Product of a positive diagonal matrix with a skew symmetric matrix yields a matrix with imaginary eigenvalues.

Question

For a skew symmetric matrix $A=-A^T$, and a diagonal matrix $D=diag(d_{ii})$ such that $d_{ii}\in (0,\,1]$.

Since $A$ has imaginary eigenvalues or a zero eigenvalue, will $DA$ also have imaginary eigenvalues?

I have until now is that $DA$ has similar eigenvalues as $D^{.5}AD^{-.5}$.

It is in general not true when $d_{ii}$ can be negative. For Example $A=\begin{bmatrix}0 & 1\\-1 &0\end{bmatrix}$ and $D={\rm diag}([1,\,-1])$, we have $DA=\begin{bmatrix}0 & 1\\1 &0\end{bmatrix}$ which has eigenvalues $-1,+1$.

Is this proof correct?

Suppose $DA$ has real eigenvalues, then $DA$ will have both positive and negative eigenvalues since $tr(DA)=0$. With $D\succ 0$ where $\succ$ denotes positive definiteness. $DA\succeq 0$ as $\mathbf x^T A \mathbf x=0$. This contradicts with our assumption that $DA$ has real eigenvalues.

Answer 1

Suppose $DA$ has real eigenvalues, then $DA$ will have both positive and negative eigenvalues since $tr(DA)=0$. With $D\succ 0$ where $\succ$ denotes positive definiteness. $DA\succeq 0$ as $\mathbf x^T A \mathbf x=0$. This contradicts with our assumption that $DA$ has real eigenvalues.

Answer 2

There is a general property that the eigenvalues of $ST$ and $TS$ coincide. Therefore the eigenvalues of $DA$ conicide with the eigenvalues of $D^{1/2}AD^{1/2}.$ The new matrix is skew symmetric so its eigenvalues are purely imaginary.