Redundant sigma points in Unscented Kalman Filter?

According to the Unscented Transform equations in an Unscented Kalman Filter, sigma points are chosen via:

$chi^{[0]}=mu$

$chi^{[i]}=mu+left(sqrt{(n+lambda)Sigma}right)_i;;;i=1,…,n$

$chi^{[i]}=mu-left(sqrt{(n+lambda)Sigma}right)_{i-n};;;i=n+1,…,2n$

where $mu$ is a vector of variable means, $Sigma$ is the covariance matrix of the variables, and $n,lambda$ are tuning parameters.

In order to take the square root of $Sigma$, the literature suggests performing a Cholesky LL’ transformation and using the resulting L as the matrix "square root".

With L being a lower triangle matrix, the upper triangle is obviously all zeros. This ultimately populates the $chi$ matrix (sigma points) with several redundant copies of the mean $mu$.

These redundant sigma points would obviously be a poor choice, as:

• they wouldn’t spread out to sample the nonlinear function well
• they waste repeated computation on the same values

A simple example:

Assume $(n+lambda)=1$

Using Cholesky LL decomposition:

$Sigma=LL^*$

$sqrtSigma=L=begin{bmatrix}L_{0,0}&0&0\L_{1,0}&L_{1,1}&0\L_{2,0}&L_{2,1}&L_{2,2}end{bmatrix}$

Substituting this back in to the equations for $chi$:

$chi=
begin{bmatrix}
mu_0&mu_0+L_{0,0}&mu_0&mu_0&mu_0-L_{0,0}&mu_0&mu_0\
mu_1&mu_1+L_{1,0}&mu_1+L_{1,1}&mu_1&mu_1-L_{1,0}&mu_1-L_{1,1}&mu_1\
mu_2&mu_2+L_{2,0}&mu_2+L_{2,1}&mu_2+L_{2,2}&mu_2-L_{2,0}&mu_2-L_{2,1}&mu_2-L_{2,2}
end{bmatrix}$

As you can see, the top two variables (rows) have sigma points that are made up of redundant values of the variable mean.

Am I doing something wrong in my calculations, or does the use of Cholesky decomposition really result in a poor choice of sigma points?