MHB Change the form of equation of surface

[mathmari]

Hey!

We consider the surface $S$ of the space $\mathbb{R}^3$ that is defined by the equation $2(x^2+y^2+z^2-xy-xz-yz)+3\sqrt{2}(x-z)=1$.

I want to find (using symmetric matrices) an appropriate orthonormal system of coordinates $(x_1, y_1, z_1)$ for which the above equation has the form $ax_1^2+by_1^2+cz_1^2=d$, for some $a,b,c,d\in \mathbb{R}$.
I have done the following:

The eqquation of the surface $S$ can be written in the form $$\vec{x}^T\begin{pmatrix}2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2\end{pmatrix}\vec{x}+\begin{pmatrix}3\sqrt{2} & 0 & -3\sqrt{2}\end{pmatrix}\vec{x}=1$$ Since that matrix is symmetric it is diagonalizable with orthonormal basis.

Then we have to write the matrix $\begin{pmatrix}2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2\end{pmatrix}$ is the form $PDP^{-1}$, or not?

What do we do next?

(Wondering)

 

[I like Serena]

The eqquation of the surface $S$ can be written in the form $$\vec{x}^T\begin{pmatrix}2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2\end{pmatrix}\vec{x}+\begin{pmatrix}3\sqrt{2} & 0 & -3\sqrt{2}\end{pmatrix}\vec{x}=1$$ Since that matrix is symmetric it is diagonalizable with orthonormal basis.

Then we have to write the matrix $\begin{pmatrix}2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2\end{pmatrix}$ is the form $PDP^{-1}$, or not?

Hi mathmari!

Yes, and moreover $P$ is orthogonal so that $P^{-1}=P^T$ (Spectral theorem).
So we have:
$$\vec{x}^TPDP^T\vec{x}+\begin{pmatrix}3\sqrt{2} & 0 & -3\sqrt{2}\end{pmatrix}\vec{x}=1 \quad\Rightarrow\quad
(P^T\vec{x})^TD(P^T\vec{x})+\begin{pmatrix}3\sqrt{2} & 0 & -3\sqrt{2}\end{pmatrix}\vec{x}=1$$
Now suppose we substitute $\vec y = P^T\vec{x}$? (Wondering)

 

[mathmari]

Yes, and moreover $P$ is orthogonal so that $P^{-1}=P^T$ (Spectral theorem).
So we have:
$$\vec{x}^TPDP^T\vec{x}+\begin{pmatrix}3\sqrt{2} & 0 & -3\sqrt{2}\end{pmatrix}\vec{x}=1 \quad\Rightarrow\quad
(P^T\vec{x})^TD(P^T\vec{x})+\begin{pmatrix}3\sqrt{2} & 0 & -3\sqrt{2}\end{pmatrix}\vec{x}=1$$
Now suppose we substitute $\vec y = P^T\vec{x}$? (Wondering)

Then we get $$
\vec{y}^TD\vec{y}+\begin{pmatrix}3\sqrt{2} & 0 & -3\sqrt{2}\end{pmatrix}P\vec{y}=1 \ \ \ \ (\star)$$

We have the matrices \begin{equation*}P=\begin{pmatrix}1 & -1 & -1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{pmatrix} \ \text{ and } \ D=\begin{pmatrix}0 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{pmatrix}\end{equation*}

Then we get from $(\star)$ :
\begin{align*}&\begin{pmatrix}x_1 & y_1 & z_1\end{pmatrix}\begin{pmatrix}1 & -1 & -1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{pmatrix}\begin{pmatrix}x_1 \\ y_1 \\ z_1\end{pmatrix}+\begin{pmatrix}3\sqrt{2} & 0 & -3\sqrt{2}\end{pmatrix}\begin{pmatrix}1 & -1 & -1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{pmatrix}\begin{pmatrix}x_1 \\ y_1 \\ z_1\end{pmatrix}=1 \\ & \Rightarrow \begin{pmatrix}x_1 & y_1 & z_1\end{pmatrix}\begin{pmatrix}0 \\ 3y_1 \\ 3z_1\end{pmatrix}+\begin{pmatrix}3\sqrt{2} & 0 & -3\sqrt{2}\end{pmatrix}\begin{pmatrix}x_1-y_1-z_1 \\ x_1+z_1 \\ x_1+y_1\end{pmatrix}=1 \\ &\Rightarrow 3y_1^2+3z_1^2+3\sqrt{2}(x_1-y_1-z_1)-3\sqrt{2}(x_1+y_1)=1 \\ & \Rightarrow 3y_1^2+3z_1^2+3\sqrt{2}x_1-3\sqrt{2}y_1-3\sqrt{2}z_1-3\sqrt{2}x_1-3\sqrt{2}y_1=1 \\ & \Rightarrow 3y_1^2+3z_1^2-6\sqrt{2}y_1-3\sqrt{2}z_1=1 \end{align*}

Is everything correct so far?

How could we continue?

(Wondering)

 

[I like Serena]

It looks correct to me.
(You substituted P instead of D though, although afterwards you did calculate as if you had substituted D.)

How about we try to write it as $3(y_1 -B)^2 + 3(z_1-C)^2 = d$? (Wondering)

 

[mathmari]

It looks correct to me.
(You substituted P instead of D though, although afterwards you did calculate as if you had substituted D.)

Oh yes (Blush)

How about we try to write it as $3(y_1 -B)^2 + 3(z_1-C)^2 = d$? (Wondering)

We have $$3(y_1^2 -2By_1+B^2) + 3(z_1^2-2Cz_1+C^2) = d\Rightarrow \ldots \Rightarrow 3y_1^2+3z_1^2-6By_1-6Cz_1=d-3B^2-3C^2$$
Since this has to be equal to $$3y_1^2+3z_1^2-6\sqrt{2}y_1-3\sqrt{2}z_1=1$$ we get $$B=\sqrt{2},\ C=\frac{\sqrt{2}}{2}, \ d=\frac{17}{2}$$ So for $$u:=y_1-\sqrt{2}, \ v:=z_1-\frac{\sqrt{2}}{2}$$ we get $$3u^2+3v^2=\frac{17}{2}$$ Is the way correct? (Wondering)

 

[I like Serena]

Yep. It looks correct to me. (Nod)

 

[mathmari]

Yep. It looks correct to me. (Nod)

Great! Thanks a lot! (Smile)