Given a matrix $A\in\mathrm R^{n\times n}$, what are the sufficient conditions on $A$ that

$v^*Av=0\Rightarrow Av=0$

holds for all $v\in\mathrm C^n\setminus 0$?

Clearly, it is necessary that either of the following holds:

$A$ is rank-deficient, for if $0$ is in the field of values of $A$, but $A$ has full rank, there exists a $v$ such that $Av\neq0$, but $v^*Av=0$.

The symmetric part of $A$ is positive definite.

But are they also sufficient?

  • 1
    $\begingroup$ Why is it necessary for $A$ to be singular? $I$ satisfies your condition but has full rank... $\endgroup$
    – user7530
    Commented Jan 26, 2013 at 0:39
  • $\begingroup$ You're right, I forgot to say that $v\neq 0$. I'll edit the original question. $\endgroup$
    – user59697
    Commented Jan 26, 2013 at 0:40
  • $\begingroup$ $I$ still satisfies your condition... ;) $\endgroup$
    – user7530
    Commented Jan 26, 2013 at 0:46
  • $\begingroup$ I can see you grinning from over here :) I've edited the necessary conditions to include the trivial case. $\endgroup$
    – user59697
    Commented Jan 26, 2013 at 0:52

2 Answers 2


Let $A_S$ be the symmetric part of $A$, and $A_N$ the antisymmetric part.

Then your property holds if and only if $A_S$ is semidefinite, with $\operatorname{rank} A_S = \operatorname{rank} A$.

Lemma: If $A_S$ is semidefinite, $\ker A \subset \ker A_S$.

Suppose $Aw = 0$ for some vector $w$. Then $w^H A w = w^H A_S w = 0$, so since $A_S$ is semidefinite, $w \in \ker A_S$.

Now for the main result:

  1. Suppose $A_S$ is (positive or negative) semidefinite, $\operatorname{rank} A_S = \operatorname{rank} A$, and $v^H A v = 0$. By the lemma, $\ker A \subset \ker A_S$, and in fact $\ker A = \ker A_S$ since both matrices have the same rank. Then since $v^H A_S v = 0$ and $A_S$ is semidefinite, $A_Sv=0$ and so $Av=0$.

  2. Suppose $A_S$ is semidefinite but $\operatorname{rank} A_S \neq \operatorname{rank} A$. Then by the lemma, $\operatorname{rank} A_S < \operatorname{rank} A$, and there is a vector $v$ in the kernel of $A_S$ but not in the kernel of $A$. Immediately we have $v^HAv=0$ and $Av\neq 0$.

  3. Lastly, suppose $A_S$ is indefinite. Let $w_1, w_2$ be two unit eigenvectors of $A_S$ with eigenvalues $\lambda_1 > 0$ and $\lambda_2 < 0$, respectively.

Let $$v = \sqrt{-\lambda_2}w_1 + i\sqrt{\lambda_1}w_2.$$ Then $$v^HAv = v^HA_Sv = -\lambda_1\lambda_2 + \lambda_1\lambda_2 = 0.$$ Moreover \begin{align*} w_1^HAv &= w_1^HA_Sv + w_1^H A_Nv\\ &= \lambda_1\sqrt{-\lambda_2} + i\sqrt{\lambda_1} w_1^H A_N w_2\\ &\neq 0, \end{align*} so $Av\neq 0$.

Interestingly, it seems the characterization of your property is much less simple when $v$ is restricted to be real, or $A$ is complex.


Original poster here. Looks like I've lost my cookie and can't accept your answer. I'm sorry. Is there any way to fix this (like reposting the question and then accepting your answer)?

I wonder what makes it so difficult to extend it to $v\in\mathbb R$.

  • $\begingroup$ Please follow the instructions given here to merge your new user with the old one. $\endgroup$ Commented May 10, 2013 at 8:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .