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Firstly, I want to thank @adam W gives a good clue to solve my homework problem. I have a set of quadratic matrix need to solve(not one equation) according to the following form:

$\alpha_{k}H(2n)\alpha^{T}_{k} = \delta(n),$

where $H(2n)$ is a symmetric matrix and the number of equations depend on the number of $\alpha_{k}$. For example, if $k=2$, the matrix can be extended in to following forms:

$[\alpha_{1} \alpha_{2}] * H(0) * [\alpha_{1} \alpha_{2}]^{T} = 1 ,$
$[\alpha_{1} \alpha_{2}] * H(2) * [\alpha_{1} \alpha_{2}]^{T} = 0 ,$ where $H(2n)$ has $2 \times 2$ matrix form.

Is there any algorithm for solving such kinds of quadratic matrix ?

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Just so you know, I do not get pinged from a new question unless I have commented or answered on the new question already. I have responded to your other also. – adam W Jan 20 '13 at 15:30

Look first (using $\mathbf{\alpha} = [\alpha_1,\alpha_2]$) at $$\mathbf{\alpha} H_2\mathbf{\alpha}^T=0$$ With a suitable change of basis, this will give $$\mathbf{x}D\mathbf{x}^T = 0$$ This will lead to two equations representing lines through the origin. The following shows where that comes from, using $D=\pmatrix{d_0 & 0 \\ 0 & d_1}$ \begin{align} d_0x_0^2 + d_1x_1^2 &= 0 \\ x_0^2 &= \frac{-d_1}{d_0}x_1^2 \\ x_0 &= \pm\sqrt{\frac{-d_1}{d_0}}x_1 \\ x_0 &= \pm r x_1 \end{align}

So you have two lines from the equation $x_0 = \pm r x_1$. More conveniently for plugging into the other equation, you will have $$x_0^2 = r^2x_1^2$$ $$x_0x_1 = \pm r x_1^2$$ When this is plugged in, it will be linear in the term $x_1^2$ (since $x_0$ is substituted and removed), though two solutions become possible depending on the sign for $r$, and yet that doubles again in the next square root. This may help to see all the (one) cross terms: $$\pmatrix{x_0 & x_1}\pmatrix{A & B \\ C & D\\}\pmatrix{x_0 \\ x_1} = Ax_0^2 + (B+C)x_0x_1 + Dx_1^2$$ Which using the previous substitution gives $$Ar^2x_1^2 + (B+C)(\pm r) x_1^2 + Dx_1^2$$ and is easily solved for $x_1^2$

With your solutions in hand, just apply the inverse change of basis. Your four solutions are the intersection points of the ellipse described by the first equation, and the two lines described by the second equation.

Now if you want for higher dimensions, $k > 2$, things are not as simple. In that case, $$x_n^2 = \frac{-1}{d_n}\sum_{i\ne n}d_ix_i^2$$ which does not give a simple square in the cross terms: $$x_nx_m = \sqrt{\frac{-1}{d_n}\sum_{i\ne n}d_ix_i^2}\sqrt{\frac{-1}{d_m}\sum_{i\ne m}d_ix_i^2}$$

I would be interested if someone has a better answer for you in the more general case of $k>2$.

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Thx for the comment, adam. But I have one question, that is , how do you obtain the equation that $xDx^{T}=0$, where $D$ is a diagonal matrix? Actually it is impossible from my equation. Can you show the method for obtaining $D$? – DavidBear Jan 21 '13 at 6:40
Look again at the implied symmetry from your previous question where $G$ in $\mathbf{\alpha}G\mathbf{\alpha}^\top$ can always be assumed to symmetric. A symmetric matrix (or Hermitian matrix if working with complex values) always has a change of basis to get $\mathbf{x}D\mathbf{x}^\top$ as was described there. Use $\mathbf{y} = \mathbf{x}L^{-1}$ (and careful as I use row vectors, where column vectors are the usual convention.) Are your $H$ matrices symmetric already? If not try everything from "implied symmetry" on to the end of that answer. – adam W Jan 21 '13 at 13:29
Hey adam. I have read your comments. But I have some question based on your comment. Based on your comment, we know $D$ is a diagonal matrix. The situation is this: since $H$ is a symmetric matrix, which can be expressed as $H = Q\Lambda Q$, where Q is a eigenvector of H, $\Lambda$ is a diagonal matrix, where eigenvalue is its diagonal value. Therefore, we can express the equation above as $\alpha Q \Lambda Q^{T} \alpha^{T}$, which is similar to your equation, $xDx^{T}$, i.e., $x = \alpha Q$. – DavidBear Jan 22 '13 at 13:21
However, the vector of $x$ do not contain only one parameter, it contains a combination of $\alpha_{k}$. So, I am still wondering how you can get that equation. Btw, if I have done something wrong, please tell me asap. Thx~~ – DavidBear Jan 22 '13 at 13:25
if $x=\alpha Q$ then the equation is $x\Lambda x^T$. Once you solve for $x$, which has the nice diagonal in the equation ( so you need to make the same substitution in the other equation as well) then use $\alpha = xQ^T$ – adam W Jan 22 '13 at 13:29

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