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Why is that if every row of a matrix sums to $1$ then the rows of its inverse matrix sum to $1$ too?

For example, consider

$$A=\begin{pmatrix} 1/3 & 2/3 \\ 3/4 & 1/4 \end{pmatrix}$$

then its inverse is

$$A^{-1}=\begin{pmatrix} -3/5 & 8/5 \\ 9/5 & -4/5 \end{pmatrix},$$

which satisfies the condition. Is it true for every such matrix?

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1 Answer 1

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Let $v = (1, 1, \ldots , 1)'$ be a column vector of all $1$s. Then the rows of $A$ adding to $1$ is equivalent to saying $Av = v$. So when $A$ is invertible, we will have $$A^{-1}v = A^{-1}Av = v$$ Thus $A^{-1}$ has rows summing to $1$ as well. (Note that $A$ will not always be invertible.)

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    $\begingroup$ To extend the assertion we can say that property of row sum equal 1 holds also for any matrix $A^k$ ($A$ invertible) when $k$ is an integer $(...,-2,-1,0,1,2, ..)$ $\endgroup$
    – Widawensen
    Jan 25, 2018 at 10:33
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    $\begingroup$ The name of such a matrix is a right stochastic matrix (en.wikipedia.org/wiki/Stochastic_matrix), and it has a few more interesting properties. $\endgroup$ Jan 25, 2018 at 16:40
  • $\begingroup$ @VictorMay The condition here is more general than that of stochastic. These entries don't need to be non-negative. $\endgroup$ Jan 25, 2018 at 17:25
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    $\begingroup$ @VictorMay This is probably one of those cases where different definitions can be found in different sources, e.g.: en.wiktionary.org/wiki/right_stochastic_matrix . Some authors require it, some don't. I suspect that most do (since it is frequently defined in the context of Markov chains where the elements are probabilities) but haven't done any systematic survey. I was mistaken in assuming that the definition that I was familiar with was universal. $\endgroup$ Jan 25, 2018 at 23:27
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    $\begingroup$ @David That meshes with my impression. Perhaps I was trying too hard to be diplomatic. I did find a couple of definitions online which didn't explicitly contain the phrase "nonnegative" (which caused me to think that Victor's use is not unprecedented) but I suspect now that those cases were just somewhat sloppy and assumed nonnegative as being clear from context. Nonnegative is both natural and at least nearly universal in the definition. $\endgroup$ Jan 30, 2018 at 20:45

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