We will call a matrix positive matrix if all elements in the matrix are positive, and we will denote the largest eigenvalue with $\lambda_{\max}$, what is exist because of the Perron–Frobenius theorem.

Theorem. Let $A$ be a positive square matrix. Then the minimal row sum is a lower bound and the maximal row sum is an upper bound of $\lambda_{\max}$.

My questions.

  1. Is there a name for this theorem and can anybody say books or papers what refer to it?

  2. How to prove it?


Let us denote by $r$ the Perron-root of the positive matrix $A \in \mathbb{R}^{n \times n}$. Then by the Collatz-Wielandt formula we have:

$$\max_{x \in S}\min_{\substack{i=1, \ldots,n}} \frac{(Ax)_i}{x_i} = r = \min_{x \in S}\max_{\substack{i=1, \ldots,n}} \frac{(Ax)_i}{x_i}, $$ where $S := \{x \in \mathbb{R}^n\setminus\{0\}: x_i > 0, \forall i=1,\ldots,n\}$. Now it is clear that $A$ and $A^T$ have same eigenvalues, since $\det(M)=\det(M^T)$ and for every $\lambda \in \mathbb{R}$ we have $$\det(A-\lambda I)=\det((A-\lambda I)^T)= \det(A^T-\lambda I).$$ Furthermore $A$ strictly positive implies $A^T$ strictly positive, thus this formula also holds for $A^T$. It follows that we have $$\max_{x \in S}\min_{\substack{i=1, \ldots,n}} \frac{(A^Tx)_i}{x_i} = r = \min_{x \in S}\max_{\substack{i=1, \ldots,n}} \frac{(A^Tx)_i}{x_i}.$$ This clearly implies that for every $y \in S$ we have $$\min_{\substack{i=1, \ldots,n}} \frac{(A^Ty)_i}{y_i} \leq r \leq \max_{\substack{i=1, \ldots,n}} \frac{(A^Ty)_i}{y_i}.$$ Choose $y = (1,1,\ldots,1)$ to get your bounds. Note also that using the same trick on $A$ directly you will get the same upper/lower bound but with the columns instead of the rows.

For reference, I recommend (in increasing order of technicality/generality):

  1. "Matrix analysis" by Horn and Johnson
  2. "Nonnegative Matrices in the Mathematical Sciences" by Bermann and Plemmons
  3. "Nonlinear Perron-Frobenius theory" by Nussbaum and Lemmens

There are even more general versions discussed in recent literature, e.g. in this paper

  • $\begingroup$ Consider $A^T=\begin{bmatrix} 3 & 1 & 1\\ 4 & 1 & 1 \\ 5 & 1 & 1 \end{bmatrix}$ and $y=(1, 0, 0)$. It can be checked that $r$ is greater than $5$, but $(A^Ty)_i/y_i=3$. The upper bound in the last inequality does not hold. $\endgroup$
    – kswim
    Oct 27 '21 at 6:51
  • $\begingroup$ Please see item 7 of "positive matrices" in Section "Statements" in en.wikipedia.org/wiki/Perron–Frobenius_theorem $\endgroup$
    – kswim
    Oct 27 '21 at 6:53
  • $\begingroup$ In the reference you mentioned (I think you would refer to "Matrix analysis and applied linear algebra" by Carl D. Meyer), as far as I could determine, the upper bounds are not stated. $\endgroup$
    – kswim
    Oct 27 '21 at 6:56
  • $\begingroup$ Yes. But, in your answer, $S$ contains non-negative vectors. $\endgroup$
    – kswim
    Oct 27 '21 at 7:16
  • 1
    $\begingroup$ I found out that this led another questions about whether the formula is correct or not. So, I wanted to make things correct. Thanks for the correction! $\endgroup$
    – kswim
    Oct 27 '21 at 7:34

See propositions 2.3 and 2.4 of the book "Banach Lattices and Positive operators" of H.H. Schaefer.

  • $\begingroup$ Can you maybe offer a good ebook version of it, or send me an image about the pages what you talking about? It could really help me. $\endgroup$
    – user153012
    Aug 26 '14 at 21:05
  • 1
    $\begingroup$ Or just write it here. $\endgroup$
    – user153012
    Aug 26 '14 at 21:21
  • 1
    $\begingroup$ See this link. books.google.com.br/…. I could not find a better look inside this book. I forgot to say that the page is 7. The number $r(A)$ is the spectral radius of $A$, which is the biggest positive eigenvalue for this type of matrix. $\endgroup$
    – Daniel
    Aug 27 '14 at 1:25

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