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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 $A$?

Rodrigo de Azevedo
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Andres
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    It is not. As a counterexample, take $$ A = \pmatrix{2&1\1&2}, $$ which has spectral radius $3$. – Ben Grossmann Jul 15 '21 at 18:06
  • @BenGrossmann Thanks. Are there non-trivial conditions on A such that which one can bound its spectral radius by the diagonal elements? – Andres Jul 15 '21 at 18:32
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    None that I can think of. Do you have any reason to expect this to hold in the context of your application? – Ben Grossmann Jul 15 '21 at 18:36
  • @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 $\rho(A) < 1$. – Andres Jul 15 '21 at 20:55
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    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 $D^{1/2}AD^{1/2}$. – Ben Grossmann Jul 15 '21 at 21:09
  • @BenGrossmann I have posted the problem here: https://math.stackexchange.com/q/4186355/165163 In this special case considered here, the matrix $D$ is just the identity matrix. The hope is that if I can show that the elements of the diagonal of matrix $J(v,x)$ are all less than one then somehow I could then conclude (with some additional work) that the spectral radius is also less than one. – Andres Jul 15 '21 at 23:40

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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 $i$.

user1551
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