Show that the equation represents an ellipsoid.

Question

Show that the equation $$\frac{4}{3}{x^2} + \frac{4}{3}{y^2} + \frac{4}{3}{z^2} + \frac{4}{3}{xy} + \frac{4}{3}{xz} + \frac{4}{3}{yz} = 1$$ represents an ellipsoid. Find the position and lengths of its principal half-axes.

Workings $$(x+y)^2+(y+z)^2+(z+x)^2=\frac32\,.$$ Common factor $\left(\dfrac32\right)^{\frac12}$ not sure how to find length and position half-axes??

Answer 1

We have $$\frac32 = (x+y)^2 + (y+z)^2 + (x+z)^2 = 2x^2 + 2y^2 + 2z^2 + 2xy + 2yz + 2xy$$

so set $A =\begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2\end{bmatrix}$ and the equation can be rewritten as $$\left\langle A\begin{bmatrix} x \\ y \\ z\end{bmatrix},\begin{bmatrix} x \\ y \\ z\end{bmatrix}\right\rangle = \frac32$$

Now diagonalize $A$:

$$A= P^TDP = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{3}} \\ 0 & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \\\end{bmatrix}^T \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 4\end{bmatrix}\begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{3}} \\ 0 & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \\\end{bmatrix}$$

so we have

$$\frac32 = \left\langle A\begin{bmatrix} x \\ y \\ z\end{bmatrix},\begin{bmatrix} x \\ y \\ z\end{bmatrix}\right\rangle = \left\langle P^TDP\begin{bmatrix} x \\ y \\ z\end{bmatrix},\begin{bmatrix} x \\ y \\ z\end{bmatrix}\right\rangle = \left\langle DP\begin{bmatrix} x \\ y \\ z\end{bmatrix},P\begin{bmatrix} x \\ y \\ z\end{bmatrix}\right\rangle$$

Therefore we can define new coordinates $$\begin{bmatrix} x' \\ y' \\ z'\end{bmatrix} = P\begin{bmatrix} x \\ y \\ z\end{bmatrix} = \begin{bmatrix} \frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}+\frac{z}{\sqrt{3}} \\ \frac{z}{\sqrt{3}}-\frac{x}{\sqrt{2}} \\ \frac{z}{\sqrt{3}}-\frac{y}{\sqrt{2}} \\ \end{bmatrix}$$

so we get $$\frac32 = \left\langle D\begin{bmatrix} x' \\ y' \\ z'\end{bmatrix},\begin{bmatrix} x' \\ y' \\ z'\end{bmatrix}\right\rangle = x'^2 + y'^2 + 4z'^2$$

or $$\left(\frac{x'}{\sqrt{\frac32}}\right)^2 + \left(\frac{y'}{\sqrt{\frac32}}\right)^2 + \left(\frac{z'}{\frac12\sqrt{\frac32}}\right)^2 = 1$$

Answer 2

Sometimes it can be less work to find eigenvectors first. In this case, you can save yourself some effort by observing that the equation is invariant under permutations of the variables, therefore $(1,1,1)^T$ is an axis of symmetry. The line with this direction vector is thus one of the principal axes of this ellipsoid. Moreover, it is rotationally symmetric about this axis (why?), so it is a spheroid and we can choose any pair of orthogonal lines in the plane $x+y+z=0$—i.e., lines orthogonal to the symmetry axis—as its other principal axes, e.g., the lines with direction vectors $(1,-1,0)^T$ and $(1,1,1)^T\times(1,-1,0)^T=(1,1,-2)^T$.

With eigenvectors in hand, computing the corresponding eigenvalues is quite simple. The associated matrix, which I believe you’ve already worked out, is $$Q = \frac23 \begin{bmatrix}2&1&1\\1&2&1\\1&1&2\end{bmatrix}.$$ There’s a pretty well-known result about eigenvalues and eigenvectors of matrices of this type, but we’ll proceed as if we didn’t know about that. We have $Q(1,1,1)^T = \frac83(1,1,1)^T$, so the eigenvalue for this eigenvector is $\frac83$, resulting in a semiaxis length of $\sqrt{3/8}$. The other eigenvalue is repeated, so we can obtain it from the trace: $\frac12\left(4-\frac83\right) = \frac23$, for semiaxis lengths of $\sqrt{3/2}$.