Inverse of rank-one perturbation of identity

**Mollier** · December 11th 2009, 03:00 AM

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.

**HallsofIvy** · December 11th 2009, 05:12 AM

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.)

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?

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.

**Mollier** · December 12th 2009, 12:34 AM

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.

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.

**Mollier** · December 13th 2009, 11:41 PM

A gentle bump...

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