I searched for an answer for an hour before asking the question. As usual, the answer presented itself only a moment later:

&lt;x:String&gt;Sort By HSB&lt;/x:String&gt;


The items should not be ComboBox items, they should be strings.

If anyone wants to comment and let me know more about the "x:" and why that works, the info would be appreciated. is that just a way for the XAML to access OS or language features?

Thanks.

Dave

How do I get the text, not the name of the comboboxitem, from a combobox with C# in Windows Runtime?
Is there an analogue of the Jordan Normal Form for an infinite dimensional vector space?

In general I think the answer is no.  It's been awhile since I studied it, but I'm pretty sure something would go wrong in distinguishing eigenvalues for an infinite dimensional space.  Normally, for infinite dimensional vector spaces, you have some topology that allows you to define compact operators, et cetera, that are easier to study.  But in my specific case I'm looking at an nilpotent linear map $D: F[X] \to F[X]$, where $F$ is a finite field.  I'm actually trying to find the kernel of said map.  I hope that a transformation to a suitably friendly basis would help (hence why I'm asking about jordan normal forms!).  Is $F[x]$ close enough to finite that there might be something similar to the JNF that can help?  

Some motivation:
For $f \in F_2[x]$ and $F_4 = F_2[\mu]/(\mu^2 + \mu + 1)$, the map is $Df = f(x + \mu) - f(x)$ (so actually $D: F_2[x] \to F_4[x]$, but clearly this can be thought of as an endomorphism).  I'm trying to find the kernel of this map in $F_2$.  Originally, I thought it would be spanned by elements like $x^{2^n} + x^{2^{n-2}}$, since $(x + \mu)^n$ would have a lot of non-zero terms when n isn't a power of 2.  But after trying to prove that, I found several counterexamples (exempli gratia $x^{12} + x^9 + x^6 + x^3$).  Additionally, if $\text{deg}f=n$, I have $n$ equations that the coefficients of $f$ must satisify for $f$ to be in the kernel, and I believe I can show that we can have polynomials of 8, 16, 32, et cetera, terms that aren't sums of something that has fewer terms and already is in the kernel, but that's quite a bit of work!  So I've decided to try looking for a different basis for $F_2[x]$ that would get close to diagonalizing the matrix of $D$. But as @Mariano, pointed out, this may not be any easier...

Edit before I provided and motivation, this question asked about idempotent maps and not nilpotent. But that was because I was being silly and got the terms "idempotent" and "nilpotent" confused.  I didn't catch my mistake until @Mariano reminded me that an idempotent operator is a projection when he answered the idempotent question.  The map $D$ is clearly not idempotent!
