7

A regular matrix $A$ is described as a square matrix that for all positive integer $n$, is such that $A^n$ has positive entries.

How then would I prove something is regular? I mean I can prove something is irregular if $A^2$ has some 0 or negative entries; but I cant prove regularity since I cant solve $A^n$ for all integers $n$.

My thoughts are that if a matrix $A$ is diagonalisable as $A=PD^{-1}P$ then it is 'regular,' since then all $A^k$ exist; but does this also imply all entries of $A^k$ are positive?

Any hints?

Rodrigo de Azevedo
  • 23,223
vora
  • 71
  • 6
    The definition is not right. A regular matrix is a matrix for which some power of the matrix has all positive entries. – Christopher A. Wong Oct 18 '12 at 02:04
  • 1
    @ChristopherA.Wong Definitions can't be right or wrong. –  Feb 27 '13 at 03:07
  • 3
    yes but they can be consistent! According to wikipedia there is some ambiguity about "regular matrix", the top choice is the one given by @Christopher A. Wong. Though it probably would be helpful to specify that it is a "regular stochastic matrix". – Paul Mar 05 '17 at 21:27
  • @user53153 Definitions are useful if many people agree on them. – a06e Oct 10 '17 at 16:03

2 Answers2

9

If $A$ has an entry that is $0$ or negative, then $A$ is not regular. If, on the other hand, every entry in $A$ is positive, can $A^2$ have a negative or zero entry? Can $A^3$? There’s an easy proof by induction waiting here for you to find it. Note that diagonalizability has nothing to do with the matter: if $A$ is square, $A^n$ exists for all $n\ge 0$ whether or not $A$ is diagonalizable. Diagonalizability of $A$ merely makes it easy to calculate the powers of $A$.

However, that’s not the usual definition of regular matrix. The usual definition is that a square matrix $A$ is regular if it is stochastic and there is some $n\ge 1$ such that all of the entries of $A^n$ are positive.

Brian M. Scott
  • 631,399
  • 2
    The statement that if A has an entry that is 0 or negative implies that it cannot be regular is incorrect. Consider $[ \left( \begin{array}{ccc} .9 & .5 & 0\ 0 & .5 & .4\ .1 & 0 & .6\end{array} \right)]$ for example. The square of this matrix is regular. – AnotherPerson May 10 '15 at 07:40
  • 1
    @SirJective: Read the definition of regular that the OP is using. By that definition your matrix is not regular. – Brian M. Scott May 10 '15 at 07:43
  • 1
    How do you prove that a matrix is regular, according to the usual definition? – a06e Oct 10 '17 at 16:02
  • If the downvoter actually has a reason, I’d very much like to hear it, so that I can address it. – Brian M. Scott Oct 15 '21 at 01:31
1

Linear combinations of positive numbers (with positive coefficients) will always yield positive numbers, yes? Therefore, it should suffice to note that all entries of $A$ are positive. On the other hand, if $A$ has an entry that is nonpositive, then $A$ can't be regular (since $A=A^1$ and $1>0$), so this is a sufficient condition, too.

Cameron Buie
  • 105,149