Let $\mathbb{K}$ denote a field and $A$ denote a $\mathbb{K}$-algebra. Then given a $\mathbb{K}$-subalgebra $\Delta$ of $A$, I suppose it make sense to declare that $m \in A$ is $\Delta$-diagonalizable iff $m \in u^{-1}\Delta u$ for some unit $u \in A$. In the special case where $A$ is the matrix algebra $M_n \mathbb{K}$ and $\Delta$ is the subalgebra of diagonal matrices, this recovers the usual notion of diagonalizability. Alternatively, we might define that $m$ is $\Delta$-diagonalizable iff there exists an automorphism $\varphi$ of $A$ such that $\varphi(m) \in \Delta$. I think that since $\mathbb{K}$ was assumed a field, these definitions should be equivalent whenever $A$ is a matrix algebra. (Is this right?)
Anyway, this leaves open the question of how to think about the subalgebra of diagonal matrices.
Question. How do modern algebraists think about diagonal matrices and/or the subalgebra of diagonal matrices?
For example, a good answer might take any of the following perspectives.
- The set $\Delta \subseteq A$ is uniquely defined, up to the action of automorphisms of $A$, by the following property...
- From the viewpoint of diagonalizability, the important thing about $\Delta$ is that it satisfies the following axioms/conditions...
- The subalgebra of diagonal matrices isn't really the right thing to be focusing on here; modern algebraists mostly focus on the concept [whatever]...
This is not an exhaustive list of perspectives a good answer might take.
I'm not $100\%$ I'm the right person to speak for "modern algebraists," but here's some thoughts.
Dietrich mentions semisimple elements in his comment: these generalize diagonalizeable operators to elements of an algebra in the presence of an algebra representation - either an associative algebra or alternatively a Lie algebra. Indeed, given such an element $a$ and a representation $(\rho,V)$, with a vector space $V$ over $K$, we can make $V$ into a $K[T]$-module (where $K[T]$ is the usual polynomial ring) by making $T$ act by $\rho(a)$. Then $a\in A$ is semisimple if $V$ is semisimple as a $K[T]$-module.
By the fundamental theorem of finitely-generated modules over PIDs, our $K[T]$-module $V$ decomposes as $\bigoplus K[T]/f(T)^e$ where the $f$s are irreducible factors of $\rho(a)$'s characteristic polynomial in $K[T]$. These summands correspond to blocks in $\rho(a)$'s Frobenius normal form. The summands $K[T]/f(T)^e$ are indecomposable, but only simple when $e=1$. If $K$ is algebraically closed, the irreducible polynomials in $K[T]$ are linear, so semisimplicity of $a$ is equivalent to the fact $V\cong \bigoplus K[T]/(T-\lambda)$ which means $\rho(a)$ is equivalent to ${\rm diag}(\lambda)$. This recovers the usual understanding of diagonalizable.
I believe $a$ may be called semisimple if it so in the regular representation, and that this implies $\rho(a)$ is semisimple for any representation $\rho$, but my memory is hazy.
So semisimple captures the idea of diagonalizability, which is being diagonal with respect to some decomposition of the space. More generally we can say that if $A$ acts on $V$ then $a\in A$ acts diagonally with respect to a decomposition $V=\bigoplus V_i$ if we can write $\rho(a)=\bigoplus_i\rho(a)_i$. This could potentially mean nondiagonalizable matrices are in fact diagonal with respect to certain decompositions. Indeed, for any endomorphism there is decomposition of $V$ that cannot be refined further under which this is true: one takes the summands of $V$ that correspond to the factors $K[T]/f(T)^e$ in the decomposition guaranteed by the fundamental theorem and then a matrix will act by the blocks of the Frobenius normal form on each summand of $V$.