Wednesday, December 8, 2021

A simple matrix inverse problem

inverse of a matrix

Closed form inverse of IσuuTI-\sigma uu^T for σ1\sigma \neq 1.

We are interested in the matrix A=IσuuTA = I - \sigma uu^T where uRnu\in \mathbb R^n such that u=1||u||=1.

From an answer of this post, we can easily see that A=diag(1σ,1,...,1)A=\text{diag} ( 1-\sigma, 1, ..., 1) in the basis {v1=u,v2,...,vn}\{ v_1=u, v_2,..., v_n\}.

We thus can easily see that if σ1\sigma \neq 1, then AA is invertible. An interesting question is that

What is the closed form inverse of AA?

This question is not very complicated and not simple since the construction of v2,...,vnv_2, ..., v_n is not obvious.

So this note is to provide a closed form inverse of AA, which is
A1=I+σ1σu2uuT.A^{-1} = I + \frac{\sigma}{1-\sigma ||u||^2} uu^T.

We can easily check that A.A1=IA.A^{-1} = I. But the interesting point is that how do we construct such formula for A1A^{-1}. The trick is as follows.

First, let
y=Ax=xσu,xu.y = Ax = x - \sigma \langle u, x\rangle u.
Now, if we can recover xx form yy, we will get the inverse of AA.
Consider
u,y=u,xσu,xu2.\langle u, y \rangle= \langle u, x \rangle - \sigma \langle u, x\rangle ||u||^2.
Now we get
u,x=u,y1σu2.\langle u, x\rangle= \frac{\langle u, y\rangle}{1- \sigma ||u||^2}.
Substitute this expression to the first expression of yy, we have
y=xσ1σu2uuTyy = x - \frac{\sigma}{1-\sigma ||u||^2} uu^T y
or
x=y+σ1σu2uuTy=A1y.x = y + \frac{\sigma}{1-\sigma ||u||^2} uu^T y=A^{-1} y.

We are done.

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