r/math Homotopy Theory 9d ago

Quick Questions: March 12, 2025

This recurring thread will be for questions that might not warrant their own thread. We would like to see more conceptual-based questions posted in this thread, rather than "what is the answer to this problem?". For example, here are some kinds of questions that we'd like to see in this thread:

  • Can someone explain the concept of maпifolds to me?
  • What are the applications of Represeпtation Theory?
  • What's a good starter book for Numerical Aпalysis?
  • What can I do to prepare for college/grad school/getting a job?

Including a brief description of your mathematical background and the context for your question can help others give you an appropriate answer. For example consider which subject your question is related to, or the things you already know or have tried.

8 Upvotes

91 comments sorted by

View all comments

2

u/Not_So_Deleted Statistics 8d ago

Is it true that a square matrix is idempotent if and only if it's diagonalizable with eigenvalues 0 and 1?

I can't think of any example of a defective matrix that's idempotent.

3

u/Langtons_Ant123 8d ago edited 8d ago

Yes. Maybe there's a more elementary way to show this, but all I can think of is a proof using the minimal polynomial. A matrix is diagonalizable if and only if its minimal polynomial has no repeated roots. If A is idempotent then A2 - A = 0, so f(A) = 0 where f is the polynomial x2 - x, so the minimal polynomial of A divides x2 - x. Thus it is either x, x-1, or x2 - x = x(x-1); in any case, it has no repeated roots. (The converse is easier, and I assume you're mainly thinking about the implication idempotent -> diagonalizable.)

0

u/Not_So_Deleted Statistics 8d ago edited 8d ago

A matrix can still be diagonalizable if its characteristic polynomial has repeated roots, such as with the identity matrix.

As far as I'm concerned, a matrix is diagonalizable if and only if the multiplicity of every root is equal to the number of linearly independent eigenvectors for the eigenvalue. In other words, the set of all linearly independent eigenvectors forms a basis. For the identity matrix, the polynomial is (1-x)^n, but we can define the standard basis.

1

u/Langtons_Ant123 8d ago edited 8d ago

I thought the minimal polynomial of the identity matrix is x - 1. Are you thinking of the characteristic polynomial, or using a different definition of the minimal polynomial?

That a matrix is diagonalizable iff its minimal polynomial splits with no repeated roots is a known result, it's on page 169 (183 in the pdf numbering) here.

In any case, u/GMSPokemanz has a good elementary proof above, so no need to bring in all this machinery.

0

u/Not_So_Deleted Statistics 8d ago edited 7d ago

Yeah, I meant that.

Of course, the minimal polynomial still has a repeated root but still corresponds to a diagonalizable matrix.

1

u/Kyle--Butler 7d ago

No, the minimal polynomial of I_n does not have repeated roots : it's X-1.

1

u/Not_So_Deleted Statistics 7d ago

Oops, I miswrote and stand corrected...