# Estimate a squared sum of the eigenvalues of a Hermitian matrix

Let $A$ be a $5\times 5$ Hermitian matrix: $$A=\begin{pmatrix} 1 & 0 & 0 & 0 & a_1\\ 0 & 2 & 0 & 0 & a_2\\ 0 & 0 & 3 & 0 & a_3\\ 0 & 0 & 0 & 4 & a_4\\ a_1 & a_2 & a_3 & a_4 & 5 \end{pmatrix},\ \ a_i\in\mathbb{R}.$$ Let $\lambda_1\leq \lambda_2\leq \lambda_3\leq \lambda_4\leq\lambda_5$ be the eigenvalues of $A$. Show that $$(\lambda_1-1)^2+(\lambda_2-2)^2+(\lambda_3-3)^2+(\lambda_4-4)^2+(\lambda_5-5)^2\leq 2(a_1^2+a_2^2+a_3^2+a_4^2).$$

This is an question from my homework of matrix theory, I have no idea how to start.

Hint: Let $B = A - C$ where
$$C = \operatorname{diag}[1,2,3,4,5] = \pmatrix{1\\&2\\&&3\\&&&4\\&&&&5}$$ From there, theorem III.2.8 from Bhatia's Matrix Analysis is sufficient.
• I don't understand why we have $\lambda_i(B)\geq \lambda_i(A)-\lambda_i(C)$. Dec 13, 2016 at 16:50
• You should have a statement in your textbook/notes which allows you to state that $\lambda_i(B + C) \leq \lambda_i(B) + \lambda_i(C)$. Did you recently cover the max-min theorem? Dec 13, 2016 at 16:53
• But we have $\mathrm{tr}(B+C)=\mathrm{tr}(B)+\mathrm{tr}(C)$ which implies that $$\sum_i \lambda_i(B+C)=\sum_i \lambda_i(B)+\sum_i\lambda_i(C),$$ so if $\lambda_i(B+C)\leq \lambda_i(B)+\lambda_i(C)$, then we can infer that $\lambda_i(B+C)=\lambda_i(B)+\lambda_i(C)$. Dec 13, 2016 at 17:01
• I find something similar to the inequality here. We have $$\lambda_{i+j-1}(A+B) \leq \lambda_i(A) + \lambda_j(B), \ \ \ \ \ (4)$$ whenever ${i,j \geq 1}$ and ${i+j-1 \leq n}$ Dec 13, 2016 at 17:04