{(b-\lambda)^3-(a-\lambda)^3}{3\ell}
\leq\varepsilon^2.
$$
Thus any choice $\varepsilon_n\to0$ produces <approximate eigenvectors>. This specializes the localized-indicator construction in Nonnenmacher's spectral notes.https://www.imo.universite-paris-saclay.fr/~nonnenma/enseign/Spectral%20Theory%202021%20chap1-4.pdf#page=47[Nonnenmacher, Introduction to Spectral Theory (2021), proposition 4.2.15]{ref} A hypothetical <bounded inverse> $B$ would imply $1=\|u_{\lambda,\varepsilon}\|_2\leq\|B\|\varepsilon$, impossible for arbitrarily small $\varepsilon$.
For an interior point with $\varepsilon\leq\min(\lambda,1-\lambda)$, $\ell=2\varepsilon$ and the residual is exactly $\varepsilon/\sqrt3$. At $\lambda=0$ or $1$, $\ell=\varepsilon$ and the same residual formula holds, but the amplitude is $\varepsilon^{-1/2}$ rather than $(2\varepsilon)^{-1/2}$. At $\lambda=1/2$, taking $\varepsilon=0.2,0.1,0.05$ gives amplitudes approximately $1.5811,2.2361,3.1623$ and residuals $0.11547,0.057735,0.028868$. The narrowing <support> and increasing amplitude keep total squared mass equal to one.
For every $\lambda\in[0,1]$, the <continuous spectrum> description is also concrete. Any <L2 function> $g$ can be approximated in <L2 norm> by its truncations to $|x-\lambda|\geq\delta$. Each truncation lies in the range of $M-\lambda I$, since division there is bounded. The range is therefore dense. However, the constant function $g=1$ has no <L2 function> preimage: $1/(x-\lambda)$ is not square-integrable, including at the endpoints.
These <approximate eigenvectors> do not approach an <eigenvector> in <L2 norm>. Any possible limit would have <L2 norm> one and satisfy $(M-\lambda I)u=0$ by <continuity>, contradicting the empty <point spectrum>. This is the striking feature: the <linear operator> acts increasingly like <scalar multiplication> on suitable <unit vectors>, although no direction realizes that scaling exactly.
\Image[/approximate-spectral-vectors.png]
{title=Unit L2 vectors concentrate while their spectral residual tends to zero}
A complementary example is the <diagonal operator on sequence space> $(Dx)_n=x_n/n$ on complex <l2 sequence space>, with $n=1,2,\ldots$. The coordinate <unit vectors> satisfy $De_n=(1/n)e_n$, so
$$
\sigma_p(D)=\{1/n:n\geq1\},\qquad
\sigma(D)=\{0\}\cup\{1/n:n\geq1\}.
$$
Zero is not an <eigenvalue>, since $Dx=0$ forces every coordinate to vanish. Nevertheless $\|De_n\|=1/n\to0$: these are <approximate eigenvectors> at zero. More concretely, $g=(1/n)_{n\geq1}$ belongs to <l2 sequence space>, but $Dx=g$ would require $x_n=1$ for all $n$, which does not. Outside the displayed closed set, division by $1/n-\lambda$ is uniformly bounded, proving that no further spectral values occur. Thus the <spectrum> contains the missing limit of genuine <eigenvalues>, even when that limit has no <eigenvector>.
Back to article page
