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[SOLVED] Reducing Quadratic Form to Principle Axes

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Hi everyone, :)

Here's a question with the summary of my method of how to solve it. I would really appreciate if you could go through it and let me know if there are any mistakes with my approach. Also are there any easier methods?

Problem:

Find an orthogonal transformation that reduces the following quadratic form to the principal axes.

\[q(x_1,\,x_2,\,x_3)=6x_{1}^{2}+5x_{2}^{2}+7x_{3}^{2}-4x_1 x_2+4x_1 x_3\]

Solution:

We reduce this to get rid of the cross terms as explained >>here<<.

\[q(x_1,\,x_2,\,x_3)=3\left(\frac{2x_1+2x_2-x_3}{3}\right)^2+9\left(\frac{-2x_1+x_2-2x_3}{3}\right)^2+6\left(\frac{-x_1+2x_2+2x_3}{3}\right)^2\]

So the matrix of the orthogonal transformation will be,

\[\begin{pmatrix}\frac{2}{3}&\frac{2}{3}&-\frac{1}{3}\\-\frac{2}{3}&\frac{1}{3}&-\frac{2}{3}\\-\frac{1}{3}&\frac{2}{3}&\frac{2}{3}\end{pmatrix}\]

Am I correct? :)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Hi everyone, :)

Here's a question with the summary of my method of how to solve it. I would really appreciate if you could go through it and let me know if there are any mistakes with my approach. Also are there any easier methods?

Problem:

Find an orthogonal transformation that reduces the following quadratic form to the principal axes.

\[q(x_1,\,x_2,\,x_3)=6x_{1}^{2}+5x_{2}^{2}+7x_{3}^{2}-4x_1 x_2+4x_1 x_3\]

Solution:

We reduce this to get rid of the cross terms as explained >>here<<.

\[q(x_1,\,x_2,\,x_3)=3\left(\frac{2x_1+2x_2-x_3}{3}\right)^2+9\left(\frac{-2x_1+x_2-2x_3}{3}\right)^2+6\left(\frac{-x_1+2x_2+2x_3}{3}\right)^2\]

So the matrix of the orthogonal transformation will be,

\[\begin{pmatrix}\frac{2}{3}&\frac{2}{3}&-\frac{1}{3}\\-\frac{2}{3}&\frac{1}{3}&-\frac{2}{3}\\-\frac{1}{3}&\frac{2}{3}&\frac{2}{3}\end{pmatrix}\]

Am I correct? :)
Yes, it is correct! :cool:

The eigenvalues of the matrix A (from your reference) are 3, 6, and 9, with corresponding eigenvectors (-2,-2,1), (-1,2,2), (2,-1,2).
As expected these are orthogonal (since the spectral theorem for real symmetric matrices applies).

So the orthogonal transformation matrix must contain these 3 vectors (normalized to unit length and possible negative) in an arbitrary order.
Your matrix does not contain them as column vectors.

Of course it's also possible you have specified the other orthogonal transformation matrix, which is the inverse, and since it's orthogonal, the transpose.
And yes, they do correspond! :D
 
Last edited:

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
I think it is almost correct. :eek:
The eigenvalues of the matrix A (from your reference) are 3, 6, and 9, with corresponding eigenvectors (-2,-2,1), (-1,2,2), (2,-1,2).
As expected these are orthogonal (since the spectral theorem for real symmetric matrices applies).

So the orthogonal transformation matrix must contain these 3 vectors (normalized to unit length and possible negative) in an arbitrary order.
However, your matrix does not seem to contain them as column vectors.

Of course it's also possible you have specified the other orthogonal transformation matrix, which is the inverse, and since it's orthogonal, the transpose.
Then I would expect the rows to match with the eigenvectors.
But that is also not the case...
I think the rows do match with your eigenvectors aren't they? But note that I have taken the eigenvectors as, (2,2,-1), (-2,1,-2) and (-1,2,2) which is essentially the same thing. Isn't? :)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
I think the rows do match with your eigenvectors aren't they? But note that I have taken the eigenvectors as, (2,2,-1), (-2,1,-2) and (-1,2,2) which is essentially the same thing. Isn't? :)
I had just noticed my mistake, and I had already edited my previous post.
So yes, it is correct!
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
I had just noticed my mistake, and I had already edited my previous post.
So yes, it is correct!
Thanks very much for the confirmation. :) That means even if I didn't find the principle axes formula I would have still obtained the orthogonal transformation matrix by considering the eigenvectors of the matrix \(A\) where \(q=z^t A z\). Isn't?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Thanks very much for the confirmation. :) That means even if I didn't find the principle axes formula I would have still obtained the orthogonal transformation matrix by considering the eigenvectors of the matrix \(A\) where \(q=z^t A z\). Isn't?
Which formula did you want to find?

You've ascertained that $q(\mathbf x)=1$ is an ellipsoid centered at the origin, with axes that are aligned with the eigenvectors, and that have half-axis-lengths of respectively $1/\sqrt 3$, $1/\sqrt 6$, and $1/3$.
See the wiki article Quadric.

What more did you want to find?
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Which formula did you want to find?

You've ascertained that $q(\mathbf x)=1$ is an ellipsoid centered at the origin, with axes that are aligned with the eigenvectors, and that have half-axis-lengths of respectively $1/\sqrt 3$, $1/\sqrt 6$, and $1/3$.
See the wiki article Quadric.

What more did you want to find?
I am sorry, what I meant was, I used,

\[q(x_1,\,x_2,\,x_3)=3\left(\frac{2x_1+2x_2-x_3}{3}\right)^2+9\left(\frac{-2x_1+x_2-2x_3}{3}\right)^2+6\left(\frac{-x_1+2x_2+2x_3}{3}\right)^2\]

to get the orthogonal transformation. Since,

\[\begin{pmatrix}y_1\\y_2\\y_3\end{pmatrix}= \begin{pmatrix}\frac{2}{3}&\frac{2}{3}&-\frac{1}{3}\\-\frac{2}{3}&\frac{1}{3}&-\frac{2}{3}\\-\frac{1}{3}&\frac{2}{3}&\frac{2}{3}\end{pmatrix} \begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}\]

I knew that the orthogonal transformation should be,

\[\begin{pmatrix}\frac{2}{3}&\frac{2}{3}&-\frac{1}{3}\\-\frac{2}{3}&\frac{1}{3}&-\frac{2}{3}\\-\frac{1}{3}&\frac{2}{3}&\frac{2}{3}\end{pmatrix}\]