This is a problem from Horn and Johnson's Matrix Analysis. I've tried to follow the problem but I can't find a way to lead to the conclusion the problem is suggesting. Any solutions, hints, or suggestions would be greatly appreciated.

Suppose that $A=[a_{ij}] \in M_n$ in Hermitian, has smallest and largest eigenvalues $\lambda_1$ and $\lambda_n$, and for some $i \in$ {$1, \dots, n$}, either $a_{ii} = \lambda_1$ or $a_{ii} = \lambda_n$. Use (4.3.34) to show that $a_{ik}=a_{ki}=0$ for all $k=1, \dots, n, k\neq i$. Does anything special happen if a main diagonal entry of $A$ is an eigenvalue of $A$ different from $\lambda_1$ and $\lambda_n$?

Consider $$ \begin{pmatrix} 0 & i & 1 \\ -i & 0 & 1 \\ 1 & 1 & 0 \\ \end{pmatrix} $$


Let $A=[a_{ij}] \in M_n$ be Hermitian, partitioned as

$$A= \begin{pmatrix} B & C \\ C* & D \\ \end{pmatrix} $$ , $B \in M_m$, $D \in M_{n-m}$, $C \in M_{m,n-m}$, and let the eigenvalues of $A$ be ordered in increasing order. Then

$a_{11}+a_{22}+ \cdots a_{mm} \ge \lambda_1(A)+ \cdots + \lambda_m(A)$


$a_{11}+a_{22}+ \cdots a_{mm} \le \lambda_{n-m+1}(A)+ \cdots + \lambda_n(A)$

If either inequality is an equality, then $C =0$.


1 Answer 1


Suppose $a_{ii} = \lambda_1$.

Rearrangement of a matrix doesn't change matrix eigenvalues, so we can get matrix $A'$ from $A$ with $a'_{11} = a_{ii}$ by exchanging of first and $i$th row and first and $i$th column. Then $a'_{ii} = \lambda_1$. Consequently, as an inequality from 4.3.33 with $m = 1$ is an equality, we get $C = 0$, so $a'_{1k} = a'_{k1} = 0$. Doing opposite rearrangement from $A'$ to $A$ we get that $a_{ik} = a_{ki} = 0$ holds.

For $a_{ii} = \lambda_n$ we can use similar approach for matrix $-A$.


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