# Error in Kac & Ulam re: diagonalizable matrices?

In Mathematics and Logic: Retrospect and Prospects (Kac & Ulam, 1969) I came across the following curious claim on p. 79:

One can show that in an $n$-dimensional vector space there are $n$ linearly independent vectors $E_1, E_2, \dots , E_n$ of this kind. They are called eigenvectors of the transformation $T$ and the corresponding scalars $\lambda_1, \lambda_2, \dots , \lambda_n$ are called the eigenvalues.

The claim seems to be that every matrix can be diagonalized. (More context doesn't help.) This is clearly false -- so what is going on here? Am I misunderstanding the claim, or is this just an error?

• This is just an error. – Qiaochu Yuan Sep 2 '16 at 19:46
• it is some attempt at popularization; note that they throw in homology groups. One would hope for something in an afterword saying, Oh, by the way, there is something called Jordan Normal Form, we talked about only the easiest case but other things can happen. – Will Jagy Sep 2 '16 at 19:56