Missing a trick: 3D Brownian Motion from a covariance matrix Construct a three-dimensional Brownian motion with covariance matrix
$M = \begin{bmatrix} a^2+b^2 & 0 & 2ab \\ 0 & c^2 & 0 \\ 2ab & 0 & a^2+b^2 \end{bmatrix}$
where $a$, $b$ and $c$ are real constants.
I know that $M$ has the spectral factorisation of the form $M = QDQ^T$ and I need to find $M^\frac{1}{2}$, which has the form $M^\frac{1}{2}=QD^\frac{1}{2}Q^T$. I know that I can do this by finding the eigenvalues and eigenvectors, with the eigenvalues making up the diagonal of $D$, and that $Q$ is the orthogonal matrix of eigenvectors.
I know that I can do all this, but I'm acutely aware that this would take a long time by hand and would be particularly arduous using the characteristic polynomial method. I also know that these values have been chosen very carefully so that we can spot something to speed up the calculation.
Can somebody please tell me what it is that I am missing from this question and offer any advice on what I should be looking out for? I have a very limited background in linear algebra so any advice would be very much appreciated.
 A: All of the eigenvalues and eigenvectors of this matrix can be found by inspection.  
In general, for a symmetric matrix you only need to find two eigenvalue/eigenvectors pairs since you get the third pair “for free:” the trace of the matrix is equal to the sum of its eigenvalues, and you can just take the cross product of the two eigenvectors you’ve found to get a third. It’s also sometimes easier to look for eigenvectors and deduce the corresponding eigenvalues from them instead of going at it the other way around.  
For this matrix in particular, observe that the second column only has a non-zero entry in its second row. This means that $M(0,1,0)^T$ will be some multiple of $(0,1,0)^T$, so there’s one eigenvector/eigenvalue pair. You know that the other eigenvectors are orthogonal to this, which means that they must be of the form $(x,0,z)^T$. So, to find other eigenvectors play around with linear combinations of the first and third columns. An easy thing to try first is their sum and difference.  
I should add that finding the eigenvalues via the characteristic polynomial isn’t all that much work for this matrix. Expanding the determinant along the second row eliminates most of the terms and gives you one of the eigenvalues immediately: $$(\lambda-c^2)((\lambda-a^2-b^2)^2-4a^2b^2)=0.$$ Simplifying and factoring the remaining term doesn’t look too terrible.
A: Start with a standard 3-dimensional Brownian motion (covariance matrix the identity) $Y=(Y_1,Y_2,Y_3)$ now define $X_1=aY_1+bY_2$, $X_2=cY_2$, $X_3=bY_1+aY_2$. Then $X=(X_1,X_2,X_3)$ has the required covariance.
