Identify the quadratic form of the given equation

[lesdes]

<Moderator's note: Moved from a technical forum and thus no template.>

Hello I am given the following problem to solve.

Identify the quadratic form given by $-5x^2 + y^2 - z^2 + 4xy + 6xz = 5$.
Finally, plot it.

I cannot seem to understand what I have to do. The textbook chapter on quadratic forms simply just forms the matrix of the given equation. Is that what I am expected to do?

It would be nice if someone could point me in the correct direction or link me to a relevant document/article.

 

[fresh_42]

Hello I am given the following problem to solve.

Identify the quadratic form given by $-5x^2 + y^2 - z^2 + 4xy + 6xz = 5$.
Finally, plot it.

I cannot seem to understand what I have to do. The textbook chapter on quadratic forms simply just forms the matrix of the given equation. Is that what I am expected to do?

It would be nice if someone could point me in the correct direction or link me to a relevant document/article.

I assume you should say what it is: a hyperboloid, a sphere or whatever which form it has. The matrix is a first step, to bring it into a form without mixed terms is the next.

 

[lesdes]

I assume you should say what it is: a hyperboloid, a sphere or whatever which form it has. The matrix is a first step, to bring it into a form without mixed terms is the next.

Okay so let's see if I am on the right path. What I have done is that I have written $−5x2+y2−z2+4xy+6xz=5$ in the form
$$
(x y z)
\begin{pmatrix}
-5 & 2 & 3 \\
2 & 1 & 0 \\
3 & 0 & -1
\end{pmatrix}
\begin{pmatrix}
x \\
y \\
z
\end{pmatrix}
= 5
$$

as suggested by the book:

is then

.

Next, I used the Principal Axes Theorem

and formed the determinant
$\begin{vmatrix} -5 -\lambda & 2 & 3 \\ 2 & 1 -\lambda & 0 \\ 3 & 0 & -1 -\lambda \end{vmatrix}$

which then gives the eigenvalies $\lambda_1 = -7$, $\lambda_2 = 2$ and $\lambda_3 = 0$. Then, the eigenvectors are $\vec v_1 = (-4, 1, 2)$, $\vec v_2 = (1, 2, 1)$, and $\vec v_3 = (1, -2, 3)$. Then, I formed the matrix

$P = \begin{pmatrix} -4 & 1 & 1 \\ 1 & 2 & -2 \\ 2 & 1 & 3 \end{pmatrix}$

Then, I calculated $P^{T}AP$ such that
$\begin{pmatrix} -147 & 0 & 0 \\ 0 & 12 & 0 \\ 0 & 0 & 0 \end{pmatrix}$.
And finally,
$(x_1 y_1 z_1) \begin{pmatrix} -147 & 0 & 0 \\ 0 & 12 & 0 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} -x_1\\ y_1 \\ z_1 \end{pmatrix}$

gave me $-147x_1^2 + 12y_1^2 = 5$. Am I correct so far?

And what next? Do I have to say what this last equation is geometrically? How do I find that out?

 

[fresh_42]

I haven't gone through the details yet, but the plot looks o.k. But you still need a $z-$component. It cannot be only a two dimensional object, as it was a three dimensional at the start. And isn't there a sign error in the eigenvalue matrix?

 

[lesdes]

I haven't gone through the details yet, but the plot looks o.k. But you still need a $z-$component. It cannot be only a two dimensional object, as it was a three dimensional at the start. And isn't there a sign error in the eigenvalue matrix?

Yes there is a sign error in the eigenvalue matrix. But I only made the error when typing it in this thread. I will correct that now.

You do not have to check whether the eigenvalues, eigenvectors and the multiplication $P^TAP$ is correct. I let Mathematica do the computations.
But where do I get the third component from? As it is not there anymore due to the last multiplication. Also in the book example, after applying this procedure the last equation was two dimensional. Is this a mistake?

 

[Dick]

You do not have to check whether the eigenvalues, eigenvectors and the multiplication $P^TAP$ is correct. I let Mathematica do the computations.
But where do I get the third component from? As it is not there anymore due to the last multiplication. Also in the book example, after applying this procedure the last equation was two dimensional. Is this a mistake?

The third component is hidden. It's $0 {z_1}^2$ since there's a zero eigenvalue. It's a type of cylinder.

 

[fresh_42]

I let Mathematica do the computations.

I guess Mathematica also needs a correct matrix, but maybe you have a special version.

 

[vela]

Then, the eigenvectors are $\vec v_1 = (-4, 1, 2)$, $\vec v_2 = (1, 2, 1)$, and $\vec v_3 = (1, -2, 3)$. Then, I formed the matrix

$P = \begin{pmatrix} -4 & 1 & 1 \\ 1 & 2 & -2 \\ 2 & 1 & 3 \end{pmatrix}$

You need to normalize the eigenvectors before forming the matrix P. The diagonalized matrix will then have the eigenvalues on the diagonal. In fact, the theorem you quoted stated this. So once you found the eigenvalues, you knew enough to figure out the geometric shape.

 

[Ray Vickson]

Yes there is a sign error in the eigenvalue matrix. But I only made the error when typing it in this thread. I will correct that now.

You do not have to check whether the eigenvalues, eigenvectors and the multiplication $P^TAP$ is correct. I let Mathematica do the computations.
But where do I get the third component from? As it is not there anymore due to the last multiplication. Also in the book example, after applying this procedure the last equation was two dimensional. Is this a mistake?

Something is wrong: for the matrix
$$A = \begin{bmatrix} -5 & 2 & 3 \\ 2 & 1 & 0 \\3 & 0 & -1 \end{bmatrix}$$
I let Maple do the computations, and got eigenvalues $-7, 2, 0$ and the associated eigenvectors. You need to remember that when forming the matrix $P$ you must use normalized eigenvectors (that is, forming an orthonormal set). Doing that in Maple I get
$$P = \begin{bmatrix}
-4/\sqrt{21} & 1/\sqrt{6} & 1/\sqrt{14} \\
1/\sqrt{21} & \sqrt{2/3} & -\sqrt{2/7} \\
2/\sqrt{21} & 1/\sqrt{6} & 3/\sqrt{14}
\end{bmatrix}
$$
This gives
$$P^T A P = \begin{bmatrix} -7 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$
as it should.

 

[lesdes]

Something is wrong: for the matrix
$$A = \begin{bmatrix} -5 & 2 & 3 \\ 2 & 1 & 0 \\3 & 0 & -1 \end{bmatrix}$$
I let Maple do the computations, and got eigenvalues $-7, 2, 0$ and the associated eigenvectors. You need to remember that when forming the matrix $P$ you must use normalized eigenvectors (that is, forming an orthonormal set). Doing that in Maple I get
$$P = \begin{bmatrix}
-4/\sqrt{21} & 1/\sqrt{6} & 1/\sqrt{14} \\
1/\sqrt{21} & \sqrt{2/3} & -\sqrt{2/7} \\
2/\sqrt{21} & 1/\sqrt{6} & 3/\sqrt{14}
\end{bmatrix}
$$
This gives
$$P^T A P = \begin{bmatrix} -7 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$
as it should.

Ah yes now I see. I have the same result through mathematica.

So now I have $-7x^2 + 2y^2 = 5$. This result should be equivalent to $−5x^2+y^2−z^2+4xy+6xz=5$ as I understand it? Hence if I graph both in mathematica, then I should receive the same graph?

 

[lesdes]

So now I've plotted both equations in mathematica and I have the following results

It might be that I did not understand the quadratic form purpose. I thought the purpose was to write a more complicated equation into a simpler form. I am a little bit confused now.

 

[Ray Vickson]

Ah yes now I see. I have the same result through mathematica.
View attachment 220952

So now I have $-7x^2 + 2y^2 = 5$. This result should be equivalent to $−5x^2+y^2−z^2+4xy+6xz=5$ as I understand it? Hence if I graph both in mathematica, then I should receive the same graph?

The variables will be different: using
$$\pmatrix{x'\\y'\\z'} = P \pmatrix{x\\y\\z}, $$
your equation in the "primed" variables will be
$$-7 x'^2 + 2 y'^2 = 5,$$
which is a cylinder parallel to the $z'$-axis in $(x',y',z')$-space. That will be a rotated cylinder in $(x,y,z)$-space.