# Change the form of equation of surface

• MHB
• mathmari
In summary, the conversation is about finding an appropriate orthonormal system of coordinates for a surface in $\mathbb{R}^3$ defined by a given equation, using symmetric matrices and the spectral theorem to transform the equation into a simplified form. The speakers discuss the steps taken to reach this simplified form and confirm the correctness of their calculations.
mathmari
Gold Member
MHB
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)

mathmari said:
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)

I like Serena said:
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)

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)

I like Serena said:
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)
I like Serena said:
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)

Yep. It looks correct to me. (Nod)

I like Serena said:
Yep. It looks correct to me. (Nod)

Great! Thanks a lot! (Smile)

## What is the purpose of changing the form of an equation of a surface?

The purpose of changing the form of an equation of a surface is to make it easier to understand and manipulate mathematically. By changing the form, we can isolate specific variables, identify key features of the surface, and make predictions about its behavior.

## How can the form of an equation of a surface be changed?

The form of an equation of a surface can be changed through mathematical operations such as factoring, completing the square, or using trigonometric identities. These operations can help simplify the equation and bring it into a more recognizable and manageable form.

## What are the benefits of changing the form of an equation of a surface?

Changing the form of an equation of a surface can provide valuable insights into its properties and behavior. It can also make it easier to solve for specific variables or make predictions about the surface's behavior under different conditions.

## Are there any limitations to changing the form of an equation of a surface?

Yes, there are limitations to changing the form of an equation of a surface. Not all equations can be simplified or manipulated into a more desirable form. Additionally, some forms may be more useful for certain applications than others, so it's important to consider the context and purpose of the equation.

## Can changing the form of an equation of a surface change its meaning or interpretation?

Yes, changing the form of an equation of a surface can change its meaning or interpretation. For example, transforming a Cartesian equation into a polar equation can reveal insights about the surface's symmetry and curvature. It's important to consider the implications of changing the form and ensure that it accurately reflects the intended meaning of the equation.

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