A is a nonnegative matrix; the only principal submatrix having spectral radius above 1 is A itself - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T20:31:02Z http://mathoverflow.net/feeds/question/110436 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110436/a-is-a-nonnegative-matrix-the-only-principal-submatrix-having-spectral-radius-ab A is a nonnegative matrix; the only principal submatrix having spectral radius above 1 is A itself Ben Golub 2012-10-23T14:50:56Z 2012-10-23T18:59:55Z <p>Let $\rho(M)$ denote the spectral radius (modulus of the largest eigenvalue) of a square matrix $M$.</p> <p>I am looking for a characterization or anything else interesting about the set of matrices $A$ having the following properties:</p> <p>(i) $A$ is (entrywise) nonnegative;</p> <p>(ii) $A$ is irreducible;</p> <p>(iii) $\rho(A) \geq 1$;</p> <p>(iv) if $B$ is obtained from $A$ by deleting row $i$ and column $i$ (for any $i$), then $\rho(B) &lt; 1$.</p> http://mathoverflow.net/questions/110436/a-is-a-nonnegative-matrix-the-only-principal-submatrix-having-spectral-radius-ab/110459#110459 Answer by Robert Israel for A is a nonnegative matrix; the only principal submatrix having spectral radius above 1 is A itself Robert Israel 2012-10-23T18:59:55Z 2012-10-23T18:59:55Z <p>Let's assume the matrix $A$ has all its row sums equal to $\lambda$, the largest eigenvalue. We can rescale the rows and columns of any other nonnegative irreducible matrix by a similarity transformation $A \to D^{-1} A D$ where $D$ is diagonal with positive diagonal entries (namely the entries of the Perron eigenvector of $A$). This makes all row sums equal to $\lambda$, and preserves the eigenvalues of $A$ and of the matrices obtained by removing the $i$'th row and column.</p> <p>Let $B_k$ be the matrix obtained from $A$ by removing row and column number $k$. The row sums of $B_k$ are $\lambda - a_{ik}$ for $i \ne k$. Thus the largest eigenvalue of $B_k$ is at most $\lambda - \min_{i \ne k} a_{ik}$ and at least $\lambda - \max_{i \ne k} a_{ik}$. So a necessary condition is $$\lambda \ge 1 > \lambda - \min_k \max_{i \ne k} a_{ik}$$ while a sufficient condition is $$\lambda \ge 1 > \lambda - \min_k \min_{i \ne k} a_{ik}$$ </p>