woodbury matrix identity
2014/6/20
【转载请注明出处】http://www.cnblogs.com/mashiqi
http://en.wikipedia.org/wiki/Woodbury_matrix_identity
Today I'm going to write down a proof of this Woodbury matrix identity, which is very important in some practical situation. For instance, the 40th equation of this paper" bayesian compressive sensing using Laplace priors" applied this identity. Now let me give the details of it.
The Woodbury matrix identity is:
${(A + UCV)^{ - 1}} = {A^{ - 1}} - {A^{ - 1}}U{({C^{ - 1}} + V{A^{ - 1}}U)^{ - 1}}V{A^{ - 1}}$
where 
, 
, 
and 
are both assumed reversible.
Proof:
We denote 
with 
, namely 
.So:
[M{A^{ - 1}} = I + UCV{A^{ - 1}}]
By multiply U with both side we get:
[egin{array}{l}
M{A^{ - 1}}U = U + UCV{A^{ - 1}}U = U(I + CV{A^{ - 1}}U)\
{kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} = UC({C^{ - 1}} + V{A^{ - 1}}U)
end{array}]
is reversible, 
we get:
But how could we deal with this nasty 
term? We should notice that this term, which may not square, is coming from 
itself, which is right a square and reversible matrix. So, from formula , we make up a pleasant 
with is nasty 
:
[egin{array}{l}
M{A^{ - 1}}U{({C^{ - 1}} + V{A^{ - 1}}U)^{ - 1}}V + A = UCV + A = M\
Rightarrow M = M{A^{ - 1}}U{({C^{ - 1}} + V{A^{ - 1}}U)^{ - 1}}V + A\
Rightarrow I - {A^{ - 1}}U{({C^{ - 1}} + V{A^{ - 1}}U)^{ - 1}}V = {M^{ - 1}}A
end{array}]
And finally due to the reversibility of 
, we get the Woodbury matrix identity:
[{M^{ - 1}} = {(A + VCU)^{ - 1}} = {A^{ - 1}} - {A^{ - 1}}U{({C^{ - 1}} + V{A^{ - 1}}U)^{ - 1}}V{A^{ - 1}}]
Done.
We should notice that if 
and 
are identity matrix, then Woodbury matrix identity can be reduced to this form:
[{(A + C)^{ - 1}} = {A^{ - 1}} - {A^{ - 1}}{({C^{ - 1}} + {A^{ - 1}})^{ - 1}}{A^{ - 1}}]
,which is equivalent to:
[{(A + C)^{ - 1}} = {C^{ - 1}}{({C^{ - 1}} + {A^{ - 1}})^{ - 1}}{A^{ - 1}}]
This is because:
[egin{array}{l}
{(A + C)^{ - 1}} = {A^{ - 1}} - ( - {C^{ - 1}} + {C^{ - 1}} + {A^{ - 1}}){({C^{ - 1}} + {A^{ - 1}})^{ - 1}}{A^{ - 1}}\
{kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} {kern 1pt} = {A^{ - 1}} + {C^{ - 1}}{({C^{ - 1}} + {A^{ - 1}})^{ - 1}}{A^{ - 1}} - ({C^{ - 1}} + {A^{ - 1}}){({C^{ - 1}} + {A^{ - 1}})^{ - 1}}{A^{ - 1}}\
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end{array}]