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:
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.