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I'm trying to transform a (2D) quadratic equation $5x^2-24xy-5x=13$ into its standard form.

First, I write it as matrix form using homogeneous coordinate vector:

$$[x,y,1]\begin{bmatrix} 5&-12&-5/2\\ -12&0&0\\ -5/2&0&-13 \end{bmatrix}\begin{bmatrix} x\\ y\\ 1 \end{bmatrix}=0$$

Then, I diagonalize the matrix $A=QDQ^{-1}$:

$$\begin{bmatrix} 5&-12&-5/2\\ -12&0&0\\ -5/2&0&-13 \end{bmatrix}=\begin{bmatrix} 1.215 & -0.468 & -9.147 \\ 0.909 & -0.475 & 11.124 \\ 1 & 1 & 1 \end{bmatrix}\begin{bmatrix} -16.037 & 0 & 0 \\ 0 & -11.830 & 0 \\ 0 & 0 & 9.867 \end{bmatrix}\begin{bmatrix} 0.368 & 0.275 & 0.303 \\ -0.324 & -0.329 & 0.692 \\ -0.044 & 0.053 & 0.005 \end{bmatrix}$$

The original equation represents a 2D hyperbola with rotated and translated center (maybe also sheared?). Now, with the transformation $[x,y,1]Q$, the 2D points get projected onto a 3D coordinate system with eigenvectors Q as axes.

How to understand the third coordinate pops up in this transformation?

what is the amount of rotation and translation of the curve in 2D space?

Also, what is the meaning, geometrically, of having two different transformation matrices,

$[x,y,1]Q$ and $Q^{-1}\begin{bmatrix} x\\ y\\ 1 \end{bmatrix}$,

for the row vector and the column vector which are not transpose of each other?

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  • $\begingroup$ shouldn' A be symmetric ? $\endgroup$
    – Digitallis
    Dec 17, 2023 at 23:50
  • $\begingroup$ A used in this question is symmetric as you can see. But it is not orthogonal. $\endgroup$
    – techie11
    Dec 18, 2023 at 0:25
  • $\begingroup$ I meant not orthonormal. $\endgroup$
    – techie11
    Dec 18, 2023 at 1:17
  • $\begingroup$ It is not symmetric. That is what I wzs trying to point out. You have 12 vs -12. $\endgroup$
    – Digitallis
    Dec 18, 2023 at 7:57
  • $\begingroup$ it was a typo. I corrected it $\endgroup$
    – techie11
    Dec 18, 2023 at 17:54

1 Answer 1

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The equation you have describes a curve in a two dimensional space. There is usually no need to pass to a 3 dimensional space to solve these kinds of problems. Here you have an equation of the form $x'Ax + p'x = 13$ where $x \in \mathbb R^2$ and $$A =\begin{pmatrix} 5 & -12 \\ -12 & 0 \end{pmatrix} , p = \begin{pmatrix} -5 \\ 0 \end{pmatrix} $$

You can then diagonalise the matrix $A$ to obtain $A = ODO'$ and substitute into the initial equation to get

$$ x'OAO'x + p'x = 13 \iff (O'x)'D(O'x) + p'O(O'x) = 13$$ Then you introduce the new variables $u = O'x$ to get $$ u'Du + (p'O)u = 13$$ This is the rotation you spoke of. Since $D$ is diagonal we have that the equation is of the form

$$ (\sum_i \lambda_i u_i^2) + \text{linear combination of }u_i = 13$$

and you can make an appropriate substitution of the form $u_i \to (u_i - c_i)$ which is a translation to get rid of the linear terms.


If you want to pass to a higher dimensional space then what you did consists in looking at the intersection of $5x^2 - 24xy - 5xz - 13 z^2 = 0$ with the plane $z = 1$. That is, at solutions of

$$ \begin{cases} 5x^2 - 24xy - 5xz - 13 z^2 = 0 \\ z = 1 \end{cases}$$

When your are diagonalizing the matrix $A$ you are doing a rotation in 3D space which explains why you have three eigenvectors.

In practice you have a $3 \times 3$ matrix $A$ with the system

$$ \begin{cases} x'Ax = 0 \\ p'x = 1 \end{cases}$$ where $p' = (0,0,1)$. You diagonalise and do a change of variable as we did above. However you must be careful to also make the change of variable in the equation $p'x = 1$. That is, geometrically we are rotating both the curve $5x^2 - 24xy - 5xz - 13 z^2 = 0$ and the plane $z = 1$.

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  • $\begingroup$ The diagonalization of A: $A=ODO′$ does not hold when A is not orthonormal. it is $A=ODO^{-1}$ instead $\endgroup$
    – techie11
    Dec 18, 2023 at 17:29
  • $\begingroup$ @techie11 You can use the Gram-schimdt process to obtain an orthonormal basis of eigenvectors. So you may assume WLOG that the matrices O is orthogonal (i.e. that $O^{-1} = O'$. The fact that $A$ is not orthonormal is irrelevant $\endgroup$
    – Digitallis
    Dec 18, 2023 at 17:32

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