0
$\begingroup$

I know that for every permutation matrix, there is an integer number $k$ such that $P^k = P$ (I am curious only about $k > 2$ case obviously as for 1 every matrix passes the test and $k=2$ is idempotent matrix). I believe that permutation matrix is not the only matrix with such property and after some random attempts to find such matrices I found one

$$\begin{pmatrix} 0 & 0 & -i\\ i & 0 & 0\\ 0 & 1 & 0 \end{pmatrix}$$ which is true for $k = 3$.

I have not yet succeed to find real/integer valued matrices with such property. Do they exist? Is there any name for these matrices and do they have any interesting properties (except of obvious ones with determinants and eigenvalues)?

$\endgroup$
2
  • 1
    $\begingroup$ For $k=2$, they are called idempotent matrices. There are examples of idempotent matrices over any field and of any size. See here: en.wikipedia.org/wiki/Idempotent_matrix $\endgroup$
    – Guy
    Oct 20 '16 at 9:05
  • $\begingroup$ @Guy sorry, I actually want to write $k > 2$. Will change it $\endgroup$ Oct 20 '16 at 9:09
1
$\begingroup$

Rotation matrices also have this property. For instance, we have $$ \begin{bmatrix}0&-1\\1&0\end{bmatrix}^5=\begin{bmatrix}0&-1\\1&0\end{bmatrix} $$

$\endgroup$
1
$\begingroup$

Whenever a matrix $A$ is such that $A^{k-1}=I$, the identity matrix, then of course, $A^k=A$ and that gives you an example.

Matrices such that that a power of them is the identity are called matrices of finite order. There are many examples. Permutation matrices, for example (and these have integer entries) More generally, there are known conditions that allow us to see what orders can integral matrices of a fixed size have.

$\endgroup$
0
$\begingroup$

Involutory matrices are their own inverse, so any odd power of one will be the matrix itself. So if $A\in\Bbb R^{n\times n} $ satisfies $A=A^{-1}$ then for $k\in\Bbb N$:$$A^{2k+1}=A^{2k}A=(A^2)^kA=I^kA=A $$

$\endgroup$
0
$\begingroup$

The projection matrix in statistics is a very important example with this property. For example, if $X$ is a $n \times p$ matrix of real numbers, then $$ X(X^{T}X)^{-1}X^{T} $$ satisfies the desired property, which is exploited often in the theory of linear models.

$\endgroup$

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.