let $A,B\in M_n$, suppose
- $A>0$ (i.e, all $a_{ij}>0$)
- $\rho (A) = \max \{ \left| \lambda \right|:\lambda $ is eigenvalue of $A$ $\}$
- $B>0$
- $A-B>0$
Why does $\rho (A) - \rho (B) > 0$?
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Bart Michels
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We could also prove this using some results from Horn and Johnson that are weaker than the PF theorem. – Ben Grossmann Dec 25 '15 at 05:28
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EDITED: Let $A(t) = B + t (A-B)$ interpolate linearly between $B$ at $t=0$ and $A$ at $t=1$. Thus $A(t)$ is a positive matrix and so is its derivative $A'(t)$. By Perron-Frobenius, $\rho(A(t)) = \lambda(t)$ is a simple eigenvalue, with left and right eigenvectors $v^T(t)$, $u(t)$ having positive entries; we can normalize them so that $v(t)^T u(t) = 1$. Moreover, $\lambda(t)$, $v^T(t)$, $u(t)$ are differentiable (in fact real-analytic) as functions of $t$. Differentiating $v(t)^T u(t) = 1$ we get $v(t)^T u'(t) + v'(t)^T u(t)= 0$. Now differentiating the equation $\lambda(t) = v(t)^T A(t) u(t)$ we get $$ \eqalign{\lambda'(t) &= v'(t)^T A(t) u(t) + v(t)^T A'(t) u(t) + v(t)^T A(t) u'(t)\cr &= \lambda(v'(t)^T u(t) + v(t)^T u'(t)) + v(t)^T A'(t) u(t)\cr &= v(t)^T A'(t) u(t) > 0}$$
Robert Israel
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1I have a question: If the derivative of the $\lambda (t)$ is positive wouldn't this imply that $\lambda (1) > \lambda ( 0)$, i.e. $\rho(B) > \rho(A)$ – math635 Dec 24 '15 at 18:29
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Why is it that we're able to normalize $v(t)^Tu(t)=1$ here? – Shivashriganesh Mahato Dec 10 '21 at 18:40
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@ShivashriganeshMahato Because a nonzero multiple of an eigenvector is an eigenvector. – Robert Israel Dec 10 '21 at 19:44
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Hints : Let $x$ be the Perron vector of $A$ , so that $Ax>Bx$ and ......
Nokia
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Although the Question may require a bit of parsing to make sense of it, I think this "hint" does not go into as much detail as it should to clarity the situation. – hardmath Dec 24 '15 at 19:18
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