Is the largest number of zero eigenvalues at least 2

Question

For any $x\le3$, is the largest (or maximal) number of zero eigenvalues of $diag\left(1,2,3\right)-U\cdot diag\left(4,5,x\right)\cdot U^{T}$ for orthogonal matrix $U$ at least $2$?

Answer 1

No, it can't be $2$. Note that $A = U \pmatrix{4 & 0 & 0\cr 0 & 5 & 0\cr 0 & 0 & x \cr} U^{-1}$ can be any hermitian matrix with eigenvalues $4$, $5$ and $x$. Let $B = \pmatrix{1 & 0 & 0\cr 0 & 2 & 0\cr 0 & 0 & 3\cr}$, which has eigenvalues $1$, $2$ and $3$, We have $B \le 3 I$ (i.e. $3 I - B$ is positive semidefinite, so $A - B \ge A - 3 I$, which has eigenvalues $1$, $2$ and $x-3$. By the Min-Max principle, the second-greatest eigenvalue of $A-B$ is at least $1$. Thus there can be at most one zero eigenvalue (counted by multiplicity).