How to construct the normalized set of random variables? I have read the paper Narrow Reliability Bounds for Structural Systems.
The author stated on page 458 that there exists a transformation such that a $n$-set of random variables $X=(X_1,...,X_n)$ can be normalized. Here, the term 'normalized' means transforming $X$ into a set of variables $Y=(Y_1,...,Y_n)$ with all means equal to zero, all standard deviations equal to one, and all covariances equal to zero.
The author further stated that

It is well known from the general theory of random variables that such a transformation exists if the covariance matrix of $X$ is regular.

$X$ in the above quotation is the original $n$-set of random variables before normalization.
I was wondering how to construct such a transformation to make all covariances equal to $0$. Besides, I was also wondering what the term 'regular' means in the above quotation.
Thank you for your kind help.
 A: This is true whenever $Cov(X,X)$ is invertible.
Assuming $E[X_i^2]<\infty$ for all $i \in \{1, ..., n\}$, we know $Cov(X,X)$ is a symmetric and positive semidefinite $n \times n$ matrix.  It has the decomposition into a product of three $n\times n$ matrices:
$$ Cov(X,X) = U\Lambda U^{\top}$$
where $UU^{\top}=U^{\top}U=I$ (with $I$ being the $n \times n$ identity matrix) and $\Lambda$ is purely diagonal with nonnegative diagonal elements $\lambda_1, ..., \lambda_n$. If $Cov(X,X)$ is invertible then $\lambda_i>0$ for all $i \in \{1, ..., n\}$. Assuming $Cov(X,X)$ is invertible we can define the following invertible $n \times n$ matrix:
$$ A = VU^{\top}$$
where $V$ is a purely diagonal $n\times n$ matrix with entries $1/\sqrt{\lambda_1}, ..., 1/\sqrt{\lambda_n}$ on the diagonal. This makes sense since $\lambda_i>0$ for all $i$. Define $$Y=A(X-E[X])$$
Then $E[Y]=0$ and
$$Cov(Y,Y) =Cov(AX,AX) = ACov(X,X)A^{\top} = V \Lambda V = I$$

The sequence $Y_1, Y_2, ..., Y_n$ is called the "innovation sequence" and can also be obtained by the Gram-Schmidt procedure of linear algebra.
