Is eigenvalue rounding monotonic?

Question

Let ${\bf S}_n$ be the set of $n \times n$ symmetric matrices. A function $f: {\bf S}_n \rightarrow {\bf S}_n$ is called nondecreasing if $A \succeq B$ implies $f(A) \succeq f(B)$.

Consider the functions which round down, or up, every eigenvalue of $A$. More precisely, if $$A = U~ {\rm diag}~(d_1, \ldots, d_n) ~U^T$$ (for simplicity, let us assume $A$ has distinct eigenvalues so that this decomposition is unique) consider the functions $\lceil A \rceil$ and $\lfloor A \rfloor$ defined as $$\lceil A \rceil = U ~{\rm diag}( \lceil d_1 \rceil, \ldots, \lceil d_n \rceil)~ U^T$$ and

$$\lfloor A \rfloor = U~ {\rm diag}( \lfloor d_1 \rfloor, \ldots, \lfloor d_n \rfloor) ~U^T$$ where $\lceil x \rceil$ rounds $x$ up to the next integer and $\lfloor x \rfloor$ rounds $x$ down to the previous integer.

Are these functions nondecreasing?

Answer 1

Let $B=\begin{bmatrix}2&\\&1\end{bmatrix}$ and $A=B+\epsilon\begin{bmatrix}1&1\\1&1\end{bmatrix}$. For sufficiently small $\epsilon$ both $\lfloor A\rfloor$ and $\lfloor B\rfloor$ have the same eigenvalues, namely $2$ and $1$, but different eigenvectors. So if we call their difference $D=\lfloor A\rfloor-\lfloor B\rfloor$, we have $D\ne0$ but $\operatorname{tr}D=\operatorname{tr}\lfloor A\rfloor-\operatorname{tr}\lfloor B\rfloor=0$, hence $D\not\succeq0$.

For example, when $\epsilon = 0.1$ we have $$\lfloor A\rfloor-\lfloor B\rfloor\approx\begin{bmatrix}-0.0097 & 0.0981 \\ 0.0981 & 0.0097\end{bmatrix}.$$