$AD$ has exactly one negative eigenvalue if $v^T A v > 0$ and $D = \mbox{diag}(-1,1,1)$ 
Let $A$ be $3 \times 3$ real matrix (which is not necessarily symmetric or diagonalizable) such that $v^T A v>0$ for every $v \in \mathbb R^3 - \{0\}$. Show that $AD$ has exactly one negative eigenvalue, where $D = \mbox{diag}(-1,1,1)$.


I can prove that $AD$ has a negative eigenvalue. If $\det(A) \leq 0$, then characteristic polynomial $f(t) = \det(tI-A)$ satisfies $f(0) \geq 0$. Since $f$ is polynomial of degree $3$ and
$$\lim_{t \to -\infty} f(t) = -\infty$$
we can find a eigenvalue $\lambda \leq 0$ of $A$ with eigenvector $v$. Then $v^TAv=\lambda v^Tv \leq 0$, contradiction. Therefore $\det(AD) = \det(A) \det(D)<0$. let $g(t)$ be characteristic polynomial of $AD$. Then $g(0) = - \det(AD)>0$ so same argument produce a result.
However, I cannot solve uniqueness part. How to solve it?
 A: Consider size $n \times n$ case with $D=\mbox{diag}(-1,1,\dots,1)$. As @user1551 write in his answer, $AD$ has at least one negative eigenvalue.
Suppose $\lambda \neq \eta$ is two negative eigenvalues of $AD$ with eigenvectors $v, w$, respectively; i.e. $ADv=\lambda v, ADw=\eta w$. Since $v$ and $w$ are linearly independent, so are $Dv$ and $Dw$. For every $s,t \in \mathbb R$ $sDv+tDw$ is nonzero unless $s^2+t^2=0$. It follows that $(sDv+tDw)^TA(sDv+tDw)>0$. Expand this yields $$ s^2(\lambda v^T Dv) + st(\lambda+\eta)v^T D w+ t^2 (\eta w^T D w) >0$$
Deduce that $v^TDv<0$ and $(w^TDw)(v^TDv)>(v^TDw)^2$. Define a symmetric matrix $B$ by $$B=Dvv^TD-(v^TDv)D$$
Then $Bv=0$. In other words, $v$ is an eigenvector of $B$ with eigenvalue $0$. Consider the subspace $U$ of $\mathbb R^n$, given by the intersection of the orthogonal complements of subspaces generated by $v$ and $e_1=(1,0,\dots,0)$; i.e. $U=\langle v \rangle^\perp \cap \langle e_1 \rangle ^\perp$. Check that $\dim U \geq n-2$. For all $u \in U$ we have $Bu = -(v^T D v)u $, because $v^Tu=0$ and $Du=u$. Finally, observe that $$\mbox{tr}(B)=v^Tv+(n-2)(-v^T Dv) $$
This shows that $B$ is positive semi-definite. Thus $$(v^TDw)^2-(w^TDw)(v^TDv)=(w^TDv)^2-(w^TDw)(v^TDv) = w^T B w \geq 0$$
contradiction.
A: Let us deal with the case where $A$ is $n\times n$ for some $n\ge2$ and $D=\operatorname{diag}(-1,1,\ldots,1)$. Since $v^TAv>0$ for all nonzero $v$, every real eigenvalue of $A$ is positive. Hence $\det(A)>0,\,\det(AD)<0$ and $AD$ has at least one negative eigenvalue. We claim that $AD$ has exactly one negative eigenvalue.
Suppose the contrary that $AD$ has at least two negative eigenvalues. By perturbing the real Jordan form of $AD$, we can pick a real matrix $B$ that is sufficiently close to $AD$, such that $B$ has at least two negative eigenvalues and is diagonalisable over $\mathbb C$. Let $J=V^{-1}BV$ be the real Jordan form of $B$. Then
$$
BD=VJV^{-1}D=VJ\left(V^{-1}D(V^{-1})^T\right)V^T=:VJEV^T,\tag{1}
$$
where $E=V^{-1}D(V^{-1})^T$ is real symmetric. Let us write
$$
J=\pmatrix{\Lambda&0\\ 0&\ast}\ \text{ and }\ E=\pmatrix{F&\ast\\ \ast&\ast},
$$
where $\Lambda$ is a $2\times2$ negative diagonal matrix and $F$ has the same size.
As $v^TAv>0$ for all nonzero $v$, $A$ has a positive definite symmetric part. As $B$ is close to $AD$, $BD$ is close to $A$. Hence $BD$ also has a positive definite symmetric part. Since $BD$ is congruent to $JE$ (by $(1)$) and $JE$ contains a principal submatix $\Lambda F$, $\Lambda F$ must have a positive definite symmetric part. It follows that all eigenvalues of $\Lambda F$ have positive real parts. By matrix similarity, the eigenvalues of $(-\Lambda)^{1/2}(-F)(-\Lambda)^{1/2}$ have positive real parts too. But $(-\Lambda)^{1/2}(-F)(-\Lambda)^{1/2}$ is also real symmetric. Hence it is positive definite. So, by matrix congruence, $-F$ is positive definite and $F$ is negative definite. However, as its parent matrix $E$ has only one non-positive eigenvalue, Cauchy's interlacing inequality dictates that $F$ can have at most one non-positive eigenvalue. Hence we arrive at a contradiction and $AD$ must have exactly one negative eigenvalue at the beginning.
A: Remark: Actually, the proof works for general case $D=\mathrm{diag}(-1,1,\dots,1)$.
Let $\lambda \in \mathbb{R}_{< 0}$.
Then $A - \lambda I$ is invertible.
Let $a = [1, 0, 0]^\mathsf{T}$. Then $AD = A - 2Aaa^\mathsf{T}$.
We have
\begin{align*}
 \det (AD - \lambda I) &= \det (A - 2Aaa^\mathsf{T} - \lambda I) \\
 &= \det (A - \lambda I) \det( I - (A - \lambda I)^{-1}\cdot 2Aaa^\mathsf{T})\\
 &= \det (A - \lambda I) \cdot \left(1 - 2a^\mathsf{T}(A - \lambda I)^{-1}Aa\right) \tag{1}
\end{align*}
where we have used $\det (I + uv^\mathsf{T}) = 1 + v^\mathsf{T}u$ for all real vectors $u, v$.
Let $$f(\lambda) := 1 - 2a^\mathsf{T}(A - \lambda I)^{-1}Aa .$$
Using $\frac{\partial Y^{-1}}{\partial x}
= - Y^{-1}\frac{\partial Y}{\partial x}Y^{-1}$, we have
$$f'(\lambda) = - 2a^\mathsf{T}(A - \lambda I)^{-1}(A - \lambda I)^{-1} Aa.$$
(Note: Actually, it is not difficult to get the expression $f'(\lambda)$ without using $\frac{\partial Y^{-1}}{\partial x}
= - Y^{-1}\frac{\partial Y}{\partial x}Y^{-1}$.)
Let $B := (A - \lambda I)^{-1}(A - \lambda I)^{-1} A$.
Since $A$ is invertible, $B$ is also invertible with $B^{-1} = A^{-1}(A - \lambda I)(A - \lambda I)
= A + \lambda^2 A^{-1} - 2\lambda I$.
We have
\begin{align*}
 a^\mathsf{T} B a &= (a^\mathsf{T} B a)^\mathsf{T}\\
 &= (B a)^{\mathsf{T}} B^{-1} (Ba)\\
 &= (B a)^{\mathsf{T}} (A + \lambda^2 A^{-1} - 2\lambda I) (Ba)\\
 &= (B a)^{\mathsf{T}} A (Ba)
 + \lambda^2(B a)^{\mathsf{T}}  A^{-1} (Ba) - 2\lambda (B a)^{\mathsf{T}} (Ba)\\
 &= (B a)^{\mathsf{T}} A (Ba)
 + \lambda^2\left[(B a)^{\mathsf{T}}  A^{-1} (Ba)\right]^\mathsf{T} - 2\lambda (B a)^{\mathsf{T}} (Ba)\\
 &= (B a)^{\mathsf{T}} A (Ba)
 + \lambda^2(A^{-1}B a)^{\mathsf{T}}  A (A^{-1}Ba) - 2\lambda (B a)^{\mathsf{T}} (Ba)\\
 &> 0
\end{align*}
where we have used $Ba \in \mathbb{R}^3 - \{0\}$.
Thus, $f'(\lambda) < 0$ for all $\lambda < 0$.
Note also that $f(-\infty) = 1$ and $f(0) = -1$. Thus, the equation $f(\lambda) = 0$ has exactly one
negative real root. From (1),
$\det (AD - \lambda I) = 0$ has exactly one negative real root.
We are done.
