Given the point cloud below and that the ratio of its eigenvalues is 3, could someone guide me to find the covariance matrix of the point-cloud below?

What I figured so far: - From the diagram, I think the two eigenvectors should be [-1, 1] (along greatest variance) and [1, 1] (orthogonal to [-1, 1]).

  • And using the Eigenvector formula, Ax = λx, I should get the following equations:
a) A[-1 1]' = 3[-1 1]'
b) A[1 1]' = 1[1 1]'

Is my thought-process correct? If yes, how should I carry on further?


enter image description here


This is apparently a uniform distribution over a rectangular region whose dimensions are not specified. Turn the region around the origin by $-45^{\circ}$. If this is the case then the distribution will look like this:

enter image description here So, we have two independent random variables both uniformly distributed over $[-a,a]$ and $[-b,b]$, respectively.

The covariance matrix is easy to calculate now:

$$\begin{bmatrix}\frac{a^2}3&0\\ 0&\frac{b^2}3\end{bmatrix}.$$

The eigenvectors are $$\begin{bmatrix}1\\ 0\end{bmatrix} \text{ and } \begin{bmatrix}\ 0\\ 1\end{bmatrix}$$ and te corresponding eigenvalues are $$\frac{a^2}3\text{ and } \frac{b^2}3.$$

Utilizing the fact that the ratio of the eigenvalues is $3$ we can tell that the covariance matrix is

$$\begin{bmatrix}\frac{a^2}3&0\\ 0&a^2\end{bmatrix}.$$

But this is the rotated covariance matrix. We have to turne back the experiment by $45^{\circ}$. The rotation matrix is


So, the covariance matrix is

$$\begin{bmatrix}\frac1{\sqrt2}&-\frac1{\sqrt2}\\\frac1{\sqrt2}&\frac1{\sqrt2}\end{bmatrix}\begin{bmatrix}\frac{a^2}3&0\\ 0&a^2\end{bmatrix}=a^2\begin{bmatrix}\frac1{3\sqrt2}&-\frac1{\sqrt2}\\\frac1{3\sqrt2}&\frac1{\sqrt2}\end{bmatrix}.$$

$a$ is still unknown. Notice that only one equation was givan and there were two unknowns.

  • $\begingroup$ Wow! Thanks for the clear logic. Please bear with me for my 2 follow-up questions: If the covariance matrix is \begin{bmatrix} x^2 & xy \\ xy & y^2 \end{bmatrix}, then (a) why do we need to divide both x^2 and y^2 by 3 at the very beginning and (b) why is xy = 0? $\endgroup$ – glendon Nov 25 '17 at 1:54

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