Matrix algebra: The "magical inverse" trick

Solution 1:

There is a way to get from (1)-(3) to (5)-(7) without the trick of assembling the matrices $u,v,w,x$ into a block matrix.

The special ingredient is the identity $$ (1+rs)^{-1} = 1 - r(1+sr)^{-1}s, \tag{*} $$ which is easy to verify. Admittedly, this too may seem "magical" if you're unfamiliar with it. At least, though, it's a universal kind of magic, in the sense that it's an identity in all rings in general, and not just for matrices. (For a reference about this identity, see mathoverflow.net/questions/31595, where it appears with a tiny difference in sign.)

Using this identity, we will derive equation (5) from (1)-(3). First, note that $u$ and $x$ are invertible. (This follows from (1) and (2), although I'm glossing over the details.) Now write (3) as $w^\dagger (x^{-1})^\dagger = u^{-1} v$. Multiply that equation by its conjugate transpose $x^{-1} w = v^\dagger (u^{-1})^\dagger$ to get $$ w^\dagger (x^{-1})^\dagger x^{-1} w = u^{-1} v v^\dagger (u^{-1})^\dagger. $$ The right-hand side, applying (1), is \begin{align} u^{-1} v v^\dagger (u^{-1})^\dagger &= u^{-1} (u u^\dagger - I) (u^{-1})^\dagger \\ &= I - u^{-1} (u^{-1})^\dagger \\ &= I - (u^\dagger u)^{-1}, \end{align} while the left-hand side, applying (2) and the identity (*), is \begin{align} w^\dagger (x^{-1})^\dagger x^{-1} w &= w^\dagger (x x^\dagger)^{-1} w \\ &= w^\dagger (I + w w^\dagger)^{-1} w \\ &= I - (I + w^\dagger w)^{-1}. \end{align} This shows that $u^\dagger u = I + w^\dagger w$ (equation (5)).

Equation (6) can be derived in the same way.

Equation (7) is easy to derive once (1)-(6) are all available: \begin{align} u^\dagger v &= u^\dagger u u^{-1} v \\ &= (I + w^\dagger w) w^\dagger (x^{-1})^\dagger \\ &= w^\dagger (I + w w^\dagger) (x^{-1})^\dagger \\ &= w^\dagger x x^\dagger (x^{-1})^\dagger \\ &= w^\dagger x. \end{align}

This way, we've proven (5)-(7) from (1)-(3), without going up to the higher dimensional "big picture" of the generalized unitary matrix $B$.