0
$\begingroup$

The question is:
If A is a square matrix such that $A^2=A$ then $A^n =A$ for all natural numbers $n$ greater than one. What is $A$ if $A \ne 0$ and $A \ne I$.

I figured out an answer but I can't tell if that's the only answer. Let's say that $a_{kk}$ is a value in $A$. Every value in $A$ is $0$ except for $a_{kk}$ and $a_{00}$, they're $1$. I haven't gotten this answer mathematically.

I've tried a few approaches but ended up with the identity matrix.

$\endgroup$
  • 4
    $\begingroup$ Hint: If $A^2=A$, then what follows about the eigenvalues? $\endgroup$ – celtschk Nov 12 '16 at 21:30
  • $\begingroup$ It would be very good if you could prove that is necessary (sufficiency should be trivial). $\endgroup$ – Jacob Wakem Nov 12 '16 at 21:34
  • $\begingroup$ I have no idea, haven't learned about eigenvalues yet. @celtschk $\endgroup$ – E.Bob Nov 12 '16 at 21:35
  • $\begingroup$ This question (find idempotent matrices) is treated in (en.wikipedia.org/wiki/Idempotent_matrix) $\endgroup$ – Jean Marie Nov 12 '16 at 21:37
  • $\begingroup$ There are maaany such matrices and there is no simple way of describing their entries (in particular, their entries can be different from 0 and 1) $\endgroup$ – Mariano Suárez-Álvarez Nov 12 '16 at 21:38
2
$\begingroup$

Perhaps one more example, besides $A=0$ and $A=I$ may be insightful: take the block matrix $$ A=\begin{pmatrix} I & 0 \cr 0 & 0 \end{pmatrix} $$ Of course, $A=A^2$, but $A\neq 0,I$. The (square) blocks can be of any size, so we obtain several examples. Up to similarity, these are the only ones, too. See "canonical forms" in the wikipedia article.

$\endgroup$
  • $\begingroup$ This is not very different from the example the OP has found by him/her self. $\endgroup$ – Jean Marie Nov 12 '16 at 21:41
  • $\begingroup$ @JeanMarie Yes, indeed. But since it said" I haven't gotten this answer mathematically", I wanted to give a clarification. In addition I pointed out that all projection matrices are similar to these. $\endgroup$ – Dietrich Burde Nov 12 '16 at 21:44
  • $\begingroup$ You are right, because an idempotent matrix is diagonalizable, and that its eigenvalues are $0$ or $1$. $\endgroup$ – Jean Marie Nov 12 '16 at 21:52
1
$\begingroup$

Such a matrix is called idempotent. Here are some examples and properties.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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