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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.

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1 Answer 1

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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.

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  • $\begingroup$ I don't understand why we have $\lambda_i(B)\geq \lambda_i(A)-\lambda_i(C)$. $\endgroup$
    – Xiang Yu
    Commented Dec 13, 2016 at 16:50
  • $\begingroup$ 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? $\endgroup$ Commented Dec 13, 2016 at 16:53
  • $\begingroup$ 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 $\lambda_i(B+C)=\lambda_i(B)+\lambda_i(C)$. $\endgroup$
    – Xiang Yu
    Commented Dec 13, 2016 at 17:01
  • $\begingroup$ 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 ${i+j-1 \leq n}$ $\endgroup$
    – Xiang Yu
    Commented Dec 13, 2016 at 17:04
  • $\begingroup$ I think that is what I really had in mind: Weyl's inequality $\endgroup$ Commented Dec 13, 2016 at 17:08

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