Hi, Ted,
Now I understand the problem. Thank you for the explanation. It's very
helpful. I appreciate it.
Cindy
On Mon, Aug 10, 2009 at 3:58 PM, Ted Harding
<ted.hard...@manchester.ac.uk>wrote:
> On 10-Aug-09 22:36:03, cindy Guo wrote:
> > Hi, Ted,
> > Thanks for the sample code. It is exactly what I want. But can
> > I ask another question? The matrix for which I want the negative
> > square root is a covariance matrix. I suppose it should be positive
> > definite, so I can do 1/sqrt(V) as you wrote. But the covariance
> > matrix I got in R using the function cov has a lot of negative
> > eigenvalues, like -5.338634e-17, so 1/sqrt(V) generates NA's. Can
> > you tell what's the problem here.
> >
> > Thanks,
> > Cindy
>
> Cindy,
> If that -5.338634e-17 is typical of the "lot of negative eigenvalues",
> then what you are seeing is the result of R's attempt to calculate
> zero eigenvalues, but defeated by the inevitable rounding errors.
> In other words, your covariance matrix is singular, and the variables
> involved are not linearly independent.
>
> The only thing that is guaranteed about a covariance matrix is that
> it is positive semi-definite (not positive definite); in other words
> all eigenvalues are positive or zero (mathematically).
>
> For example, if Y=X, var(X) = var(Y) = 1, then
> cov(X,Y) = 1 1
> 1 1
> which is singular (eigenvalues = 2, 0).
>
> The result of attempting to compute them is subject to rounding errors,
> which (for zero eigenvalues) can be slightly negative.
>
> So the covariance matrix in your case would not have an inverse,
> still less a negative square root!
>
> The basic problem is that you have luinear dependence between the
> variables. To make progress, you would need to find a maximal linearly
> independent set (or possibly find the principal components with
> nozero weights).
>
> Ted.
>
> --------------------------------------------------------------------
> E-Mail: (Ted Harding) <ted.hard...@manchester.ac.uk>
> Fax-to-email: +44 (0)870 094 0861
> Date: 10-Aug-09 Time: 23:58:00
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