I am looking for an example of a linear space $V$ over $\mathbb{R}$ and an endomorphism $\phi : V \to V$ such that each $a \in \mathbb{R}$ is an eigenvalue of $\phi$. One relatively simple example would be an endomorphism given by the matrix

$$\begin{bmatrix} a_1 & 0 & 0 & \cdots \\ 0 &a_2& 0 & \cdots \\ 0 & 0 & a_3&\cdots \\ \vdots & \vdots & \vdots & \ddots \end{bmatrix}$$ where $a_1, a_2, \dots$ are all eigenvalues associated with unit vectors $e_1, e_2, \dots$. We could possibly have infinitely many such eigenvalues. My problem here is that this still seems to be slightly complicated as the material I am going over has not introduced infinite dimensional vector spaces. Is there a more straightforward finite dimension example I am somehow missing?

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    $\begingroup$ In the finite dimensional case, you will have only finitely many eigenvalues $\endgroup$ – D_S Mar 6 '17 at 18:45
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    $\begingroup$ Eigenvectors associated to distinct eigenvalues are linearly independent, so this sort of construction is the best you can do; the dimension of the underlying vector space is an upper bound on the number of distinct eigenvalues, even in the infinite-dimensional case. $\endgroup$ – Qiaochu Yuan Mar 6 '17 at 18:55
  • $\begingroup$ @D_S One question. It bothers me that in my example, the set of eigenvectors $E$ (and the corresponding set of eigenvalues) satisfies $\lvert E \rvert = \lvert \mathbb{N} \rvert < \lvert \mathbb{R} \rvert$. Doesn't this mean that we can't claim that every $a \in \mathbb{R}$ is an eigenvalue? $\endgroup$ – Zelazny Mar 7 '17 at 7:34
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    $\begingroup$ Right, you only have countably many eigenvalues. In general, the cardinality of eigenvalues cannot exceed the dimension of your vector space. $\endgroup$ – D_S Mar 7 '17 at 9:39
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    $\begingroup$ Your example can be modified. Let $$V = \bigoplus\limits_{x \in \mathbb{R}} \mathbb{R}$$ and for $x \in \mathbb{R}$, let $e_x$ be the element with $1$ in the $x$ place, $0$ elsewhere. This is a basis for $V$. You can define a linear operator $T$ on $V$ by sending $e_x$ to $xe_x$. $\endgroup$ – D_S Mar 7 '17 at 9:41

As others have said, we need to consider something over an infinite dimensional space, however you slice it. As you've said, however, infinite matrices are a bit bizarre.

Here's a more "familiar" example: let $V$ denote the vector space of all smooth (infinitely differentiable) functions. Consider $\phi:V \to V$ to be the "derivative map", defined by $\phi(f) = f'$. We note that for any $\lambda \in \Bbb R$, the equation $$ \phi(f) = \lambda f $$ has the solution $f(x) = e^{\lambda x}$. So, every real number is an eigenvalue of $\phi$.

  • $\begingroup$ There are a lot of very strange things about my choice of vector space and linear map from a functional analysis perspective. However, it certainly does the job. $\endgroup$ – Omnomnomnom Mar 6 '17 at 20:00
  • $\begingroup$ This is a great example and it doesn't seem to suffer from the problem I mentioned above. Thanks! $\endgroup$ – Zelazny Mar 7 '17 at 8:14

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