Four-dimensional Brownian motion from covariance matrix Construct a four-dimensional Browninan motion with covariance matrix
$\begin{bmatrix} 25&0&24&0\\0&4&0&0\\24&0&25&0\\0&0&0&2 \end{bmatrix}$
This must be obtained numerically by finding the symmetric positive definite matrix square-root $M^\frac{1}{2}$.
 A: To find a square root of a matrix, one usually proceeds by decomposing it in some way. For this matrix, eigenvectors and eigenvalues are fairly easy to find by inspection, so an eigenvalue diagonalization seems a natural way to proceed.  
One can follow the usual process of finding the roots of the characteristic polynomial and then computing associated eigenspaces, but the structure of this matrix lets us find eigenvectors easily without going through that. Recalling that right-multiplying a matrix by a column vector produces a linear combination of the columns of the matrix, we can immediately find two eigenvectors from the two isolated diagonal elements of $M$: $(0,1,0,0)^T$ and $(0,0,0,1)^T$, with eigenvalues $4$ and $2$, respectively. Now observe that the sum of the first and third columns is $(49,0,49,0)^T=49(1,0,1,0)^T$. Multiplying by $(1,0,1,0)^T$ adds the first and third columns, so we have another eigenvector/eigenvalue pair. Similarly, multiplying by $(1,0,-1,0)$ subtracts the third column from the first, so that’s an eigenvector of $1$.  
Normalizing these vectors to make the change of basis matrix orthogonal so that its inverse is its transpose, we have $$M=\pmatrix{1/\sqrt2&0&0&1/\sqrt2\\0&1&0&0\\1/\sqrt2&0&0&-1/\sqrt2\\0&0&1&0}\pmatrix{49&0&0&0\\0&4&0&0\\0&0&2&0\\0&0&0&1}\pmatrix{1/\sqrt2&0&1/\sqrt2&0\\0&1&0&0\\0&0&0&1\\1/\sqrt2&0&-1/\sqrt2&0}^T.$$ Taking the principal square root of the diagonal matrix and putting it all back together produces $$M^{\frac12}=\pmatrix{4&0&3&0\\0&2&0&0\\3&0&4&0\\0&0&0&\sqrt2}.$$ From there, you’re on your own since you haven’t given any details of the construction that you’re meant to use. Typically, you’ll be multiplying some normally-distributed random vector by $M^{\frac12}$ as part of the construction.
