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I have a symmetric variance/covariance matrix $A$ which is of size (27 x 27). I know that it's rank deficient (rank = 21).

I also know that the units of $A$ are $m^2$. I am trying to use Singular Value Decomposition (SVD) to reduce the dimensionality of the variance/covariance matrix so that it is non-singular and I can use it as a proper variance/covariance matrix to draw random samples from (given a mean $X_0$ of size (27 x 1)).

My plan is to reduce the dimensionality of $A$ to size (21 x 21) ($A^*$) and use this matrix. I know have created a matrix $u$ of size (27 x 1) and $s$ of size (21 x 21) such that $A = u*s*u^T$.

Now, I think that the rectangular matrix $u$ is a mapping of 27 variables in the domain to the 21 variables comprising the subspace. If this is true, I should be able to take $u^T*A*u$ to get $A^*$. This turns into a diagonal (21 x 21) matrix.

My question is:

1.- is this correct until now?

2.- Assuming this is correct, what are the units of $u$ and $s$? I am inclined to believe that $u$ is unit-less (because of my assumption that it is a mapping from one domain to another) and that the units of $s$ are the same as $A$, but I am not sure.

Sergio Sarmiento
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drjrm3
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  • Regarding the question about units, the eigenvalues of a matrix $X^TX$ are the covariances $\Sigma_i^2$ while the singular values of symmetric matrix X are $\Sigma_i$ (taken from this question). I would say then that the units are contained in the diagonal matrix of both decompositions. – Stefano Oct 01 '21 at 19:32

1 Answers1

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$A$ is a map from a vector space $V$ to $W$. Supposing that $V$ has units $v$ and you say $A$ has units $m^2$, it follows that $Av=\sigma u$ for any singular value, so that $\sigma u$ must have units $m^2v$, or if $v$ is unitless, $\sigma$ has units of $m^2$.

Alex R.
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