Classifying certain types of matrices

Question

How many similarity classes of nilpotent $4 \times 4$ matrices over $\mathbb{C}$ are there?

I suspect the answer is connected to minimal polynomials, but I'm not sure. Any suggestions?

Answer 1

An $n \times n$ matrix is nilpotent if and only if it has no eigenvalues except $0$, i.e. its characteristic polynomial is $\lambda^n$. It is similar to its Jordan canonical form, which then consists of blocks with diagonal $0$. So the similarity classes in the $4 \times 4$ case correspond to the possible Jordan forms, with blocks of the following sizes: $$ \matrix{1,1,1,1\cr 2,1,1 \cr 2,2 \cr 3,1 \cr 4\cr}$$

Answer 2

This is not a separate answer but a long comment (addendum) to Robert Israels execellent answer. I only type it in the 'answer' box because of the length. This is something I learned years ago from a representation theorist at Zagreb university (unfortunately I forgot their name) and found very helpful.

As Robert Israel explains the similarity classes of nilpotent $n \times n$ matrices correspond to partitions of $n$.

Now a standard way of 'visualizing' a partition that shows up in many areas of mathematics is as Young diagram, where the rows represent the parts in the partition. Now the beautiful thing is that this representation has an actual meaning in this context:

Application of $A$ corresponds to deleting the left-most column

What I mean by this: $A$ maps $\mathbb{C}^n$ into a lower dimensional subspace and on this subspace $A$ is still nilpotent, hence representable by a Young diagram. This diagram (and subsequently the dimension of the subspace) is what you get when you delete the leftmost column in the original diagram!

So we can see the 'step by step killing off' that nilpotent operators do life in action here, one column at the time. The smallest number $k$ such that $A^k = 0$ is the number of steps $A$ needs to destroy the entire vector space. We see here that it equals the number of columns in the Young diagram, hence the length of the top row or the dimension of the largest Jordan block. Of course in this last formulation it is obvious that this is true. The more interesting information is in what happens at the intermediate steps.

Of course the reason that this works is exactly what Robert Israel writes, but I hope you enjoy this visual version as much as I did, when I first learned it!