# Lower and upper bound for the largest eigenvalue

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:
$$\max_{x \in S}\min_{\substack{i=1, \ldots,n\\ x_i \neq 0}} \frac{(Ax)_i}{x_i} = r = \min_{x \in S}\max_{\substack{i=1, \ldots,n\\ x_i \neq 0}} \frac{(Ax)_i}{x_i},$$ where $S:= \{x \in \mathbb{R}^n\setminus\{0\}: x_i \geq 0, \forall i=1,\ldots,n\}$. See here page 667/668 for a reference. 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 $$\max_{x \in S}\min_{\substack{i=1, \ldots,n\\ x_i \neq 0}} \frac{(A^Tx)_i}{x_i} = r = \min_{x \in S}\max_{\substack{i=1, \ldots,n\\ x_i \neq 0}} \frac{(A^Tx)_i}{x_i}.$$ This clearly implies that for every $y \in S$ we have $$\min_{\substack{i=1, \ldots,n\\ y_i \neq 0}} \frac{(A^Ty)_i}{y_i} \leq r \leq \max_{\substack{i=1, \ldots,n\\ y_i \neq 0}} \frac{(A^Ty)_i}{y_i}.$$ 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.
• 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. – Daniel Aug 27 '14 at 1:25