Inverse of rank-one perturbation of identity

__Problem statement:__

If $\displaystyle u$ and $\displaystyle v$ are m-vectors, the matrix $\displaystyle A=I+uv^*$ is known as a rank-one perturbation of the identity. Show that if $\displaystyle A$ is nonsingular, then its inverse has the form

$\displaystyle A^{-1}=I+\alpha uv^*$ for some scalar $\displaystyle \alpha$.

- give an expression for $\displaystyle \alpha$

- for what $\displaystyle u$ and $\displaystyle v$ is $\displaystyle A$ singular?

- if singular, what is $\displaystyle null(A)$

__Attempt:__

*Expression for $\displaystyle \alpha$:*

$\displaystyle \textbf{A}\textbf{A}^{-1}=(\textbf{I}+\textbf{u}\textbf{v}^*)(\textbf{I}+ \alpha\textbf{u}\textbf{v}^*)$

$\displaystyle \Rightarrow \textbf{I}+\alpha\textbf{u}\textbf{v}^*+\textbf{u} \textbf{v}^*+\alpha\textbf{u}\textbf{v}^*\textbf{u }\textbf{v}^*$

$\displaystyle \Rightarrow \textbf{I}+\textbf{u}\textbf{v}^*[\alpha(1+\textbf{v}^*\textbf{u})+1]$

$\displaystyle \textbf{A}\textbf{A}^{-1}=\textbf{I}$ if $\displaystyle \alpha(1+\textbf{v}^*\textbf{u})+1=0$

$\displaystyle \alpha = -\frac{1}{1+\textbf{v}^*\textbf{u}}$

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

__For what $\displaystyle u$ and $\displaystyle v$ is $\displaystyle A$ singular:__

A is singular when $\displaystyle u$ is a unit vector on one of the axis, and $\displaystyle v$ is a unit vector in the opposite direction of $\displaystyle u$.

Ex. $\displaystyle \textbf{u}=(0,i),\textbf{v}=(0,-i)$

__If singular, what is $\displaystyle null(A)$:__

$\displaystyle I$ has rank m, and $\displaystyle uv^*$ has rank 1.

If $\displaystyle A$ is singular $\displaystyle I+uv^*$ has rank $\displaystyle m-1$.

$\displaystyle A$ has then one free variable and its nullspace has rank 1.

It is a line in $\displaystyle C^m$

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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** $\displaystyle \textbf{I}+\alpha\textbf{u}\textbf{v}^*+\textbf{u} \textbf{v}^*+\alpha\textbf{u}\textbf{v}^*\textbf{u }\textbf{v}^*$

$\displaystyle \textbf{I}+\textbf{u}\textbf{v}^*[\alpha(1+\textbf{v}^*\textbf{u})+1]$

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?