# Homework Help: Nonsingular perturbed identity matrix

1. Jan 19, 2009

### Codezion

1. The problem statement, all variables and given/known data
If A = I + uv*, where u and v are m vectors and A is known to be nonsingular, show that the inverse of A = I + $$\alpha$$uv* where $$\alpha$$ is a scalar value

2. Relevant equations

3. The attempt at a solution
Since A is nonsingular, we know the rank of A is m.
Since both u and v are vectors of dimension m, we know that uv* is a square matrix

From the above statements, we conclude that A is an orthogonal matrix which imples:

A$$^{-1}$$
= A $$^{*}$$
= (I + uv*)*
= I + (uv*)*
= I + u*v
= I + [tex]\alpha[\tex], since u*v is a scalar value

As you can see, I am arriving at the wrong conclusion. What am I missing? Thank you!!

2. Jan 19, 2009

### Hurkyl

Staff Emeritus
I note two problems:

(1) I have no idea how you conclude A should be orthogonal.
(2) (AB)* = B*A*

3. Jan 19, 2009

### Codezion

(1) - I think one of the definition of an Orthogonal matrix is a square matrix, whose columns are all linearly independent so that they make the basis for an m space. One of the properties of a nonsingular matrix is that it has the maximum rank (in this case m since it is a square matrix). These statements led me to conclude our matrix A is orthogonal!

(2) - That is a correct property...did I miss something?

4. Jan 19, 2009

### Hurkyl

Staff Emeritus
You forgot a key word in that characterization: they have to make an orthonormal basis.

(Lemme switch to S and T instead of A and B, to eliminate confusion)

Yes. The property is (ST)*=T*S*, but you used it in your calculation as (ST)*=S*T*.

5. Jan 19, 2009

### Codezion

Wow! Thank you I missed that! I was assuming (AB)* = A*B* all along - even after I read your post! Thank you.

If A is not an orthogonal matrix, then I guess I am back to square one :(

6. Jan 19, 2009

### Hurkyl

Staff Emeritus
Simple is often best, even if it's just to get ideas. Rather than looking for a clever argument, why not test directly whether or not this alledged inverse to A really is?

7. Jan 19, 2009

### Dick

The matrices are inverses if (I+uv*)(I+alpha uv*)x=x for any vector x. Expand it out and try to figure out a value of alpha that will make that true.

8. Jan 19, 2009

### Codezion

Thank you! Figured it out :)

(I + uv*) (I + alpha uv*)x = x
(I + alpha Iuv* + Iuv* + alpha uv*uv*)x = x
v*u is a scalar and let it be beta
(I + alpha Iuv* + Iuv* + alpha beta uv*)x = x
[I + Iuv*(alpha + 1 + alpha beta)]x = x
this implies (alpha + 1 + alpha beta) = 0!
alpha = -1/(1+beta)!!!

9. Jan 19, 2009

### Hurkyl

Staff Emeritus
Note that x was really just a placebo. You could have gone straight to the definition of inverse, and put an I on the right hand side, without any of the x's, and done exactly the same thing.

10. Jan 19, 2009

### Codezion

I like the expression (placebo). Yup, thanks Hurkyl. Thanks so much.

11. Jan 19, 2009

### Codezion

If beta = -1 (which would imply A is singular), what would the nullspace of A be? I got only as far as writing the equation as:

(I+uv*)x = 0
x + uv*x = 0....

Do you have any suggestions?

12. Jan 19, 2009

### Hurkyl

Staff Emeritus
Well, if it's going to be at all easy to figure out, it's probably going to be some expression involving u or v. (I+uv*)x = 0 is a linear equation, so I would start by plugging in a few things, and see if I can combine them to make 0.

13. Jan 19, 2009

### Dick

Sure you don't want to take a crack at this without a hint? It's pretty easy. Remember beta=v*u=(-1).

14. Jan 19, 2009

### Codezion

Thanks a lot for your help guys.
Dick, I think it could be the pressure:), but I still cannot see it. Here is what I did

v*x is some constant beta,
x + uv*x = x + beta u
=> x = -beta u
Therefore, I concluded x is a scalar multiple of u.

This is what I came up with, but it looks dodgy because I somewhat defined beta with x itself..???

15. Jan 19, 2009

### Codezion

OH!!!! I see it now :) if x = u, then

x + uv*x = u + uv*u = u + -u = 0

You are right Dick, it was easy! :)

Thanks a bunch!!!

16. Jan 19, 2009

### Hurkyl

Staff Emeritus
I think you confused yourself, because you now have two variables named beta! One of them was the inner product of u with v, and the other is the inner product of x with v.

But no matter; you can test directly if u is a right null-vector or not (I now see you've done that). Of course, you'll have to clean up your proof if you want to claim that the right nullspace is one-dimensional.

(p.s. as an alternative, there is a rank-counting argument that tells you that A cannot be less than n-1 dimensional. Can you see it? Hint: it might be easier to think instead of the equation I = A - uv*)

Last edited: Jan 19, 2009