Consider a diagonally dominant matrix $A$ with all positive entries. Is it true that the spectral radius of $A$ is lower than or equal to the maximum diagonal entry of $A$?
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2$\begingroup$ It is not. As a counterexample, take $$ A = \pmatrix{2&1\\1&2}, $$ which has spectral radius $3$. $\endgroup$– Ben GrossmannJul 15, 2021 at 18:06
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$\begingroup$ @BenGrossmann Thanks. Are there non-trivial conditions on A such that which one can bound its spectral radius by the diagonal elements? $\endgroup$– AndresJul 15, 2021 at 18:32
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1$\begingroup$ None that I can think of. Do you have any reason to expect this to hold in the context of your application? $\endgroup$– Ben GrossmannJul 15, 2021 at 18:36
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$\begingroup$ @BenGrossmann In our application we have a matrix $J=AD$ with $A$ having all nonnegative elements and D being a nonnegative diagonal matrix. Numerically we see that $\rho(J)<1$, but we don't know how to prove that. However, if $D$ has $d_{ii} =0,\forall i\neq k$ then $\rho(AD) = a_{kk} d_{kk}$ and we may be able to show that $ a_{kk} d_{kk} < 1 , \forall k$. So I was hoping that we could then uuse this to construct our proof that $\rho(A) < 1$. $\endgroup$– AndresJul 15, 2021 at 20:55
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1$\begingroup$ Maybe you should post a new question with the details of this problem. I'd be interested in knowing if $A$ is symmetric and how exactly the numbers $d_{ii}$ are chosen. Sometimes, it's helpful to note that $AD$ has the same eigenvalues as $D^{1/2}AD^{1/2}$. $\endgroup$– Ben GrossmannJul 15, 2021 at 21:09
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1 Answer
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No. Quite the contrary, the spectral radius is always greater than the largest diagonal element. Diagonal dominance is irrelevant here. Let $A$ be positive and $x$ be its Perron vector. By considering the $i$-th entry on both sides of $Ax=\rho(A)x$, we obtain $a_{ii}x_i<\sum_ja_{ij}x_j=\rho(A)x_i$. Hence $a_{ii}<\rho(A)$ for each $i$.
