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I am working with sampling of multivariate normally distributed numbers. I have a very fundamental question regarding the eigendecomposition of the $k \times k$ covariance matrix $\Sigma = \mathbf{U}\mathbf{\Lambda}\mathbf{U}^{-1} = \mathbf{U}\mathbf{\Lambda}\mathbf{U}^{T}$, where $U = [\mathbf{u}_1 \mathbf{u}_2 \dots \mathbf{u}_k]$ is an orthonormal (or at least orthogonal) matrix with eigenvectors and $\mathbf{\Lambda}$ is a diagonal matrix with the corresponding eigenvalues on the diagonal.

The eigendecomposition can also be further decomposed into $\Sigma = \mathbf{U}\mathbf{\Lambda}\mathbf{U}^{T} = ({\mathbf{U}\Lambda}^{1/2}){({\mathbf{U}\Lambda}^{1/2})}^{T}$.

But there is also the Cholesky factorization: $\mathbf{\Sigma} = \mathbf{L}{\mathbf{L}}^{T}$.

For real, symmetric, and positive-definite matrices $\mathbf{\Sigma}$ - what is the general relationship between $\mathbf{U}{\Lambda}^{1/2}$ and $\mathbf{L}$?

What I want to do, is to use $\mathbf{U}{\Lambda}^{1/2}$ (or $\mathbf{L}$, but I believe it might be wrong) to generate normally distributed numbers. Let's say that I have a $n \times k$ matrix $\mathbf{X}$ with $n$ points $\mathbf{x}_i \in \mathbb{R}^k$ as rows. These points are drawn from $N(\mathbf{0}, \mathbf{I})$ (done with a pseudo-random multivariate generator).

The following relationship exists: $X \sim N(\mathbf{\mu}, \mathbf{\Sigma}) \iff X \sim \mathbf{\mu}+\mathbf{U}{\Lambda}^{1/2}N(\mathbf{0}, \mathbf{I}) \iff X \sim \mathbf{\mu}+\mathbf{U}N(\mathbf{0}, \mathbf{\Lambda})$ for random variable $X$ taking its value from a multivariate normal distribution with mean $\mathbf{\mu}$ and covariance $\mathbf{\Sigma}$.

Using this knowledge I then apply $\mathbf{U}{\Lambda}^{1/2}$ as a scaling and rotation operator on $\mathbf{X}^T$ and my data are then distributed as $N(\mathbf{\mu}, \mathbf{\Sigma})$ (as far as I understand).

Can I use $\mathbf{L}$ here instead as the operator? And what does it mean if I do?


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You can use either $L$ or $U \Lambda^{1/2}$ to generate you multivariate normal. They are equivalent, but have different interpretions. The $U \Lambda^{1/2}$ gives you the nice picture of a orthogonal rotation plus a scaling of the axis. The Cholesky $L$ factorization can be related to a Gram–Schmidt decomposition (more so if seen as the reverse transformation, from the arbitrary gaussian to the canonical one) or, if we regard the components as being indexed in time (like a stochastic process) it can be see as a "causal" filter, in the sense that each $x_i$ is obtained as a linear combination of the "previous" values from the canonical normal, say:

$x_1 = L_{1,1} z_1$

$x_2 = L_{1,2} z_1 + L_{2,2} z_2$



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Thanks - I believe the $U{\Lambda}^{1/2}$ is the most appropiate for my work right now. I am trying to understand how to use the Mahalanabis distance for supervised image segmentation. But just wanted to understand how normally distributed numbers are really created and how the change of basis works (the size of the ellipse volume is tricky though - but that is another question I think). – Ole Thomsen Buus Jul 6 '11 at 19:47
Yes, that is the most appropiate for Mahalanobis distance, as this can be related to the hiperellipsoids (level curves of the gaussian), which are evident in the $U{\Lambda}^{1/2}$ form. – leonbloy Jul 6 '11 at 19:52
BTW: note that $L$ can be obtained numerically in finite steps (Cholesky), but the other -in general- not. – leonbloy Jul 6 '11 at 19:53
Level curves of the Gaussian... That is actually the concept that I am working with. These levels will exist at constant values of the Mahalanobis distance. The hyperellipsoid will have the following semiaxes: $c\sqrt{{\lambda}_i}\mathbf{e}_i$ where $c^2={\chi}_{\nu}^{2}(\alpha)$ (the inverse cdf of the Chi-square distribution). Finally I am now able to segment my images based in this distance measure - it works :) – Ole Thomsen Buus Jul 8 '11 at 7:40

Generally, $U\Lambda^{1/2}$ is the same as $LV$ for some orthogonal matrix $V$. Either of them can be used to generate the normal distribution with the given covariance matrix $\Sigma$ from independent standard normals.

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Thanks for answer. It was mentioned in a comment in the other answer, that $L$ can be found in finite steps. That is very interesting. But what about $V$. Would $U$ work since it is orthogonal? Can $V$ (and/or $U$) also be found in finite steps? – Ole Thomsen Buus Jul 8 '11 at 7:51

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