# Does positive row scaling matrix don't change a sign of matrix's eigenvalue?

Consider a positive diagonal matrix $$D = \textrm{diag}{[a_1,a_2,\ldots,a_n]} \in \mathbb{R}^{n \times n}, \textrm{with } a_i\ge 0$$ Consider a diagonalizable matrix $$A \in \mathbb{R}^{n \times n}$$ such that it has an eigendecomposition $$A = T^{-1} \Lambda T$$

Now, I only care about the whether the sign of eigenvalue of $$A$$ and $$DA$$ will differ.

I tested on some case, it looks to be true.

## Special case

Clearly, if the sign of eigenvalue already indicated by Gershgorin circle theorem (diagonal dominanted), then simply scaling it by any positive factor, won't change the side of circle.

The question, as put, has no answer since eigenvalues can be complex. For example, $$\begin{pmatrix}2&-1\\2&2\end{pmatrix},\qquad\begin{pmatrix}2&0\\0&\frac{1}{2}\end{pmatrix}\begin{pmatrix}2&-1\\2&2\end{pmatrix}=\begin{pmatrix}4& -2 \\ 1 & 1\end{pmatrix}$$ have eigenvalues $$2\pm i\sqrt2$$ and $$1,1$$ respectively.
Moreover, if $$a_i=0$$ then the signs can obviously change. But if you allow only real eigenvalues and $$a_i>0$$, the answer is that the sign can still change in special cases. For example,
$$\begin{pmatrix}20&-31\\20&-30\end{pmatrix},\qquad\begin{pmatrix}2&0\\0&\frac{1}{2}\end{pmatrix}\begin{pmatrix}20&-31\\20&-30\end{pmatrix}=\begin{pmatrix}40 & -62 \\ 10 & -15\end{pmatrix}$$ have eigenvalues $$-5\pm\sqrt5=-5\pm2.2$$ and $$(25\pm\sqrt{545})/2=12.5\pm11.7$$ respectively.