Necessity of Commuting in Jordan Decomposition

Question

Suppose $A$ is some matrix with Jordan decomposition $A = D + N$, where $D$ is diagonal and $N$ is nilpotent. The Jordan decomposition requires also that $[D,N] = DN-ND=0.$ Is there an example of a matrix $A$ such that $A = D+N$, but we don't have the commutativity condition? What is the significance of this commutativity condition? That is, why is the Jordan decomposition not simply defined to be the decomposition of $A$ into the sum of $D+N$?

Answer 1

The best way to think about Jordan decompositions, I think, is to think about it as a decomposition of $V$, the vector space $A$ is acting on, into generalized eigenspaces for $A$.

There are many situations where you can understand the action of a linear map by looking at what happens in the different generalized eigenspaces, prove some results there and then conclude something about what is happening in the total space.

The point of the whole construction is the following:

Restricted to the generalized eigenspace at eigenvalue $\lambda$ (let's call it $E_\lambda \subset V$) the operator $A$ acts as $\lambda I + N_\lambda$ where $I$ is the identity (not on all of $V$ but on the smaller space $E_\lambda$) and $N_\lambda$ is a nilpotent operator on this small space.

In fact you can take the definition of $E_\lambda$ to be the biggest subspace of $V$ on which $N_\lambda := (A - \lambda I)$ acts nilpotently. (The notion of biggest is well defined here because whenever $A - \lambda I$ acts nilpotently on two subspaces $E_1$ and $E_2$ of $V$ it also does so on $E_1 + E_2$.)

It is important to realize that the several generalized eigenspaces of $A$ have intersection $\{0\}$ and togehter span all of $V$ so:

$$V = \bigoplus_{\lambda \in \sigma(A)} E_\lambda$$

(Here $\sigma(A)$ denotes the set of eigenvalues of $A$. Alternatively we could sum over all of $\mathbb{C}$ using that for almost all choices of $\lambda$ we have that $E_\lambda = \{0\}$.)

This decomposition holds true regardless how beautiful or ugly $A$ is. (I do assume $V$ to be finite dimensional, though)

Now to get back to the Jordan decomposition of $A$, rather than the decomposition of $V$, we extend the operators $\lambda I$ and $N_\lambda$ on $E_\lambda$ to operators on all of $V$ by demanding them to act as zero on the other generalized eigenspaces. (Since together the generalized eigenspaces span all of $V$ this actually is a definition of the resulting operator.)

Let's introduce some (non-standard) notation: let $D_\lambda: V \to V$ be the operator that acts on $E_\lambda$ as $\lambda I$ and on $E_\mu$ as zero, for $\mu \neq \lambda$ and let $N_\lambda: V \to V$ be the operator that acts as $N_\lambda = A|_{E_\lambda} - \lambda I$ on $E_\lambda$ and as zero on $E_\mu$, for $\mu \neq \lambda$.

Now the Jordan decomposition $A = D + N$ you are talking about is just this:

$$D = \sum_{\lambda \in \sigma(A)} D_\lambda$$ $$N = \sum_{\lambda \in \sigma(A)} N_\lambda$$

It is very easy to see that $D$ is diagonalizable (you write diagonal but that is only the case when you picked your basis very clever from the beginning, where clever exactly means that each basis element is an element of a generalized eigenspace for $A$). It is also really easy to see that $N$ is nilpotent.

The closest thing to an answer to your question I can give though is the following:

For this choice of $D$ and $N$ we have that $D$ and $N$ commute, as can be seen by looking at what happens within each generalized eigenspace. Also this is the best choice.

Other choices of $D$ and $N$ (subject to $D + N = A$, $D$ diagonalizable and $N$ nilpotent) aren't just as useful because it is the decomposition of $V$ into the $E_\lambda$ and the action of $A$ there (expressed by the eigenvalues and the 'small' nilpotent operators $N_\lambda$) that really give us a deeper understanding of $A$.