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Every linear map on a finite dimensional complex vector space has an eigenvalue. Not so in the infinite case.

I'm interested in nice counterexamples anyone might have.

Here's one:

Consider the vector space $\mathbb C^\infty$ of sequences and the right shift map $R$ defined by

$$R(a_1, a_2, a_3, ...) = (0, a_1, a_2, a_3, ...)$$

$R$ has no eigenvalue (using the usual convention that there must be a non-trivial eigenvector).

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Let $C[0,1]$ be the linear space of continuous functions on $[0,1]$; this linear space is a complete normed linear space (i.e., a Banach Space) when given the max norm $\|f\|=\max_{t \in [0,1]}|f(t)|$. The Volterra operator $$ (Vf)(x) = \int_{0}^{x}f(t)dt $$ maps continuous functions to continuous functions. $V$ has trivial null space because $(Vf)(x)=0$ for all $x$ implies $f(x)=0$ for all $x$ by the Fundamental Theorem of Calculus. So $V$ does not have $0$ as an eigenvalue. If $Vf=\lambda f$ for a non-zero $\lambda$, then $f=\frac{1}{\lambda}Vf$ is continuously differentiable with $f'=\frac{1}{\lambda} f$ and $f(0)=0$, which has only the trivial solution $f\equiv 0$.

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    $\begingroup$ Hmm I just realized how similar the Volterra operator is to a shift operator. If one considers a basis of monomials, then it acts as a right shift (with extra factors thrown in). These are a basis by Stone Weierstrass $\endgroup$ – Cameron Williams Mar 3 '15 at 14:50
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In the space $\mathbb C^{\omega}$ the right shift operator sending $(a_1,a_2,\ldots)\mapsto (0,a_1,a_2,\ldots)$ has no eigenvalue. To see this suppose $i$ is the first index with $a_i$ nonzero. Then after shifting index $i$ is 0 but index $i+1$ is not, so the vector cannot be a multiple of the original.

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  • $\begingroup$ This is my original example, yes? $\endgroup$ – Simon S Mar 3 '15 at 14:52
  • $\begingroup$ @SimonS indeed it is. I didn't notice, sorry. $\endgroup$ – Matt Samuel Mar 3 '15 at 14:53
  • $\begingroup$ No worries. Your proof there is no eigenvalue is elegant and better than the one I had in my head. $\endgroup$ – Simon S Mar 3 '15 at 14:53
  • $\begingroup$ Isn't $\mathbb{C}^\omega$ different from $\mathbb{C}^\infty$? Not that it makes any difference in this case. $\endgroup$ – Brian Bi Mar 3 '15 at 21:11
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    $\begingroup$ @simons using $\infty$ usually implies all but finitely many coordinates are 0, while there is no such restriction for $\omega$. $\endgroup$ – Matt Samuel Mar 4 '15 at 0:16
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Let $\mathcal{H}=L^2([0,1])$. Given $f \in C([0,1])$, define the multiplication operator $M_f : \mathcal{H} \to \mathcal{H}$ by $$M_fh(x) = f(x)h(x).$$ Then $M_f$ is a bounded linear operator on $\mathcal{H}$. But if we take $f(x) = x$ (for example), then $M_f$ has no eigenvalues.

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Example : In the spaces of polynomials $\mathbb{C}[X]$, the anti-derivative $P \mapsto Q$ with $Q' = P$ and $Q(0) = 0$ has no eigenvalue.

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  • $\begingroup$ New exemple, this time correct (used the antiderivative instead of the derivative) $\endgroup$ – Tryss Mar 3 '15 at 14:46
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There is an easy generalisation of the example in the question. In the polynomial ring $\Bbb C[X]$, the $\Bbb C$-linear operator $\phi: Q\mapsto PQ$ of multiplication by a fixed non-constant polynomial $P$ cannot have any eigenvalues. This is simply so since if $Q\neq0$ were an eigenvector for$~\lambda$, then one would lave $(P-\lambda)Q=0$, contradicting that $\Bbb C[X]$, is an integral domain. The example in the question is just the case $P=X$.

You can generalise this further using other $\Bbb C$-algebras that are integral domains, of which there are many.

It might be worth noting, and I did in this answer, that though there are no eigenvalues, one can make (the image of) any nonzero vector become an eigenvector for any desired eigenvalue by taking a suitably chosen quotient of $\Bbb C[X]$ by a $\phi$-stable subspace.

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  • $\begingroup$ Interesting generalization. Thank you. $\endgroup$ – Simon S Mar 3 '15 at 18:40

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