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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 $M^\frac{1}{2}$.

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    OK, so what are you stuck on? – 2017-01-02
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    @amd, if I'm honest I don't even know where to start. Could you please give me a hint? – 2017-01-04
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    Start by computing the square root of $M$: diagonalize it into $M=V^{-1}DV$, set $S$ to the diagonal matrix consisting of square roots of elements of $D$ and then compute $VDV^{-1}$. Since $M$ is symmetric, you can find an orthonormal set of eigenvectors, so that $V^{-1}=V^T$. It’s probably easiest to find some eigenvectors first and compute their associated eigenvalues. Both $(0,1,0,0)$ and $(0,0,0,1)$ are obviously eigenvectors with eigenvalues $4$ and $2$, respectively. The similarity of the first and third rows should make it easy to find another two orthogonal eigenvectors. – 2017-01-05
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    @amd Thanks. I will try tackling it using that approach. – 2017-01-07
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    Oops. There’s a typo above: it should say “compute $V^{-2}DV$.” – 2017-01-08
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    @amd Okay. So I've calculated the eigenvalues: $\lambda_1 = 49, \lambda_2 = 4, \lambda_3 = 2$ and $\lambda_4 = 1$. Their corresponding eigenvectors are: $v_1 = (-1,0,-1,0), v_2 = (0,1,0,0), v_3 = (0,0,0,1)$ and $v_4 = (-1,0,1,0)$. This took me a long time to do and this makes me think there is a quicker way of doing this as it is a past exam question. I've read online that $M^\frac{1}{2} = \alpha I + \beta \bf u \bf u^T$. Do you know a way of doing this using the above? Many thanks for your help so far. – 2017-01-08
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    D’oh! That should be $V^{-1}$ above, of course. I’ve not seen this decomposition before. It writes the square root as a linear combination of the identity and projection onto some unit vector $\mathbf u$. This only works if $M$ itself can be so decomposed ($(\alpha I+\beta\mathbf u\mathbf u^T)^2=\alpha^2I+(\beta^2+2\alpha\beta)\mathbf u\mathbf u^T$), but it can’t in this case. If you try to solve $M=\gamma I+\delta\mathbf u\mathbf u^T$, you end up with $4=2$ at some point. Eigenvectors and eigenvalues of this particular matrix $M$ are actually pretty easy to find. See my answer, below. – 2017-01-10

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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.

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    That's great. Really clear explanation. Thanks. – 2017-01-11