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?

  • $\begingroup$ @RodrigodeAzevedo: I think that since $A$ being symmetric would imply that $A$ is positive definite which is what OP doesn't want. $\endgroup$ – Koro Jul 15 '20 at 13:33
  • $\begingroup$ Is $A$ allowed to be diagonizable? $\endgroup$ – Koro Jul 15 '20 at 13:34
  • $\begingroup$ Did u try writing $A$ with all its elements and then note that first columns of $AD$ and that of $A$ have opposite signs. Then write their characteristic polynomials. Note that $AD$ being 3 by 3 can either have 1 negative root or all roots negative ( since det(AD) is negative). Assume on the contrary, that $AD$ has 3 negative eigenvalues. From there, you can try to get a contradiction $\endgroup$ – Koro Jul 15 '20 at 13:38
  • $\begingroup$ Sorry for my writing. I edited it. $\endgroup$ – Kim Jul 15 '20 at 14:57
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    $\begingroup$ Possibly useful: $v^TAv>0$ for $v \neq 0$ tells you that the eigenvalues of $A$ all have positive real part. $\endgroup$ – Ben Grossmann Jul 15 '20 at 15:37

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 $$(v^TDw)^2-(w^TDw)(v^TDv)=(w^TDv)^2-(w^TDw)(v^TDv) = w^T B w \geq 0$$


  • $\begingroup$ Nice answer. I have considered $|\lambda||\eta|(v^TDv)(w^TDw)>\frac{(|\lambda|+|\eta|)^2}{4}(v^TDw)^2$ too, but somehow also have confused it with $\frac{(|\lambda|+|\eta|)^2}{4}(v^TDv)(w^TDw)>|\lambda||\eta|(v^TDw)^2$, and so I didn't deduce that $|\lambda||\eta|(v^TDv)(w^TDw)>\frac{(|\lambda|+|\eta|)^2}{4}(v^TDw)^2 \ge |\lambda||\eta|(v^TDw)^2$. $\endgroup$ – user1551 Jul 17 '20 at 8:56
  • $\begingroup$ By the way, since $AD$ can have a repeated negative eigenvalue of geometric multiplicity $1$, do you need a continuity argument to guarantee that $AD$ has two linearly independent eigenvectors $v$ and $w$? $\endgroup$ – user1551 Jul 17 '20 at 9:55
  • $\begingroup$ @user1551 I think I don't need continuity argument if I'm correct. I supposed that $AD$ has two distinct (negative) eigenvalues and get a contradiction. Isn't it true that eigenvectors $v$ and $w$ are linearly independent if $ADv=\lambda v$ and $ADw=\eta w$ for distinct $\lambda, \eta$? $\endgroup$ – Luxerhia Jul 17 '20 at 11:25
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    $\begingroup$ I think the OP wants to prove that $AD$ has exactly one negative eigenvalue, counting multiplicity. When it has more than one negative eigenvalues, there is no guarantee that those negative eigenvalues are distinct. It may happen that they are the same repeated eigenvalue, but the eigenspace is only one-dimensional. $\endgroup$ – user1551 Jul 17 '20 at 11:37
  • $\begingroup$ @user1551 Thank you for explanation. I'll think about that. $\endgroup$ – Luxerhia Jul 17 '20 at 11:44

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.


Fact 1: If $B$ (not necessarily symmetric) is a $3\times 3$ (real) positive definite matrix, then $B^{-1}$ is also positive definite.
Proof: For non-zero $x\in \mathbb{R}^3$, since $B^{-1}x \ne 0$, we have $x^\mathsf{T}B^{-1}x = x^\mathsf{T}(B^{-1})^\mathsf{T}x = (B^{-1}x)^\mathsf{T}B (B^{-1}x) > 0$. We are done.

Now, let $a = [1, 0, 0]^\mathsf{T}$. Note that $AD = A - 2Aaa^\mathsf{T}$. For $\lambda < 0$, $A - \lambda I$ is invertible, and \begin{align} \det (AD - \lambda I) &= \det (A - 2Aaa^\mathsf{T} - \lambda I) \\ &= \det (A - \lambda I) \det( I - (A - \lambda I)^{-1}2Aaa^\mathsf{T})\\ &= \det (A - \lambda I) \cdot \left(1 - 2a^\mathsf{T}(A - \lambda I)^{-1}Aa\right) \end{align} where we have used $\det (I + uv^\mathsf{T}) = 1 + v^\mathsf{T}u$ for real vectors $u, v$.

Let $f(\lambda) = 1 - 2a^\mathsf{T}(A - \lambda I)^{-1}Aa$. For $\lambda < 0$, by using $\frac{\partial Y^{-1}}{\partial x} = - Y^{-1}\frac{\partial Y}{\partial x}Y^{-1}$, we have \begin{align} f'(\lambda) &= - 2a^\mathsf{T}(A - \lambda I)^{-1}(A - \lambda I)^{-1} Aa\\ &= -2a^\mathsf{T}[A^{-1}(A - \lambda I)(A - \lambda I)]^{-1}a\\ &= -2a^\mathsf{T}(A + \lambda^2 A^{-1} - 2\lambda I)^{-1}a. \end{align} By Fact 1, $A^{-1}$ is positive definite. Thus, $A + \lambda^2 A^{-1} - 2\lambda I$ is positive definite for $\lambda < 0$. By Fact 1, $(A + \lambda^2 A^{-1} - 2\lambda I)^{-1}$ is positive definite for $\lambda < 0$. Thus, $f'(\lambda) < 0$ for $\lambda < 0$. Note also that $f(-\infty) = 1$ and $f(0) = -1$. Thus, the equation $f(\lambda) = 0$ has exactly one negative real root. As a result, $\det (AD - \lambda I) = 0$ has exactly one negative real root. (Q. E. D.)


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