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.

 A: 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.
A: I believe a more general answer is as follows:  Given the eigenvector matrix E (with eigenvectors as columns) and the eigenvalue matrix ${\mathbf{\Lambda}}$, where the diagonal elements are the scalar eigenvalues $\lambda$ and the off-diagonal elements are zero, then the covariance matrix S is given by the matrix product
S = E ${\mathbf{\Lambda}}$ E$^{\rm T}$
