# Bound spectral radius by diagonal elements of diagonally dominant matrix?

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 $$A$$?

• It is not. As a counterexample, take $$A = \pmatrix{2&1\\1&2},$$ which has spectral radius $3$. 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? Jul 15 '21 at 18:32
• None that I can think of. Do you have any reason to expect this to hold in the context of your application? 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 $\rho(A) < 1$. Jul 15 '21 at 20:55
• 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 $D^{1/2}AD^{1/2}$. Jul 15 '21 at 21:09

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 $$i$$.