The question hopefully says it all!

We have a Hermitian matrix $A=A^* \in \mathbb{C}^n$ and a quadratic form: $f(x)=x^*Ax,~x\in \mathbb{C}^n$

We want to find the solution of $f(x) = x^*Ax = 0$

When the matrix is positive semi definite (p.s.d.), the solution seems to be null-space of the matrix of $A$. This I found by diagonalising $A$. But suddenly I became helpless when $A$ has negative Eigen values too!

Please note that when the matrix is not p.s.d., the solution space contains the null space. (I.e. null space is always a solution)

  • $\begingroup$ Have you tried solving this in the case that $A$ is diagonal? $\endgroup$ – Omnomnomnom Mar 20 '15 at 17:21
  • $\begingroup$ In fact, it suffices to solve this in the case that $A$ has elements $\pm 1$ and $0$. $\endgroup$ – Omnomnomnom Mar 20 '15 at 17:23
  • $\begingroup$ @omnomnomnom Thanks for the comment. That's exactly where my solution stage is. This tri element diagonal case gives me an equation with no further insights! $\endgroup$ – Loves Probability Mar 20 '15 at 23:01
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    $\begingroup$ For example, the solution for the matrix $$\pmatrix{1\\&1\\&&-1}$$ is the cone $x^2+y^2=z^2$. Note that this is not a linear subspace. $\endgroup$ – Omnomnomnom Mar 21 '15 at 5:07

Consider $N\times N$ matrices and $N\times 1$ vectors. Let $A=U\Lambda U^H$ be its eigen-decomposition. Then \begin{align} x^HAx &=y^H\Lambda y && \{\text{where I define }y=Ux,~\forall x \} \\ &=\sum_{i=1}^{N}|y_i|^2\lambda_i \\ &= \theta^T\lambda \end{align} where $\lambda$ is the vector with all eigenvalues and $\theta$ is any vector whose entries are non-negative. Thus, for any $\theta\in\left(\mathcal{N}(\lambda)\cap\mathbb{R}_{+}^N\right)$ where $\mathcal{N}(\lambda)$ is the null-space of $\lambda$ vector (set of all vectors orthogonal to $\lambda$) and the non-negative $N$-dimensional orthant (non-negative quadrant where all entries are non-negative), we have $\theta^T\lambda=0$. Now consider the set $\mathcal{D}$ of all diagonal matrices such that the diagonal entries lie on the unit circle in the complex plane. Then your solution set is \begin{align} \mathcal{S}_x\,=\,\{U^HD\sqrt{\theta} \,\mid\,\forall D\in\mathcal{D}~,~\forall \theta\in\left(\mathcal{N}(\lambda)\cap\mathbb{R}_{+}^N\right) \} \end{align} where $\sqrt{\theta}$ is the entry-wise square root of $\theta$. $\mathcal{D}$ is needed because it doesn't matter what the phase of each entry of $y$ is. Note that $y_{i}=D_{ii}\sqrt{\theta_i}$ and $|y_i|^2=|D_{ii}|^2\theta_i=\theta_i$. The difficult part is that $\mathcal{S}_x$ is not a linear subspace and you won't have that nice properties. It is a highly non-linear transformation and I am not sure what intuitive sense you can derive out of it.

  • $\begingroup$ Thanks for the solution. This does give a good insight. But $\mathcal{N}(\lambda)\bigcap\mathbb{R}_{+}^{N}$ is not the solution of $y$ (but a solution of $y\cdot^* y$), let alone desired $x$! (The question actually asks about $x$, as the zeros of the quadratic $x^*Ax$) Ofcourse, $y$ is a simple but non-linear mapping from $\mathcal{N}(\lambda)\bigcap\mathbb{R}_{+}^{N}$. But there is another unitary transformation to be applied after this non-linear transform. What does that all finally mean, is the mystery still! $\endgroup$ – Loves Probability Mar 23 '15 at 15:22
  • $\begingroup$ You are right. It is far from stating the answer. I thought the reverse transformation is obvious. I have now explicitly stated it. $\endgroup$ – dineshdileep Mar 24 '15 at 5:17

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