# Math Help - Proofs of Invertibility

1. ## Proofs of Invertibility

I'm supposed to prove an invertible, but I'm not sure how, as the prof didn't talk about it in class, and the book is rather vague...

The question is this:

Let u and v be column vectors in R^n, and let A= I + u(v^T). Show that if (u^T)v =/= -1, then A is invertible and
A^-1 = I - (1/(1+((u^T)v)) (u(v^T))

Sorry about the last equation being so confusing, but that's it. I'm not sure how to prove this, since I don't even know how it works. I'm terrible at this stupid inverse matrix stuff...

2. Originally Posted by Hellreaver

Let u and v be column vectors in R^n, and let A= I + u(v^T). Show that if (u^T)v =/= -1, then A is invertible and A^-1 = I - (1/(1+((u^T)v)) (u(v^T))
this is a nice problem! first we need a little point:

$\boxed{1}$ if $u,v$ are two vectors in $\mathbb{R}^n,$ then $uv^Tuv^T=uv^Tu^Tv.$

Proof. note that $u^Tv$ and $v^Tu$ are both scalars and obviously $u^Tv=v^Tu.$ also a scalar commutes with a vector. thus: $uv^Tu v^T=uu^Tvv^T=uv^Tu^Tv. \ \ \ \Box$

back to your problem: we have $A^2=(I+uv^T)^2=I + 2uv^T+uv^Tuv^T=I+2uv^T+uv^Tu^Tv. \ \ \ \ \ \ \ \text{by} \ \boxed{1}$

hence: $A^2=I+uv^T + uv^T + uv^Tu^Tv=A+uv^T(1+u^Tv)=A+(A-I)(1+u^Tv),$ which gives us: $A(A-I-(1+u^Tv)I)=-(1+u^Tv)I. \ \ \ \ \ \ (*)$

now $1+u^Tv \neq 0.$ so we can divide both sides of $(*)$ by $-(1+u^Tv).$ also we have $A-I=uv^T.$ thus $(*)$ gives us: $A \left(I - \frac{1}{1+u^Tv}uv^T \right)=I.$

3. Friggen complicated enough?
I so don't understand this stuff. I was pretty good at trig proofs, but that was back when I actually had a teacher. This linear algebra stuff just doesn't make sense...

4. Originally Posted by Hellreaver
Friggen complicated enough?
I so don't understand this stuff. I was pretty good at trig proofs, but that was back when I actually had a teacher. This linear algebra stuff just doesn't make sense...
his response is very easy to follow, grab a pen and paper and write it carefully down. You should be able to understand it.

5. I understand his response perfectly enough (thanks, NonCommAlg). I just don't understand the majority of the concept behind it.