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Let $X^*$ mean the conjugate transpose of matrix $X.$

We have two matrices $H,V$ and a vector $x.$ We know that $$H^*x=V^*y\wedge V^*x=H^*z\wedge Va=x=Hb$$ for some vectors $y,z,a,b$ and also we know that $$ H^2=0\wedge V^2=0\wedge VH=HV.$$

Question. Does $x=0$ then?

This problem really bothers me. I have already posted two question (1, 2) with simplified assumptions, but users without a problem constructed counterexamples. So I post this final question, which I really hope, will have positive answer.

PS. This problem occurred when I wanted to show that Bott-Chern Laplacian is elliptic without using local coordinates.

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    [This more recent question](http://math.stackexchange.com/q/2129361/81360) is also relevant. – 2017-02-04
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    As I commented on the other post, it might help if you explained how you arrived at this problem starting from the original question of *showing that Bott-Chern Laplacian is elliptic without using local coordinates* – 2017-02-04
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    @Omnomnomnom [You do know](http://math.stackexchange.com/q/2122868) that the answer is "no" (when $V=H=$ the 2x2 nilpotent Jordan block, $x=y=z$ is an eigenvector of $V$ and $a=b=$ another generalised eigenvector), don't you? – 2017-02-04
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    @user1551 I guess I failed to see that one was an instance of the other – 2017-02-05
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    @Omnomnomnom users1551s argument is wierd, cause $V$ has no eigenvectors. Do I miss something? – 2017-02-05
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    @fallenapart $(1,0)$ is an eigenvector – 2017-02-05
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    @Omnomnomnom Associated to 0.. ok, so it is just kernel of associated map. Sorry for bothering you with such silly question. – 2017-02-05

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If $H^2=0$ and $H=V$, then $V^2=0$ and $HV=VH$, and you can let $a=b$ and $x=y=z\ne0$.

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    You are right. There is plenty of such vectors with $x\neq 0.$ – 2017-02-05