Inversion of matrix

[Sheng-Yih Guan]

My special thanks for all the email responses I got so far, especially
Mr. Vincent Broman who is with Naval Ocean Systems Center. I still have
trouble finding references of the Householder's formula mentioned
in his response. Any leads to its refereces would be highly appreciated.
Here is Mr. Broman's response:

If you write
(a b b c)
(b a c b)
M = (b c a b)
(c b b a)

and
(a b b' c)
(b a c' b)
M'= (b c a b)
(c b b' a)

then
(0 0 b'-b 0)
(0 0 c'-c 0)
M'-M = (0 0 0 0)
(0 0 b'-b 0)

= (b'-b) (0 0 1 0) = x*yT
(c'-c)
(0 )
(b'-b)

So M' = M + x*yT, where T means transpose and x and y are column vectors.
x*yT is a rank-one matrix. Now for any rank-one change in a matrix,
the corresponding change in the inverse is also rank-one and comes from
Householder's formula:

M * N = I => (M + x*yT) * (N - N*x*yT*N/(1+yT*N*x)) = I.

M' will have an inverse if the denominator ^^^^^^^^ is nonzero (a scalar)
and M has an inverse. Some other cases of 0/0 or 0*infinity
will also allow M' to have an inverse when M doesn't or the
denominator is zero.

Stanley Guan
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