Finding a positive definite matrix to satisfy the general equation of an ellipse

In summary, the conversation is about finding a matrix A that satisfies the equation $v^TAv=1$ where $v=(x, y)^T$ and $(1)$ is a given equation with certain constraints. The first step is to divide through by $\sin^2(\delta)$ to get the equation in the form $ax^2+(b+c)xy+dy^2=1$. The parameters a, b+c, and d can then be determined in terms of $a_1$, $a_2$, and $\sin(\delta)$, but b and c remain as free variables unless additional conditions are imposed on the matrix A.
  • #1
kalish1
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I am trying to find a matrix A such that $(1)$ can be written as $v^TAv=1$ where $v=(x, y)^T$.

$(1)$: $$\left(\frac{x}{a_1}\right)^2 + \left(\frac{y}{a_2}\right)^2 - 2\left(\frac{xy}{a_1a_2}\right)\cos(\delta)=\sin^2(\delta)$$ $$a_1, a_2, \sin(\delta)\neq 0.$$

I am positive that $\cos(\delta)$ should not be $\cos^2(\delta)$, as it is not even indicated in my textbook's errata.

**Here is my attempt:**

$v^TAv=1 \iff (x,y)A(x,y)^T=1 \iff A [=] 2 $x $2$ $$\iff 1 = (x \ y) \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \iff 1 = (x \ y) \begin{pmatrix} ax+by \\ cx+dy\end{pmatrix} \iff ax^2 + bxy + cxy + dy^2 = 1 \iff a\sin^2(\delta)x^2 + b\sin^2(\delta)xy + c\sin^2(\delta)xy + d\sin^2(\delta)y^2 = \sin^2(\delta)$$

So, $a\sin^2(\delta)=\frac{1}{{a_1}^2}, d\sin^2(\delta)=\frac{1}{{a_2}^2}, (b+c)\sin^2(\delta)=\frac{-2\cos(\delta)}{a_1a_2}$.

**Where I'm stuck:** Beyond this, I can't seem to separate $b$ and $c$! Is one of them just going to be a free variable?

Thanks.

By the way, I have cross-posted this question on Math Stack Exchange but did not get a satisfactory answer.

Moderator Edit: Link to the Thread at Stack Exchange: matrices - Finding a positive definite matrix to satisfy the general equation of an ellipse - Mathematics Stack Exchange
 
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  • #2
An first step to this is to divide through by sin(x) to get
[tex]sin^2(x)[/tex] so that it actually is equal to 1.

write
[tex]\begin{pmatrix}x & y \end{pmatrix}\begin{pmatrix} a & b \\ c & d\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}= \begin{pmatrix}ax+ cy & cx+ dy\end{pmatrix}\begin{pmatrix}x \\ y \end{pmatrix}[/tex][tex]= ax^2+ bxy+ cxy+ dy^2= ax^2+ (b+ c)xy+ dy^2[/tex] which we compare to [tex]\frac{1}{a_1^2sin^2(\delta)}x^2- \frac{2cos(\delta)}{a_1a_2sin^2(\delta)}xy+ \frac{1}{a_2^2sin^2(\delta)}y^2= 1[/tex].

So [tex]a= \frac{1}{a_1^2sin^2(\delta)}[/tex], [tex]b+ c= \frac{2cos(\delta)}{a_1a_2asin^2(\delta)}[/tex], and [tex]d= \frac{1}{a_2^2sin^2(\delta)}[/tex]. That has a single equation for b and c so there can be many such matrices unless you have some other condition such the matrix being orthogonal- or positive definite!
 
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FAQ: Finding a positive definite matrix to satisfy the general equation of an ellipse

How can I find a positive definite matrix to satisfy the general equation of an ellipse?

To find a positive definite matrix, you can use the Cholesky decomposition method. This involves decomposing the given matrix into an upper triangular matrix and its transpose. The diagonal entries of the upper triangular matrix will be the square roots of the eigenvalues of the original matrix, which will ensure that the matrix is positive definite.

What is the general equation of an ellipse?

The general equation of an ellipse is given by xTAx = 1, where x is a column vector of the form [x, y] and A is a positive definite matrix. This equation represents all possible points on the ellipse centered at the origin with semi-major axis a and semi-minor axis b.

Why does the matrix A need to be positive definite for the equation of an ellipse?

For the equation xTAx = 1 to represent an ellipse, the matrix A must be positive definite. This ensures that the eigenvalues of A are all positive, which in turn guarantees that the ellipse is centered at the origin and has a non-zero length in all directions. If A is not positive definite, the equation may represent a degenerate ellipse or a different conic section.

Can a non-square matrix be positive definite for the equation of an ellipse?

No, for the equation xTAx = 1 to represent an ellipse, the matrix A must be a square matrix. This is because the matrix A is used to scale the vector x in all directions, and a non-square matrix cannot scale a vector in all directions uniformly.

Are there any other methods for finding a positive definite matrix for the equation of an ellipse?

Yes, apart from the Cholesky decomposition method, you can also use the eigenvalue decomposition method or the singular value decomposition method to find a positive definite matrix for the equation of an ellipse. These methods involve decomposing the matrix into its eigenvalues and eigenvectors, and then reconstructing a positive definite matrix using these decompositions.

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