# Spectral radius of a real, symmetric, positive semi - definite matrix.

While answering a question, the OP made 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!! – Stephen Montgomery-Smith Jul 20 '15 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. Jul 20 '15 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. – Stephen Montgomery-Smith Jul 20 '15 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: $\rho(A) \gt 1.$ – thanasissdr Jul 20 '15 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. Jul 20 '15 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 $\rho(A)=\|A\|_2\ge\|Ae_j\|_2\ge|a_{ij}|>1$.