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

Prof Brian Ripley ripley at stats.ox.ac.uk
Wed Jan 24 09:02:19 CET 2007

```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

> 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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> and provide commented, minimal, self-contained, reproducible code.
>

--
Brian D. Ripley,                  ripley at 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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