Does congruence transformation preserve definiteness of a nonsymmetric matrix?

Question

Let $A$ be a nonsymmetric negative definite matrix, i.e., $x^\top (A+A^\top) x < 0$. If we invoked a congruent transformation, i.e., $DAD^\top=B$ where $D$ is a nonsingular matrix, will the resulted matrix $B$ still remain negative definite? Can it be said that $x^\top(B+B^\top)x<0$? Is there any condition on $D$ to preserve the definiteness?

Answer 1

As you said, $D$ needs to be nonsingular. But that's it.

Given nonsingular $D \in \mathbb{R}^{n\times n}$, we have

$x^T(B+B^T)x = x^T (DAD^T + DA^TD^T) x = x^T D (A+A^T) D^T x = (D^T x)^T (A+A^T) D^T x$.

Assume we have $z^T(A+A^T)z < 0$ for any $z \in \mathbb{R}^n\backslash \{0\}$. Then, since $D^T x \ne 0$ whenever $x\ne 0$ ($D$ nonsingular), $x^T(B+B^T)x = (D^T x)^T (A+A^T) D^T x < 0$ whenever $x \ne 0$. This proves that if $A$ is negative definite then $B$ is negative definite.

$A = D^{-1} B D^{-T} = D' B {D'}^T$ (Note: $(D^{T})^{-1} = (D^{-1})^{T}$, which is why I'm combining the inverse and transpose into a single symbol).

Reusing the previous result, if $B$ is negative definite then $A = D' B {D'}^T$ is negative definite.

Hence, we've shown that $A$ is negative definite if and only if $B$ is negative definite.