I forget exactly where, but I've heard/read multiple times it be mentioned that defining eigenvalues/vectors implicitly requires a choice of inner product.
Question: Is it more correct to say that "the definition of eigenvalues/vectors of a linear map $L : R^n \to R^n$ does not depend on a choice of inner product" but that "the definition of eigenvalues/vectors of a matrix $M \in R^{n \times n}$ does implicitly depend on a choice of inner product"?
Specifically, I don't see how the definition of eigenvalues/vectors of a linear function depends in any way on choosing an inner product. The statement $L(v) = c v$ makes sense / is well defined regardless of what inner product does or does not exist on the vector space $V$ such that $v \in V$ and $L: V \to V$.
I know that a choice of inner product (or more generally any non-degenerate bilinear form, inner products being positive definite special case) is equivalent to a choice of isomorphism between a vector space $V$ and its dual $V^*$. (For example compare the notion of "musical isomorphisms".)
I also know that the space of linear maps $\mathcal{L}(V,V)$ is isomorphic to $V^* \otimes V$. Taking matrices to be "fully contravariant", the space of matrices would be isomorphic to $V \otimes V$, and thus to define an isomorphism (correspondence) between linear functions $V^* \otimes V$ and matrices $V \otimes V$, one requires an isomorphism between $V^*$ and $V$, i.e. a choice of inner product (or more generally non-degenerate bilinear form).
Hence we would expect that notions like "eigenvalues/vectors of a linear map" that don't depend on a choice of inner product for $V$ end up depending on a choice of inner product for $V$ when we define their "matrix analogues".
Is this correct? Or is there a deeper / subtler way in which the definition of eigenvalues / eigenvectors depends on an implicit choice of inner product such that the dependence still exists even when defining those concepts with respect to linear functions?
Related question: Does matrix multiplication require an inner product space?
Short answer.
The definitions of eigenvalues and eigenvectors of a linear map from a vector space $V$ to itself are independent of any choice (including none) of an inner product on $V$.
If you choose a basis for $V$ then you can reason about the eigenvectors and values of the matrix corresponding to a linear map. That's still independent of any notion of an inner product.
However, choosing a basis implicitly defines an inner product since vectors "become" $n$-tuples (assuming $V$ is finite dimensional). Then it makes sense to discuss the relationship between eigenvectors and the inner product - for example, to ask whether eigenvectors are orthogonal.