Re: Inverse of this matrix

Hey tttcomrader.

Just to clarify, what are the dimensions of the matrices a and b (are they row/column or are they square)?

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!

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.

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!

Re: Inverse of this matrix

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