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Let $A$ be a $n$-by-$n$ left stochastic matrix.

The followings are the properties I found so far:

Algebraic Property

$A$ has left eigenvector $[1,1,\cdots,1]$ ($n$ 1s) with corresponding eigenvalue $1$.

Geometric Property

Let $S$ be a $(n-1)$-simplex whose vertices are at $[1,0,\cdots,0],[0,1,\cdots,0],\cdots,[0,0,\cdots,1]$, border inclusive.

Then $A$ maps $S$ within $S$. (Linearly, of course)

Question

By the geometric property, It seems intuitively clear that:

  • $A$ has a right eigenvector within $S$ with corresponding eigenvalue $1$.

  • Along with the vector above, $n-1$ vectors from the (hyper)plane that passes through the origin and is orthogonal to $[1,1,\cdots,1]^T$ complete a basis of the eigenspace of $A$.

Is this true?

Dannyu NDos
  • 2,161

1 Answers1

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The answer to your first question is yes. In general, for any square matrix $A$ (not just the stochastic ones), $A$ has a left eigenvector corresponding to an eigenvalue $\lambda$ if and only if it has a right eigenvector corresponding to the same eigenvalue. In fact, the two eigenspaces have the same dimension, because $\lambda I-A$ has the same nullity as $\lambda I-A^\top$.

The answer to your second question is negative. See Gerry Myerson's answer to another question for a counterexample.

user1551
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