0
$\begingroup$

Let $\Sigma$ be the the covariance matrix for some correlation coefficient $\rho>0$. $$\Sigma=\begin{bmatrix}1 & \rho &\rho&\rho&\rho\\\rho&1&\rho&\rho&\rho\\\rho&\rho&1&\rho&\rho\\\rho&\rho&\rho&1&\rho\\\rho&\rho&\rho&\rho&1\end{bmatrix}$$ Find the eigenvalues and eigenvectors of this covariance matrix.

I'm looking for some trick to find the eigenvalues and eigenvectors, because if I do $$det(\Sigma-\lambda I)=0$$ it will be very complicated to find the eigenvalues.

Any property that I can use in this case?

$\endgroup$
2
$\begingroup$

You can see in it the usual matrix:

$$\begin{bmatrix}\rho & \rho &\rho&\rho&\rho\\\rho&\rho&\rho&\rho&\rho\\\rho&\rho&\rho&\rho&\rho\\\rho&\rho&\rho&\rho&\rho\\\rho&\rho&\rho&\rho&\rho\end{bmatrix}=\rho J$$

where $J$ is the matrix full of 1. This matrix $J$ is an interesting one: what can be said of $J^2$? By this remarks you can find the reduction of $J$, in particular its eigenvalues and eigenvectors.

Then, what is the relations between $\rho J$ and your initial matrix? You see that:

$$\Sigma : \rho J + (1-\rho)I_n$$

and since $(1-\rho)I_n$ is a scalar matrix, it is the same in every base, so finding a base of diagonalization of $J$ is enough to gives you the complete reduction of $\Sigma$.

$\endgroup$
0
$\begingroup$

Clearly, $\sum=\rho J_n+(1-\rho)I_n$. The eigenvalues of $J_n$ are $n$ (with multiplicity $1$), $0$ (with multiplicity $n-1$). The eigenvectors corresponding to $n$ of $J_n$ is $\mathbf{1}_n$, the vector of all ones. The rest all eigenvectors of $J_n$ are orthogonal to $\mathbf{1}_n$. So one can choose them as the vectors $\varepsilon_i$, $i=2, \ldots, n$, where $\varepsilon_i(j)=\begin{cases}1, &\text{ if }j=1\\-1 & \text{ if }j=i,\\0 & \text{ otherwise.}\end{cases}$

Now, as the eigenvectors of $\sum$ are also the eigenvectors of $J_n$, the eigenvalues of $\sum$ are $\rho n+(1-\rho)=\rho(n-1)+1$ (with multiplicity $1$) and $1-\rho$ (with multiplicity $n-1$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.