Inverse of rank-one perturbation of identity

__Problem statement:__

If and are m-vectors, the matrix is known as a rank-one perturbation of the identity. Show that if is nonsingular, then its inverse has the form

for some scalar .

- give an expression for

- for what and is singular?

- if singular, what is

__Attempt:__

*Expression for :*

if

I did not actually show that the inverse has that form, and I'm not sure how to do so.

__For what and is singular:__

A is singular when is a unit vector on one of the axis, and is a unit vector in the opposite direction of .

Ex.

__If singular, what is :__

has rank m, and has rank 1.

If is singular has rank .

has then one free variable and its nullspace has rank 1.

It is a line in

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Just throwing around some ideas on the last two questions as I hate to post a question without attempt.

It seems that I can not get any further by my self,so I'm hoping some of you might spare a few moments.

Thanks.

Re: Inverse of rank-one perturbation of identity

Hi,

I am stuck on this problem, too. In particular, going between these two steps:

Quote:

Originally Posted by

**Mollier**

makes ZERO sense. How did uv* magically turn into v*u????? And what are these two expressions equal to, I?? :(

EDIT: OK, the trick for the case when A is nonsingular is to realize that (1) matrices, including vectors, associate, and (2) v*u is a scalar. That is, uv*uv* = u(v*u)v* = (v*u)uv*. So how exactly do we handle the A-is-singular case?