Inequality in the eigenvalues of $B$

Question

Suppose $B \in M_{n \times n} (\mathbb C)$ is Hermitian. Prove that $$\lambda_{min}\leq b_{kk} \leq \lambda_{max}, \; \text{for}\; 1 \leq k \leq n.$$ With the equality in both sides for some $k$ only if $b_{kj}=b_{jk}=0,$ for all $k\neq j.$ Here $\lambda_{min} \;\& \; \lambda_{max}$ are the minimum and maximum eigenvalues of $B$ and $b_{kj}$ is the $(k,j)$- entry of $B$.

Attempt: Using Rayleigh-Ritz theorem we achieved the inequalities above. How can I show the equality in both sides if $b_{kj}=b_{jk}=0,$ does it follow from Rayleigh-Ritz? Also suppose we take the matrix $\text{diag}(1,2,3)$ does the condition $b_{kj}=b_{jk}=0,$ for all $k\neq j$ imply that $b_{kk}$ is equal to either $\lambda_{min} \;\vee \; \lambda_{max}?$

Any help and hint would be much appreciated.

Answer 1

It follows from Rayleigh Ritz. In particular: part of the Rayleigh Ritz theorem states that if $x$ is a unit vector with $x^*Bx = \lambda_{\max}$, then it must hold that $x$ is an eigenvector of $B$ associated with $\lambda_\max$. Now, note that if $b_{kk} = \lambda_{\max}$, then then the vector $x = e_k$ (the $k$th standard basis vector) satisfies $x^*Bx = \lambda_\max$. Similar logic can be applied in the case that $b_{kk} = \lambda_\min$.