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If $A$ is a positive square matrix, then the Collatz–Wielandt implies that

$\min_{𝑖=1,…,𝑛;𝑦_𝑖\neq 0}\frac{(Ay)_i}{y_i}≤𝑟≤\max_{𝑖=1,…,𝑛;𝑦_𝑖\neq 0}\frac{(Ay)_i}{y_i}$, Where $r$ is the largest eigenvalue of $A$.

By replacing $y=e_j$ in the previous expression wouldn't we obtain that $a_{jj}\leq r \leq a_{jj}$ for each $j\in\{1,\ldots,n\}$? This cannot be true for a matrix $A$ that has different arguments in the diagonal.

The previous inequality is from @Surb’s answer in here: Lower and upper bound for the largest eigenvalue.

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2 Answers 2

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Collatz-Wielandt theorm only states the lower bound. In fact, a counter example can be given by taking a matrix with large off diagonal entries.

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No. In $e_j$, $j$ is not related to $i$. You would have $$ \min_i a_{ji} \leq r \leq \max_i a_{ji} $$ (... or possibly "$a_{ij}$" in both places, since I don't know whether you are using row first or column first indexing for matrices). You are pulling the minimum and maximum entries from the $j^\text{th}$ column.

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  • $\begingroup$ Thanks! However, I thought that one only takes the min and max over the arguments for which $y_i$ is positive. Am I understanding that wrong? $\endgroup$
    – Condor5
    Commented Apr 18, 2019 at 2:00
  • $\begingroup$ @user71031 : You are understanding that correctly, but $e_j$ just selects the entire $j^\text{th}$ column of $A$. Then the index $i$ selects a row (from a column vector and/or from a column in a matrix), so is selecting from all the elements in the $j^\text{th}$ column of $A$ regardless of whether they are zero or not. $\endgroup$ Commented Apr 18, 2019 at 2:05
  • $\begingroup$ I think you are mistaken. The inequality should be $\min_{i,i = j} a_{ji} \leq r \leq \max_{i,i = j} a_{ji} $. The reason is that when $y_j=0$ the term would be $a_{ij}/y_j$ which is infinity. Therefore, it is omitted from the min and the max. I'm thinking that the right hand side of the inequality in my original post is not true. Other authors write the maximum over $y>0$ not $y\geq 0$. A post that interprets the formula the same way I do is the following: math.stackexchange.com/questions/872398/…. $\endgroup$
    – Condor5
    Commented Apr 24, 2019 at 21:51

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