Re: Partitioned matrix operations
- To: mathgroup at smc.vnet.net
- Subject: [mg100664] Re: Partitioned matrix operations
- From: "Steve Luttrell" <steve at _removemefirst_luttrell.org.uk>
- Date: Wed, 10 Jun 2009 17:11:21 -0400 (EDT)
- References: <h0l467$oof$1@smc.vnet.net> <h0nug5$b6e$1@smc.vnet.net>
Your proposed use of Inverse works only if A, B and C commute (where
appropriate). You can verify that your proposed inverse is not correct by
premultiplying it by the original matrix thus:
{{A, B}, {C, 0}} . {{0, C^(-1)}, {B^(-1), -A (B C)^(-1)}}
When you expand this out the upper right element is
A C^(-1) - B A (B C)^(-1)
which does not simplify to 0 (as you would have liked) when A, B and C don't
commute (where appropriate).
Anyway, back to the original problem. Since it is only a 2x2 block matrix
you can solve it manually thus:
{{A,B},{C,0}} . {{U,V},{W,X}} = {{1,0},{0,1}}
where you have to solve for the matrices U, V, W, X, and the 1 and 0 symbols
on the r.h.s. stand for the appropriate matrices.
I won't say any more in case this is a homework problem.
--
Stephen Luttrell
West Malvern, UK
"dh" <dh at metrohm.com> wrote in message news:h0nug5$b6e$1 at smc.vnet.net...
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> Hi,
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> you may symbolically invert your matrix:
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> Inverse[M]
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> giving:
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> {{0, 1/C}, {1/B, -(A/(B C))}}
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> Now it is up to you to ensure that C,B and B C are invertible.
>
> Daniel
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>
>
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> Joe Hays wrote:
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>> Hello,
>
>> Here's a Mathematica newbie question. Say I have a matrix, M, defined as,
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>>
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>> M = {{A, B}, {C, 0}}
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>>
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>> where A is nxn, B is nxm, C is mxn, and the zero sum matrix is mxm. I
>> would
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>> like to perform an operation on the matrix M without fully defining A, B,
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>> and C and get a result in terms of A, B, and C. For example, if I wanted
>> to
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>> determine the matrix inverse of M in terms of A, B, and C.
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>>
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>> Is this possible in Mathematica? I've unfortunately not found anything in
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>> the docs that indicates that this is possible.
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>>
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>> Thx.
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>>
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>>
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