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Let $A\in\mathbb{R}^{n\times n}$ be an $n\times n$ symmetric, invertible matrix with nonnegative real entries, $\mathbf{1}$ be the all one $n$-dimensional vector, and $\mathrm{diag}(v)$, $v=[v_1,v_2,\dots,v_n]^\top\in\mathbb{R}^n$ be the diagonal matrix with diagonal entries $v_1,v_2,\dots,v_n$. Consider the following transformation of $A$ $$ g(A,t) := \mathrm{diag}((A+tI)\mathbf{1})^{-1}(A+tI), $$ where $t$ is a nonnegative real number. Notice that $g(A,t)$ is a (row) stochastic matrix.

My question. Assume that $g(A,0)$ has $k$ distinct eigenvalues $\lambda_1,\dots,\lambda_k$. Then, does $g(A,t)$ have $k$ distinct eigenvalues for all (finite) values of $t> 0$?

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    $\begingroup$ 1) Did you forget to take the inverse when defining $\Sigma(t)$? 2) How do you enumerate the eigenvalues? What is $\lambda_1$? Do you merely ask whether teo eigenvalues may become equal? $\endgroup$
    – Ilya Bogdanov
    Commented Nov 12, 2019 at 20:58
  • $\begingroup$ @IlyaBogdanov: yes thanks! I fixed the typo. I enumerate the eigenvalues in an increasing order: $\lambda_1$ is the smallest eigenvalue of $g(A,t)$. Basically, I wonder whether there are no crossings between the eigenvalues. $\endgroup$
    – Ludwig
    Commented Nov 12, 2019 at 21:42
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    $\begingroup$ I don't think $g(A,t)$ is still symmetric. $\endgroup$
    – Christian Remling
    Commented Nov 12, 2019 at 23:52
  • $\begingroup$ @ChristianRemling yes it is not symmetric (there was a typo). However $g(A,t)$ is similar to a symmetric matrix. $\endgroup$
    – Ludwig
    Commented Nov 13, 2019 at 6:32

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