# Inverse of rank-one perturbation of identity

• Dec 11th 2009, 04:00 AM
Mollier
Inverse of rank-one perturbation of identity
Problem statement:
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
$A^{-1}=I+\alpha uv^*$ for some scalar $\alpha$.
- give an expression for $\alpha$
- for what $u$ and $v$ is $A$ singular?
- if singular, what is $null(A)$

Attempt:
Expression for $\alpha$:

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

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

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

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

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

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

For what $u$ and $v$ is $A$ singular:

A is singular when $u$ is a unit vector on one of the axis, and $v$ is a unit vector in the opposite direction of $u$.
Ex. $\textbf{u}=(0,i),\textbf{v}=(0,-i)$

If singular, what is $null(A)$:

$I$ has rank m, and $uv^*$ has rank 1.
If $A$ is singular $I+uv^*$ has rank $m-1$.
$A$ has then one free variable and its nullspace has rank 1.
It is a line in $C^m$
----------------------------------------------------------------

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.
• Dec 11th 2009, 06:12 AM
HallsofIvy
Quote:

Originally Posted by Mollier
Problem statement:
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
$A^{-1}=I+\alpha uv^*$ for some scalar $\alpha$.
- give an expression for $\alpha$
- for what $u$ and $v$ is $A$ singular?
- if singular, what is $null(A)$

Attempt:
Expression for $\alpha$:

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

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

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

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

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

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

Yes, you have! The actual proof goes the other way: start by taking $A^{-1}= I-\frac{1}{1+ v^*u}$ and going "up" your list of steps. But since it is clear that you can do that, what you have shown is enough. (This is sometimes called "synthetic" proof.)

Quote:

For what $u$ and $v$ is $A$ singular:

A is singular when $u$ is a unit vector on one of the axis, and $v$ is a unit vector in the opposite direction of $u$.
Ex. $\textbf{u}=(0,i),\textbf{v}=(0,-i)$
There are NO "axes" in a general vector space. You may be thinking of basis vectors but there are an infinite number of bases for a vector space, just as you can have axes any where on a graph.

But I think you have a good starting idea. Try this in 2 dimensions first. If u= <a, b> and v= <c, d> then uv*= $\begin{bmatrix}a \\ b\end{bmatrix}\begin{bmatrix}c & d\end{bmatrix}= \begin{bmatrix}ac & ad \\ bc & bd\end{bmatrix}$ and so I+ uv*= $\begin{bmatrix}1+ ac & ad \\bc & 1+ bd$. What is the determinant of that matrix? What has to be true of a, b, c, and d so that determinant is 0?

Quote:

If singular, what is $null(A)$:

$I$ has rank m, and $uv^*$ has rank 1.
If $A$ is singular $I+uv^*$ has rank $m-1$.
$A$ has then one free variable and its nullspace has rank 1.
It is a line in $C^m$
----------------------------------------------------------------

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.
• Dec 12th 2009, 01:34 AM
Mollier
Quote:

Originally Posted by HallsofIvy
The actual proof goes the other way: start by taking $A^{-1}= I-\frac{1}{1+ v^*u}$ and going "up" your list of steps. But since it is clear that you can do that, what you have shown is enough. (This is sometimes called "synthetic" proof.)

Did not know of that, thanks. I've purchased the book "How To Prove It" in hope to gain some knowledge about proofs.

Quote:

Originally Posted by HallsofIvy
What is the determinant of that matrix? What has to be true of a, b, c, and d so that determinant is 0?

$det(A)=ac+bd+1$. $ac+bd=u^*v$, so for $A$ to be singular, the inner product $u^*v=-1$.
Is that sufficient, or do I have to specify in more detail?

You quoted my last question about $null(A)$. Should I take that as a good sign, or a bad one? :)

Thank you very much.
• Dec 14th 2009, 12:41 AM
Mollier
A gentle bump...
• Sep 4th 2014, 07:37 PM
phys251
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
$\textbf{I}+\alpha\textbf{u}\textbf{v}^*+\textbf{u} \textbf{v}^*+\alpha\textbf{u}\textbf{v}^*\textbf{u }\textbf{v}^*$

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

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?