The following question is motivated by the construction of the fermionic path/field integral, as done for example in Altland & Simons "Condensed Matter Field Theory".
Consider the vector space $\mathbb C^2$ with two standard basis vectors named $|0\rangle$ and $|1\rangle$. Furthermore, consider the linear operator $a$ defined by
$$ a |0\rangle = 0 \text{ and } a |1\rangle = |0\rangle .$$
(This is the "annihilation" operator for a single fermion). Its hermitian conjugate $a^\dagger$ (the "creation" operator) is given by
$$ a^\dagger |0\rangle = |1\rangle \text{ and } a^\dagger |1\rangle = 0 .$$
Clearly, both operators $a$ and $a^\dagger$ are nilpotent and have the following Jordan normal form
$$ a, a^\dagger \simeq \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}$$
In particular, these operators are not completely determined by their eigenvalues. (They don't commute or anticommute, though, we have $aa^\dagger + a^\dagger a = 1$.)
However, when constructing the fermionic field integral, physicists treat these operators as if they had useful eigenvalues! Namely, the eigenvalues are taken to be Grassmann-numbers, i.e. two "numbers" $\eta$ and $\bar\eta$ that anticommute with each other, $\eta \bar\eta = - \bar\eta \eta$, and that also anticommute with the operators $a$ and $a^\dagger$. Then, physicists construct the so-called "coherent state"
$$ |\eta\rangle := e^{-\eta a^\dagger} |0\rangle = (1 -\eta a^\dagger) |0\rangle $$
which behaves like an eigenvector for the annihilation operator
$$ a |\eta\rangle = \eta |\eta\rangle .$$
Together with the dual vector,
$$ \langle\eta| = \langle 0| e^{-a\bar\eta} $$
we can write the projection onto the corresponding "eigenspace" as $|\eta\rangle \langle\eta|$. These projections form a "complete set", as can be seen by summing/integrating over the Grassmann variables
$$ \int d\bar\eta d\eta\ e^{-\bar\eta \eta} |\eta\rangle \langle\eta| = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$$
Now my question: how on earth does this make sense? The basic idea is to expand the available supply of "numbers" to obtain eigenvalues. This is a common idea and can be used to construct many familiar field extensions like $\mathbb R \subseteq \mathbb C$. After all, the imaginary unit $i$ is the eigenvalue of a 90° rotation in two dimensions. The problem here is Grassmann algebras can't be fields and we're no longer dealing with vector spaces.
How can elements of a Grassmann algebra be interpreted as eigenvalues of nilpotent operators?
I guess I'm looking for representations of the matrix algebra $\mathbb C^{n\times n}$ on Grassmann modules or something like that. Probably the "natural" representation on the tensor product $\mathbb C^n \otimes \Lambda \mathbb C^n$.
Let me just retell a part of the story in slightly different language — maybe it will become less mysterious then.
The story starts with a module $M=\mathbb C[\xi]$ (free supercommutative algebra with one odd generator — aka exterior algebra on one generator) over a Clifford algebra $A=\mathbb C[\xi,\frac\partial{\partial\xi}]$ (“odd differential operators” — i.e. the only relation is $[\frac\partial{\partial\xi},\xi]_+=1$). “Physical notation”: $|0\rangle=1\in M$, $|1\rangle=\xi\in M$, $a=\frac\partial{\partial\xi}\in A$, $a^\dagger=\xi\in A$.
One can extend scalars: instead of an algebra $A$ over (the field of) complex numbers we get an algebra $\tilde A=A[\eta,\bar\eta]=R[\xi,\frac\partial{\partial\xi}]$ over (the ring) $R=\mathbb C[\eta,\bar\eta]$; and module $M$ extends to $A[\eta,\bar\eta]$-module $\tilde M=M[\eta,\bar\eta]=R[\xi]$ (cf. Koszul resolution). Over our base ring, $R$, it's a free module of dimension 2.
Now $\frac{\partial}{\partial\xi}$ has an eigenvector, $\left| \eta\rangle \right.=exp(-\xi\eta)=1-\xi\eta$ with eigienvalue $\eta$.
(Nothing too suprising here: everybody knows that $\exp(\lambda x)$ is an eigenvector of $\frac\partial{\partial x}$. One might ask, why not just use $\exp(\xi)$ then — but an easy calculation show that it's not an eigenvector of $\frac\partial{\partial\xi}$ — because, in fact, we need the argument of exponent have to be even, not odd. That's why we need to extend scalars, I believe — $\xi\eta$ is an even element, and exponentiation works fine.)
But there is still no way to extend it to an eigenbasis. The problem is, we can't divide by $\eta$ (or two exponents — say, $\left|\eta\rangle \right.$ and $\left|\bar\eta\rangle \right.$ — would generate everything). So for any free $\mathbb C[\eta,\bar\eta]$ let's define (Berezin) integral by $\int d\eta d\bar\eta\,(e+f\eta+g\bar\eta+h\eta\bar\eta)=h$. Now $\left|\eta\rangle \right.$ generates our $\tilde M$ if we allow not only summation, but also this “integration”: $$ \begin{align} 1 &=\pm\int d\eta d\bar\eta\,\eta\bar\eta|\eta\rangle;\\ \xi&=\pm\int d\eta d\bar\eta\, \bar\eta|\eta\rangle. \end{align} $$ In other words, $$ \begin{align} \left|0\rangle\langle0\right|&=\pm\int d\eta d\bar\eta\,\bar\eta\eta|\eta\rangle;\\ \left|1\rangle\langle1\right|&=\pm\int d\eta d\bar\eta\,|\eta\rangle\bar\eta\frac{\partial}{\partial\xi}. \end{align} $$ Hence $$ \operatorname{Id}= \left|0\rangle\right. \left. \langle0\right|+\left|1\rangle\langle1\right|= \int d\eta d\bar\eta(1\pm\eta\bar\eta)|\eta\rangle\langle\eta|. $$
// Note, in particular, that (unlike $|0\rangle\langle0|$ or $|1\rangle\langle1|$), $|\eta\rangle\langle\eta|$ is not really a projector on anything — (it's unipotent, so) it's an automorphism, actually!
(Oh, I appologize for mixing “mathematical” and “physical” notation freely.)
(So, as for the question in gray from the OP, it's not really a problem: much of linear algebra can be done not only in vector spaces over fields, but also in module over (super)commutative rings — and it's useful to consider not only field extensions but also ring extensions sometimes.)