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Suppose we have a normally distributed random variable $\boldsymbol{X}$ with covariance matrix $\Sigma$, which is symmetric positive definite.

Now if I multiply $\boldsymbol{X}$ by some matrix A, the resulting random variable has covariance:

cov($A \boldsymbol{X}$) = $A\Sigma A'$.

Is there any restrictions on the matrix A such that $A\Sigma A'$ is a valid covariance, ie is symmetric positive definite with full rank and non zero determinant?

Edit:

$\Sigma$ is a diagonally dominant matrix of size 10x10, with 6 in the diagonal elements and -1's in 5 of the columns for each row.

Matrix A contains data and is large 480x10. I have tried using just random generated normal data in A and get the same issue that $A\Sigma A'$ has rank 10 and zero determinant. The rank of the resulting matrix seems to depend on the rank of $\Sigma$.

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    $\begingroup$ Any linear transform of a random variable with covariance is also a random variable. It should also have a covariance. How would it be any different? $\endgroup$ – Jim Sep 16 '16 at 15:56
  • $\begingroup$ The reason I ask is because I have tried this for a particular A and get a singular covariance matrix after the transformation. Hence my question of if there is some restrictions on what A can be. $\endgroup$ – darren86 Sep 16 '16 at 16:03
  • $\begingroup$ @darren86: If you can put your problematic $X,\Sigma,A$ into the question it might be possible to understand your problem better. Covariance matrices can be singular, and in your example will be for example if $A$ has zero determinant or if $A$ has more rows than columns. $\endgroup$ – Henry Sep 16 '16 at 19:44
  • $\begingroup$ @Henry Question edited. If A has more rows than columns with the resulting matrix be singular? Is this a general result? Do you have a link? Thanks $\endgroup$ – darren86 Sep 18 '16 at 11:48
  • $\begingroup$ How can one multiply these matrices if their dimensions (480x6 and 10x10 respectively) do not fit? $\endgroup$ – Did Sep 18 '16 at 12:07
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The condition which you need on $A$ is that it $A$ has full row rank. That is, we want the rows of $A$ to be linearly independent. Assume $A$ has $d$ rows and $m$ columns. A necessary but not sufficient condition for this to hold is that $d$ is less than or equal to $m$. This condition fails in your example since $d = 480$ and $m = 10$.

To see why this condition holds, note that $x'A \Sigma A' x > 0$ for all non-zero $x$ is the condition that $A \Sigma A'$ is positive definite. If $\Sigma$ is positive definite this condition is equivalent to the condition that $x' A \Sigma^{1/2} (\Sigma^{1/2}) A' x > 0$ which is equivalent to the condition that $|x'A \Sigma^{1/2}|^2 > 0$. The matrix $A$ has full row rank if and only if $x'A$ is not an $m$-dimensional vector of zeros for all non-zero $x$ (this is an equivalent definition of the matrix $A$ having full row rank).

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