# Find the covariance matrix given its eigenvectors and eigenvalues

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?

Thanks.

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:

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

$$\begin{bmatrix}\frac1{\sqrt2}&-\frac1{\sqrt2}\\\frac1{\sqrt2}&\frac1{\sqrt2}\end{bmatrix}.$$

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.

• 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? – glendon Nov 25 '17 at 1:54