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[chessman4991]

So this is probably one of those problems where the solution is very quick and easy once you see it, but I'm just not seeing it. Been thinking about it and trying stuff out for a while now. Here's the problem:

If u and v are m-vectors, the matrix  A = I + uv*  is known as a rank-one perturbation of the identity. Show that if A is nonsingular, then its inverse has the form A^-1 = I + kuv*  for some scalar k, and give an expression for k. For what u and v is A singular? If it is singular, what is null(A)?

I know I at least need help on the first part, but I might be able to get the rest after that.

[chessman4991]

Nevermind, I figured it out, here's the solution if you're curious:

AA^-1 = I by definition
(I + uv*)(I + kuv*) = I
I + uv* + kuv* + kuv*uv* = I

I got to there before and got stuck. What I didn't realize was that v*u is a constant so it's commutative:
I + uv* + kuv* + k(v*u)uv* = I
uv* + kuv* + k(v*u)uv* = 0
(1 + k + k(v*u))uv* = 0
so then k = -1/(1+v*u)

and all those steps are reversible so the proof is just that backwards.

Then the matrix is singular for v*u = -1

And then for the null space question:

Ax = 0
(I+uv*)x = 0
x + uv*x = 0
uv*x = -x
uv*uv*x = -uv*x
...and v*u = -1 as above so
-uv*x = -uv*x
obviously true for all x so the null space = C^m