Why is the square root of Cholesky decomposition equal to the lower triangular matrix?

Question

I came across this as I was learning unscented Kalman filters.

Suppose I have a symmetric and positive definite matrix $P$. I want to take its square root. After I perform the Cholesky decomposition of $P$, I get $LL^T$. One of the resources (inaccessible on the web) I am using says that when you take the square root of $LL^T$, you get the lower triangular matrix $L$. That's why I don't understand.

My understanding was that matrix $Y$ is a square root of $X$ if the matrix product $YY$ is equal to $X$. But what seems to be acceptable is that matrix Y can be a square root of X even when the matrix product $Y^TY$ is equal to $X$.

Answer 1

The source you are quoting may be found in -- https://www.hindawi.com/journals/ijno/2011/416828/ equation 18 and the wording right after it.
Same thing from the wikipedia page, https://en.wikipedia.org/wiki/Cholesky_decomposition#Kalman_filters which effectively says the same thing.

The closest answer I've found is from -- http://ais.informatik.uni-freiburg.de/teaching/ws12/mapping/pdf/slam05-ukf.pdf page 14.

It states that Cholesky Matrix Square Root is Alternative definition of the matrix square root.

Edit:

so, if you go by the alternative definition (from page 14 of the pdf file) that $L$ is the square root of $P$ when $P = LL^T$ instead of the normal (or scalar way) matrix square root definition of $P=LL$. It comes naturally that when you take the square root of $LL^T$ you get the lower triangular matrix $L$.

Answer 2

I had the same question also in reference to the UKF sigma points. But if you take the Cov of these sigma points then you see the def of Cov has a transpose on one side so you end up with the true cov. Or said another way if you took the cov of these sigma point that where parameterized by Cov P then you could get back P when solving for their cov if you use the Cholesky decomp as the square root. Hence the Cholesky decomposion does what it has to do for this application. I am looking at "A New Extension of the Kalman Filter to Nonlinear Systems" by Julier and Uhlmann.

Or more generally does this new definition of matrix square root do what it is used for. If so, then it doesn't mater what it is call sqrt or whatever.