The space of all $A_n$ such that $\lambda_i(A_n)=\lambda_i(A_{n-1})$ has co-dimension 2 for Hermitian matrices or 1 for symmetric matrix

21 Views Asked by At

(Cauchy Interlacing Theorem) For any $n \times n$ Hermitian matrix $A_{n}$ with top left $n-1 \times n-1$ minor $A_{n-1}$ ($A_{n-1}$ is simply removing the last row and last columns from $A_n$), then $$ \lambda_{i+1}\left(A_{n}\right) \leq \lambda_{i}\left(A_{n-1}\right) \leq \lambda_{i}\left(A_{n}\right) $$ for all $1 \leq i \leq n .$ Show furthermore that the space of $A_{n}$ for which equality holds in one of the inequalities above has codimension 2 (for Hermitian matrices) or 1 (for real symmetric matrices).

I can show the main part of Cauchy interlacing theorem but I fail to see how the claim at the end that counts codimensions when equality holds is true. In fact, I don't think I fully understand what I'm supposed to prove in precise terms. This is an exercise from Terry Tao's "Topics in random matrices". Is it related to $\lambda_i(A)$ is differentiable? Any help is appreciated