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When I compute the eigenvectors of a certain matrix, the first of them is composed entirely of a single constant.

What properties of a matrix lead to this result?

Update

By "a vector composed entirely of a constant" I mean n repetitions of a constant comprising a length-n vector.

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What do you mean by a vector being "composed entirely of a single constant"? If possible, please provide a concrete example. –  Henning Makholm Oct 8 '11 at 0:10
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Each row of the matrix has the same sum. –  André Nicolas Oct 8 '11 at 0:11
    
My guess is it means something like $(3,3,3,3,3,3)$; every component is the same. –  Michael Hardy Oct 8 '11 at 0:22
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Special case of this phenomenon are stochastic matrices, which are of importance in several areas of mathematics. –  Martin Sleziak Oct 8 '11 at 8:29
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2 Answers

up vote 5 down vote accepted

Take an $n \times n$ matrix $A$, and suppose that $v$ is an eigenvector of $A$, with all entries of $v$ equal to a constant $k$. Naturally, $k \ne 0$. Let $\lambda$ be the eigenvalue of $A$ that has $v$ as an eigenvector. If $(b_1, b_2, \dots, b_n)$ is any row of $A$, then by the definition of eigenvalue and eigenvector, we have $$kb_1+kb_2+\cdots +kb_n=\lambda k,$$ from which we conclude that $b_1+b_2+\cdots+b_n=\lambda$. It follows that each row sum of the matrix is equal to $\lambda$.

Conversely, suppose that all row sums of $A$ are equal to $\sigma$. Let $v$ be the vector with all entries equal to $1$. Then $Av$ is a vector with all entries equal to $\sigma$, which means that $v$ is an eigenvector of $A$ with eigenvalue $\sigma$.

Thus $A$ has an eigenvector with all entries equal if and only if all row sums of $A$ are equal.

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Exercise: a matrix has an eigenvector with all its entries equal if and only if all rows of the matrix have the same sum.

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