0
$\begingroup$

I've encountered a problem that needs to be solved by solving the following algebraic equation

$$\mathbf{Y}^\mathsf{T}\mathbf{Q}\mathbf{Y}=\mathbf{X}^\mathsf{T}\mathbf{Q}^\mathsf{-1}\mathbf{X}$$

where $Q$ is a $4 \times 4$ symmetric matrix to be solved. $X$ and $Y$ are both known $4 \times 1$ vectors. There are enough $X$ and $Y$ inputs so that I think a nonlinear least-squares solution must be existed. However, I can not figure it out by myself. Hope you could provide me with some clues.

$\endgroup$
8
  • $\begingroup$ What do you mean by "the equation needs to be solved"? Do you need a parameterization of every possible solution? $\endgroup$ Jan 7 '20 at 12:24
  • 1
    $\begingroup$ My friend, you have one relation and $10$ unknowns!! On the other hand, it's not a linear equation in $Q$. That is not serious... $\endgroup$
    – user91684
    Jan 7 '20 at 19:04
  • $\begingroup$ Actually,Q is an ellipsoid, X is the point on the ellipsoid, and Y is the plane tangent to the ellipsoid. I hope to turn this equation into least square solution @Omnomnomnom $\endgroup$
    – sjChen
    Jan 8 '20 at 1:47
  • $\begingroup$ Actually,Q is an ellipsoid, X is the point on the ellipsoid, and Y is the plane tangent to the ellipsoid.X and Y have enough inputs so we can construct more than 10 equations.I hope to turn this equation into least square solution @loupblanc $\endgroup$
    – sjChen
    Jan 8 '20 at 1:48
  • $\begingroup$ I am not sure how to interpret "X is the point on the ellipsoid, and Y is the plane tangent to the ellipsoid". I think that you are trying to say that $X$ is a point on the ellipsoid, and that the tangent plane to the ellipsoid is the solution set for $Z$ of the equation $Y^TZ = Y^TX$. Is this correct? $\endgroup$ Jan 8 '20 at 2:22
0
$\begingroup$

I think you can use $\operatorname{vec}$ operator for constructing a least squares problem. But I'm not sure it would produce a consistent answer.

$$\begin{align*} \operatorname{vec}(y^TQy) &= \operatorname{vec}(x^TQ^{-1}x) \\ (y^T \otimes y^T) \operatorname{vec}(Q) &= (x^T \otimes x^T) \operatorname{vec}(Q^{-1}) \end{align*}$$

So, you can write this as

$$ \begin{bmatrix}(y_1^T \otimes y_1^T) & -(x_1^T \otimes x_1^T) \\ \vdots & \vdots \\ (y_N^T \otimes y_N^T) & -(x_N^T \otimes x_N^T) \end{bmatrix} \begin{bmatrix} \operatorname{vec}(Q) \\ \operatorname{vec}(Q^{-1}) \end{bmatrix} = 0$$

for $N$ data you have. From this point the problem becomes finding the null space of the known matrix and selecting the set of vectors such that $Q Q^{-1} \approx I$.

$\endgroup$
2
  • $\begingroup$ Excuse me,$Q$ is an ellipsoid,But is $Q^{-1}$ also an ellipsoid? Because if $Q^{-1}$ is represented by the parameters of $Q$, $Q^{-1}$ is very complicated.@obareey $\endgroup$
    – sjChen
    Jan 10 '20 at 2:13
  • $\begingroup$ This solution doesn't assume any relation between $Q$ and $Q^{-1}$. This is why it may produce inconsistent results, because the solution most likely is not unique. $\endgroup$
    – obareey
    Jan 10 '20 at 7:21
0
$\begingroup$

Presumably the underlying field is real. There are infinitely many solutions but I think the easiest ones are the followings:

  1. If both $X$ and $Y$ are nonzero, set $Q=\frac{\|X\|}{\|Y\|}I$.
  2. If $X=Y=0$, set $Q=I$.
  3. If $X=0\ne Y$, extend $y=\frac{Y}{\|Y\|}$ to an orthonormal basis $\{y,u,v,w\}$ of $\mathbb R^4$. A solution is given by $Q=Q^{-1}=yu^T+uy^T+vv^T+ww^T$.
  4. If $X\ne0=Y$, extend $x=\frac{X}{\|X\|}$ to an orthonormal basis $\{x,u,v,w\}$ of $\mathbb R^4$. A solution is given by $Q=Q^{-1}=xu^T+ux^T+vv^T+ww^T$.
$\endgroup$
0
$\begingroup$

a) I propose this (very elementary) method. I assume that all the matrices are real.

i) Randomly choose a real symmetric matrix $Q\in S_4$; put $Q[1,1]:=q,Q[1,2]:=r$.

ii) Calculate $p(q,r)=numer(Y^TQY-X^TQ^{-1}X)$ (the numerator of the quotient); it's a polynomial of degree $2$ wrt $q$. Then $d(r)=discrim(p,q)$ (the discriminant of $p$ wrt $q$) is a polynomial of degree $4$ wrt $r$.

You easily obtain a solution in $Q$ if there is $r_0$ s.t. $d(r_0)\geq 0$.

iii) I tested $30000$ random values of the couple $(X,Y)$. In each case, $d(0)>0$. Anyway, if your random $Q$ is not convenient, then choose another $Q$!

If you have a problem with some $(X,Y)$, then write me. Now, it's up to you to work.

EDIT. b) $\textbf{Proposition}$. If for every $i$, $Y_i\not= 0$, then there is at least a solution of $(*)$ $Y^TQY=X^TQ^{-1}X$.

$\textbf{Proof}$. We seek $Q$ in the form $Q=diag(q_i)$ where $q_i\not= 0$.

$(*)$ can be rewritten $\sum_i(q_iY_i^2-\dfrac{1}{q_i}X_i^2)=0$. We choose $q_i=X_i/Y_i$.

$\textbf{Remark}$. After reflexion, I think that there are solutions in any cases. For example, if $Y=0,X=[1,0,\cdots,0]^T$, then a solution is

$Q=\begin{pmatrix}1&1&1&1\\1&2&1&1\\1&1&3&1\\1&1&1&3/5\end{pmatrix}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.