# Multivariate normal distribution transformation

Suppose that $$X$$ has a multivariate normal distribution $$X\sim MVN (\mu, \Sigma)$$, How can I transform $$X$$ into $$Z$$ so that $$Z\sim MVN(\mu, I)$$ where $$I$$ is the identity matrix?

For instance, let $$\mu= \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$$ and the variance-covariance matrix $$\Sigma= \begin{bmatrix} 0.75 & -0.09& 0.33\\ -0.09 & 0.37& 0.10\\ 0.33 & 0.10& 0.29 \\ \end{bmatrix}$$

I tried to use the singular value decomposition (SVD) and calculate the eigenvalues $$\lambda_{1},\lambda_{2}, \lambda_{3}$$ and eigenvectors $$e_{1}, e_{2}, e_{3}$$, but I do not know how to continue or if my approach is correct.

Your approach using SVD is one way to do it. Let $$\Sigma$$ decomposed by $$UDU^{\rm T}$$, where $$D={\rm diag}(\lambda_1, \ldots, \lambda_p)$$ and $$U$$ is a matrix with its columns the eigenvectors. Then, define $$K=UD^{1/2}$$, which satisfies $$KK^{\rm T} = \Sigma$$. Now multiply the inverse matrix of $$K$$ on the original vector $$X$$, i.e. $$K^{-1}(X-\mu) \sim N(0, K^{-1}\Sigma (K^{-1})^{\rm T}).$$ By noting that $$K^{-1}\Sigma (K^{-1})^{\rm T}=I$$, we get the desired result.
• Is $K$ a Cholesky factor of $\Sigma$?
• The Cholesky decomposition is based on a lower triangular matrix, so $K$ is not a Cholesky factor. Commented Feb 13, 2020 at 21:06