How would I find the eigenvalues of this linear map? I'm only really used to finding eigenvalues of matrices, so I'm unfamiliar on how to do these kind of questions. Especially since in this case, there is two possible outcomes from the map depending on the value of $k$.
The linear map $A:V\rightarrow V$ is defined by

*

*$Ax = y$ with $y_{k} = x_k$ for $k = 1$.


*$Ax = y$ with $y_{k}$ = $x_k - 2x_{k-1}$ for $k > 1$.
How would I find the eigenvalues of A?
 A: If $V$ is finite-dimensional:
It looks like you have something like
$$
\begin{pmatrix}
1&0&0&\ldots\\
-2&1&0&\ldots\\
0&-2&1&\ldots\\
\vdots&\vdots&\vdots&\ddots
\end{pmatrix}
\begin{pmatrix}
x_1\\
x_2\\
x_3\\
\vdots
\end{pmatrix}
=
\begin{pmatrix}
y_1\\
y_2\\
y_3\\
\vdots
\end{pmatrix}
=
\begin{pmatrix}
x_1\\
x_2-2x_1\\
x_3-2x_2\\
\vdots
\end{pmatrix}
$$
No matter the (finite) dimension of $V$, apparently $A$ is a (lower) triangular matrix, and thus the diagonal values are the eigenvalues. So here, $1$ seems to be the only eigenvalue.
...
If $V$ is infinite-dimensional, we might have to switch tactics entirely. Here is my idea: Let $V$ be the space of analytic functions on the real line. Let the basis be $\{1,x,\frac{x^2}{2},\frac{x^3}{3},\ldots\}$. Then the linear map may be expressed as $L(f(x))=f(x)-2f'(x)$. The eigenvalue equation becomes $f-2f'=\lambda f$, which solves to $f(x)=ce^{\frac{1-\lambda}{2}x}$, which is the eigenfunction (eigen'vector') corresponding to the eigenvalue $\lambda$. Now, I am tempted to say that there don't seem to be any restrictions on $\lambda$, so the spectrum of $L$ is probably just $\mathbb{R}$, but this seems to run us into another problem: The basis, and hence dimension of $V$, seems countable, but here we are getting an uncountable number of eigenvalues. This is probably a problem, and I hope someone can fix this up or clarify things in the comments.
Edit: Apparently this isn't a problem, so the eigenvalues are just all real numbers. Yay!
A: Let $V$ be infinite dimensional. The operator $A$ is of the form $A=I-2S,$ where $S$ is the shift operator acting on the sequences, by
$$ S(x_1,x_2,\ldots )=(0,x_1,x_2,\ldots)$$
The operator $S$ does not admit any eigenvectors, as the infinite system of equations $\lambda x_1=0$ and $\lambda x_n=x_{n-1},$ for $n\ge 1,$ has only trivial solution. Therefore the operator $A$ has no eigenvalues as well.
If we equip the space $V$ with a norm we can look for a spectrum of $A.$ For example let $V=\ell^2.$ Then $\sigma(S)$ coincides with the closed unit disc centered at $0.$ Hence $\sigma(A)$ is equal $\{z\,:\, |z-1|\le 2\}.$ The same holds for $V=\ell^p$ with $1\le p\le \infty.$
Remark In the finite dimensional case, say $n,$ the kernel of the shift operator is  one-dimensional, as $S$ acts by
$$S(x_1,x_2,\ldots,x_n)=(0,x_1,\ldots,x_{n-1})$$ so the $n$th basis vector is sent to $0.$ This vector is the eigenvector of $A$ with eigenvalue $1,$ as $A=I-2S.$
