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Let $A$ be an $n\times n$ real matrix with entries in a fixed interval $[a_\min,a_\max]$, with $a_\min$, $a_\max>0$.

Question: Are there any upper bounds on the condition number of the eigenvector matrix associated to $A$ as $n$ increases?

The motivation of this question comes from the $n\times n$ random dense case. Indeed, with reference to this case, in [1, Page 338] it is stated that: "[...] experiments indicate that the condition number of the associated eigenvector matrix also grows linearly [as $n$ increases]".

Hence, I wonder whether any similar result can be/has been proved for the deterministic (positive) dense case.

[1] L. N. Trefethen and M. Embree. Spectra and pseudospectra: the behavior of nonnormal matrices and operators. Princeton University Press, 2005.

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  • $\begingroup$ For Hermitian matrices, is it equal to 1? $\endgroup$ Commented Apr 13, 2021 at 22:01

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No, there are no such bounds.

Let $a>0$ be a fixed real and $\varepsilon\ll a$. Take the non-diagonalizable matrix $A=\begin{bmatrix}3a & 0 & 0\\ 0 & 0 & \varepsilon \\ 0 & 0 & 0 \end{bmatrix}$. Let $V$ be a symmetric, orthogonal matrix that maps $\begin{bmatrix}\sqrt {3}\\0\\0\end{bmatrix}$ to $\begin{bmatrix}1\\1\\1\end{bmatrix}$ (for instance, a Householder reflector).

Now $M=VAV^{-1}$ has entries in $[a-O(\varepsilon),a+O(\varepsilon)]$ and is not diagonalizable. So the eigenvector condition number of any sequence of matrices $M_k$ that converges to $M$ blows up.

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  • $\begingroup$ I think you need a orthogonal V, at least the first row of V^-1 should be (1,1,1). Furthermore as there is no orthogonal matrix mapping vectors of different lengths you probably mean a scaled version and then multiply the result. $\endgroup$
    – user35593
    Commented Jan 28, 2017 at 1:27
  • $\begingroup$ @user35593 You are correct; one needs to be more careful than I was with the choice of $V$. Fortunately, Householder reflectors (which are the standard tool used to construct orthogonal decompositions) work. I have fixed my answer now. $\endgroup$ Commented Jan 28, 2017 at 7:06
  • $\begingroup$ @user35593 Thanks a lot for the fixes! $\endgroup$ Commented Jan 28, 2017 at 21:15

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