Spectral radius of a real symmetric positive semidefinite matrix

While answering the question Trace of a real, symmetric positive semi-definite matrix, the OP asked a follow-up question, that I was not able to answer at that moment. However, I came up with an intriguing (at least to me) question.

Let $$\mathcal{S_n}(\mathbb R)$$ be the class of all the real $$n\times n$$, symmetric, positive semi - definite matrices, with one entry strictly greater than $$1$$. Then, is it true that for all $$A\in \mathcal{S_n}(\mathbb R)$$ holds $$\rho(A) \ge 1?$$

• I know it is true. Because for real symmetric matrices, the spectral radius is equal to its spectral norm. And your condition clearly implies that the spectral norm is greater than or equal to 1. But I cannot think of an elementary argument!! Commented Jul 20, 2015 at 0:57
• It is true. Elementary proof: let $a_{ij}>1$. If $i=j$ then it is obvious that $e_i^TAe_i>1$, so the spectral radius $>1$. Otherwise, it is necessary for semi-definiteness that $a_{ii}a_{jj}-a_{ij}^2\ge 0$, so either of two $a_{ii}$ or $a_{jj}$ must be $>1$ and we are again in the first "if".
– A.Γ.
Commented Jul 20, 2015 at 1:05
• @A.G. Smart! But what the OP is asking is also true if the matrix is merely symmetric. Positive semi-definiteness isn't required. Commented Jul 20, 2015 at 1:14
• @A.G. Neat proof! I assume that for the first case you used the inequality $\rho(A) \ge \dfrac {x^TA x}{x^Tx}$ and as far I can understand in the case I described it holds: $\rho(A) \gt 1.$ Commented Jul 20, 2015 at 1:34
• @StephenMontgomery-Smith You are right, definiteness is not necessary. If $|a_{ij}|>1$ then $\|A\|\ge \|Ae_j\|_2=\sqrt{\text{some squares}\,+a_{ij}^2}>1$.
– A.Γ.
Commented Jul 20, 2015 at 1:36

Symmetric positive definite matrices $\subset$ Hermitian matrices $\subset$ normal matrices $\subset$ radial matrices (the set of all matrices $A\in M_n(\mathbb C)$ such that $\rho(A)=\|A\|_2$).
So, in your question, $\rho(A)=\|A\|_2$ and some $|a_{ij}|>1$. Hence $\rho(A)=\|A\|_2\ge\|Ae_j\|_2\ge|a_{ij}|>1$.