Do eigenvalues depend on the choice of basis? Suppose we have a basis $B$ for an endomorphism $f$ that has eigenvalues $\lambda_{1},\dots,\lambda_{k}$. 
Do these eigenvalues change or stay the same if we change to another basis $B'$?
 A: Recall the definition:

Let $f$ be an endomorphism of a vector space $V$, then  $\lambda$ is an eigenvalue of $f$ if there exists some non-zero $v \in V$ such that $f(v)=\lambda v$.

This does not involve a basis of the space at all. Thus it must be invariant under change of basis. 
A: No, eigenvalues are invariant to the change of basis, only the representation of the eigenvectors by the vector coordinates in the new basis changes.
Indeed suppose that
$$Ax=\lambda x$$
and let consider the change of basis $x=My$ then
$$Ax=\lambda x\implies AMy=\lambda My\implies M^{-1}AMy=\lambda y \implies By=\lambda y$$
A: The whole point of eigenvalues and eigenvectors is to produce a bunch of axes that define your skewy transformation, so that your skewy transformation becomes a scaling transformation on these axes. If anything, this gives you a nice basis (one in which your matrix is diagonal, i.e. scaling). Your eigenvalues are clearly the same in the eigenbasis as in any other basis (they're across the diagonal), so the eigenvalues are the same in all bases.
A: As @Christoph says, the definition of an eigenvalue does not involve a basis. Given a vector space $V$ and linear operator $f$, an eigenvector of $f$ is a vector v such that there exists a scalar $\lambda$ such that $f$(v) = $\lambda$ v. $\lambda$ is then an eigenvalue. A basis is a system of associating ordered tuples and vector. You take a basis set of vectors, then express every other vector as a linear combination of those vectors. You can then take those coefficients and represent the vector with an ordered tuple of those coefficients: v = $c^i$b$_i$. You can then write a matrix representing $f$ by taking $a_{ij}$ as being the coefficient of b$_i$ of $f$(b$_j$). That is, you apply $f$ to b$_j$, then look at the b$_i$ component of the answer. Doing this for all i,j gives you $A$. $A$ then represents $f$, but is not quite the same thing as $f$. Changing the basis will change what matrix represents $f$, and it will change what tuples represent the eignevectors, but it won't change what actual vectors are eigenvectors, and it won't change the eigenvalues.
Note that if you find an $A$ from one basis, and you want to use it to find out what $f$ does in terms of another basis, you have to first change the vector back to the original basis, then apply $A$, then go back to the new basis. That can be represented by 
$S^{-1}AS$v
where S is made up of the vectors of the new basis, expressed in the old. The matrix $S^{-1}AS$ is known as a conjugation of $A$ by $S$. If one matrix can be obtained from another through conjugation, then the matrices are called "similar matrices". Similar matrices have the same eigenvalues, as they can be considered to represent the same operator in different bases.
