Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $A$ is a diagonalizable $n\times n$ matrix for which the eigenvalues are $0$ and $1$, then $A^2=A$.

I know how to prove this in the opposite direction, however I can't seem to find a way prove this. Could anyone please help?

share|cite|improve this question
This is the spectral characterization of idempotent matrices. Just observe that $A^2$ and $A$ coincide on the eigenspaces $\mbox{Ker} A$ (both $0$) and $\mbox{Ker} (A-I_n)$ (both $I$). – 1015 Apr 20 '13 at 19:22
up vote 9 down vote accepted

Write $A = QDQ^{-1}$, where $D$ is a diagonal matrix with the eigenvalues, $0$s and $1$s, on the diagonal. The $A^2 = QDQ^{-1}QDQ^{-1} = QD^{2}Q^{-1}$. But $D^2 = D$, because when you square a diagonal matrix you square the entries on the diagonal and $1^2 = 1$ and $0^2 = 0$. Thus

$$A^{2} = QD^{2}Q^{-1} = QDQ^{-1} = A$$

share|cite|improve this answer
Ah, I see. I didn't have the notion that in this case $D^2=D$. Thanks a lot :D. – dreamer Apr 20 '13 at 19:19

Another approach, more theoretical but perhaps simpler and shorter:

Since a matrix is diagonalizable iff its minimal polynomial splits as a product of different linear factors, being that only $\,0,1\,$ are the only eigenvalues of the matrix, its minimal polynomial must be $\,x(x-1)=x^2-x\,$ , from where

$$A^2-A=0\implies A^2=A$$

share|cite|improve this answer
Hmm. I think in fairness we could also have $A = 0$ or $A = I$, but both of those are uninteresting. – Ben Millwood Apr 20 '13 at 20:15
The OP wrote "the eigenvalues are $\,0\,$ and $\,1\,$ ". For me this means both values appear so $\,A\neq0,I\,$ – DonAntonio Apr 20 '13 at 21:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.