Compute the inverse

Question

Compute the inverse

anonymous · Answer

a.) $I_{m} + 1_{m}(1_{m})^{T}$
b.) $I_{m} + e_{1}(1_{m})^{T}$

Where $1_{m}$ is the mx1 vector having each component equal to 1 and $e_{1}$ is the first column of the m-dimension identity matrix. 

Basically, my issue is actually proving what these inverses are. Ive calculated them for smaller dimensions and there definitely is a pattern, but I dont know how to actually show it. Is there a way that isnt just horrible brute force?

ganeshie8 · Answer

Hey

anonymous · Answer

Hello :)

ganeshie8 · Answer

$1_m(1_m)^T$ is the $m\times m$ matrix with all $1$'s, right ?

anonymous · Answer

Yep. So for the first one, you get all 1's except on the main diagonal, which would all be 2. But then computing that inverse, or showing how to compute it, seems nasty. I only messed with it for a 3x3 and 4x4. But I cant just write down an answer because I saw the pattern, so Im trying to figure out the best way to actually show what it is.

ganeshie8 · Answer

I think gauss jordan elimination may work nicely...

anonymous · Answer

Well, doing the gauss jordan does give a pattern when you finish. You end up with the main diagonal being $a_{11} = 2, a_{22} = 3/2, a_{33} = 4/3, a_{44} = 5/4, ... , a_{mm} = (m+1)/m$. But yeah, I guess if I were to explain the process of doing that for an mxm matrix that it would get really long for that arbitrary case. I was hoping there was some algebraic trick I may not be keen to so I wouldnt have to do a real long brute force calculation.

ganeshie8 · Answer

Ahh right, first we observe that the rank of $1_m(1_m)^T$ is $1$.
The inverse of $I_m$ is $I_m$ itself

ganeshie8 · Answer

So, we want to find the inverse of $(A+B)^{-1}$ such that : 
1) $A^{-1}$ exists and is known
2) $B$ has rank $1$

ganeshie8 · Answer

http://math.stackexchange.com/questions/17776/inverse-of-the-sum-of-matrices

ganeshie8 · Answer

please look at the lemma in the first answer

anonymous · Answer

Well, there ya go, haha. Yeah, I definitely wouldnt have been looking at the rank or anything. Yeah, I think that basically fixes things up for me then. Thanks for noticing that and giving the link ^_^

ganeshie8 · Answer

http://www.jstor.org/stable/2690437?seq=1#page_scan_tab_contents

ganeshie8 · Answer

np :) im gonna try and understand the proof of that lemma..

anonymous · Answer

Yeah, I gotta take care of something right now, so I cant immediately jump in and solve it, but Im going to digest it when I'm done. Thanks again :)