# Inverse of this matrix

• September 15th 2013, 06:02 PM
Inverse of this matrix
Find the inverse of the matrix $I_n+ab^T$.

I know that I'm really suppose to work a little bit on my own first before asking questions here... But I don't really know how to start this problem here other than just writing up the matrices in this manner - the (i,j) entry being $1+a_i^T*b_j$ where $a_i^T$ is the ith row of a and $b_j$ is the jth column of b.

Then the next question is finding the inverse of the matrix $D+ab^T$ where D is a diagonal matrix.

I suppose the first question will help me with this second?

Thank you!!!
• September 15th 2013, 09:11 PM
chiro
Re: Inverse of this matrix

Just to clarify, what are the dimensions of the matrices a and b (are they row/column or are they square)?
• September 15th 2013, 09:36 PM
Re: Inverse of this matrix
The problem did not state, but I assume that a is n x m and b is m x n. Thanks!
• September 16th 2013, 03:45 AM
chiro
Re: Inverse of this matrix
I would try and look at doing an eigen-decomposition and see if you can relate the eigen-decomposition of ab^t to that of ab^t + I.

Remember that I = Q*Q^t = Q*Q^(-1) if Q is an orthogonal matrix.
• September 16th 2013, 04:00 PM
So I write $ab^T=Q \Lambda Q^*$ then I have $(I-ab^T)^{-1}=(QQ^*-Q \Lambda Q^*)^{-1}=[Q(I- \Lambda )Q^*]^{-1}$ $=Q^*(I- \Lambda)^{-1}Q$
So I know that the (i,j) entry of $(I - \Lambda )^{-1}$ is $\frac {1}{1- \lambda _i}I \{ i=j \}$