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A proof of Perron-Frobenius theorem is given in "Introduction to Linear Algebra" by Gilbert Strang as:

strang

This is asked before, check Perron-Frobenius theorem proof, but the explanation given by @Kavi Rama Murthy only addresses part of the proof which is:

If $y>z$ then $y_j>z_j$ for all $j$, therefore $\sum_{j=1}^{n} a_{ij} (y_j-z_j)>0\implies A(y-z)>0$ since all elements of $A$ is positive.

ie., if $Ax\geq t_{\max}x$ is not an equality then $Ax>t_{\max}x\implies A^2x>t_{\max}Ax$

ie., $Ay>t_{\max}y$ for some positive vector $y$.

Similarly we can also say that for positive matrices, $A(y-z)>0\implies \sum_{j=1}^{n} a_{ij} (y_j-z_j)>0\implies y=z$

I think this much is clear. How can this tell me $t_{\max}$ can be increased ?

Thanks @user8675309, due to the reasoning above when $Ax\geq t_{\max}x$ is not an equality we get the strict inequality $Ay>t_{\max}y$ for a positive matrix $A$. ie., we can increase $t_{\max}$ by a small $\delta t$ till atleast one of the corresponding elements of $Ax$ and $t_{\max}x$ are equal. This leads to the conclusion that $Ax=t_{\max}x$.

But where does the assumption $x\neq 0$ being a non-negative vector affect in this proof ?

And how can one make sense of the rest of the proof ?

SOORAJ SOMAN
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    if $A\mathbf y \gt t_{\max}\cdot \mathbf y$ (component-wise), then $A\mathbf y \gt (1+\delta) \cdot t_{\max}\cdot \mathbf y$ for small enough $\delta\gt 0$. This is in every way analogous to the scalar case of $\gamma \gt t_{\max}\implies \gamma \gt (1+\delta) \cdot t_{\max}$ for small enough $\delta\gt 0$ – user8675309 Feb 05 '21 at 03:02
  • @user8675309 Okay but where will the assumtion of $x$ being a nonnegative vector affect ? – SOORAJ SOMAN Feb 05 '21 at 04:13

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