Re: [R] Matrix question: obtaining the square root of a positive definite matrix?

From: Prof Brian Ripley <ripley_at_stats.ox.ac.uk>
Date: Wed 24 Jan 2007 - 08:02:19 GMT

On Wed, 24 Jan 2007, gallon li wrote:

> I want to compute B=A^{1/2} such that B*B=A.

According to your subject line A is positive definite and hence symmetric? The usual definition of a matrix square root involves a transpose, e.g. B'B = A. There are many square roots: were you looking for a symmetric one?

For such an A,

> e <- eigen(A)
> V <- e$vectors
> V %*% diag(e$values) %*% t(V)

recovers A (up to rounding errors), and

> B <- V %*% diag(sqrt(e$values)) %*% t(V)

is such that B %*% B = A. Even that is not unique, e.g. -B is an equally good answer.

> For example

(with A = b and B = a, it seems)

> a=matrix(c(1,.2,.2,.2,1,.2,.2,.2,1),ncol=3)
>
> so
>> a
> [,1] [,2] [,3]
> [1,] 1.0 0.2 0.2
> [2,] 0.2 1.0 0.2
> [3,] 0.2 0.2 1.0
>> a%*%a
> [,1] [,2] [,3]
> [1,] 1.08 0.44 0.44
> [2,] 0.44 1.08 0.44
> [3,] 0.44 0.44 1.08
>> b=a%*%a
>
> i have tried to use singular value decomposion
>
>> c=svd(b)
>
>> c$u%*%diag(sqrt(c$d))
> [,1] [,2] [,3]
> [1,] -0.8082904 2.043868e-18 0.6531973
> [2,] -0.8082904 -5.656854e-01 -0.3265986
> [3,] -0.8082904 5.656854e-01 -0.3265986
>
> this does not come close to the original a. Can anybody on this forum
> enlight me on how to get a which is the square root of b?
>
> [[alternative HTML version deleted]]
>
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-- 
Brian D. Ripley,                  ripley@stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272866 (PA)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595

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Received on Wed Jan 24 19:09:40 2007

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