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Assume I have 2 matrices, each of size nxn with only 1 and 0 as entries in both. (n>10) The first matrix (call it A) has each row summing up to 2 (ie: on each row, it has two "1" and n-2 "0"). It is easy to see that 2 is the largest eigenvalue corresponding to the right eigenvector of all-ones The second matrix (call it B) has each row summing up to 3 (ie: on each row, it has three "1" and n-3 "0"). It is also easy to see that 3 is the largest eigenvalue corresponding to the right eigenvector of all-ones

Now , if I have a matrix similar to matrix A, but with only two rows summing up to 3 (each one) rather than 2 , is it always true that the largest eigenvalue of this matrix is between 2 and 3 (inclusive) ? How can I prove it ?

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A standard result from the theory of non-negative matrices states that if $M$, $N$, and $P$ are square non-negative matrices such that $0\leq M\leq N\leq P$ (the inequalities are meant component-wise), then their spectral radii satisfy $$\tag{☆} 0\leq\rho(M)\leq\rho(N)\leq\rho(P). $$ Now take $N$ to be your matrix with the row sums 2 or 3, $M$ to be same as $N$ with one entry deleted from each row with row sum 3, and $P$ to be same as $N$ with an entry 1 added to each row with row sum 2. Hence row sums in $M$ and $N$ are respectively all $2$ and $3$, we have $0\leq M\leq N\leq P$, and (☆) holds.

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  • $\begingroup$ For the result, see, e.g., Theorem 2.21 here. $\endgroup$ – Algebraic Pavel Apr 16 '14 at 2:19

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