Vtyjkyuk

I am reading the paper "Detection of Signals by Information Theoretic Criteria" by Wax and Kailath.

In this paper, a vector $Theta$ is defined according to:

$Theta = (lambda_1, ..., lambda_k, sigma^2, V_1^T, ..., V_k^T)$ where $lambda_i$ are the eigenvalues of a covariance matrix (which is symmetric so the eigenvalues are real), $V_i$ are complex eigenvectors of length $p$, and $sigma^2$ is a real number.

It is said that there are $k+1+2pk$ parameters in this vector, which makes sense because there are $k$ eigenvalues; $1$ parameter $sigma^2$; and there are $k$ vectors of length $p$ with $2$ parameters in each component (accounting for the real and the imaginary part). So far, so good.

Now I get confused. The authors constrain the eigenvectors to have unit norm and to be mutually orthogonal, and say that the reduction of the degrees of freedom is $2k$ due to normalization and $2 frac12k(k-1)$ due to mutual orthogonalization. I don't understand why.

I would have said that each normalization equation reduces $1$ degree of freedom, so $k$ of them (one for each eigenvector) would reduce $k$ degrees, instead of $2k$. Also, I see that the mutual orthogonalization of $k$ vectors yields $frac12k(k-1)$ equations, and the same argument as before makes me think that this is also the number of degrees reduced due to this condition. You see that in both cases I am missing a factor of $2$ , do you know where does it come from?

I think by normalization something more is intended. To see this precisely, see that if $v$ is an eigenvector of $A$ then for any $cinmathbb C$, $cv$ is also an eigenvector. Therefore if $v$ is non-zero and unit norm, then one can make an entry of the vector $v$ always to be real by proper normalization and still keeping the unit norm. After this there are only $2(p-1)$ free parameters. So the normalization is not by $|v|_2$ but by $|v|_2e^-ivarphi$ where $re^ivarphi$ is a non-zero entry of $v$.

For the orthogonal condition, this is much simpler to see. Just consider the following example:

$$
(a+ib)(c+id)^*=(ac+bd)+i(bc-ad).
$$
If the above product is zero we loose two degrees of freedom one for the real part and another one for the complex part. Therefore each orthogonal condition takes 2 degrees of freedom.