# Thread: Inverse of this matrix

1. ## Inverse of this matrix

Find the inverse of the matrix $\displaystyle I_n+ab^T$.

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 $\displaystyle 1+a_i^T*b_j$ where $\displaystyle a_i^T$ is the ith row of a and $\displaystyle b_j$ is the jth column of b.

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

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

Thank you!!!

2. ## 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)?

3. ## 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!

4. ## 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.

5. ## Re: Inverse of this matrix

So I write $\displaystyle ab^T=Q \Lambda Q^*$ then I have $\displaystyle (I-ab^T)^{-1}=(QQ^*-Q \Lambda Q^*)^{-1}=[Q(I- \Lambda )Q^*]^{-1}$$\displaystyle =Q^*(I- \Lambda)^{-1}Q$

So I know that the (i,j) entry of $\displaystyle (I - \Lambda )^{-1}$ is $\displaystyle \frac {1}{1- \lambda _i}I \{ i=j \}$

Are there any ways I can simplify this? Thank you!

6. ## Re: Inverse of this matrix

I'm not sure you can: if you can diagonalize ab^t then you can get a specific solution.