I'd like to prove the equvialence of the following statements.

For $K\in \mathbb{ C }^{ p\times q }$ :

i) ${ I }_{ p }-{ KK }^{ * }$ is invertible

ii) ${ I }_{ q }-{ K }^{ * }K$ is invertible

iii) $\begin{pmatrix}{ I }_{ p }&K \\{ { K }^{ * } }&{ I }_{ q }\end{pmatrix}$ is invertible

So I'm not really sure if my thoughts are right but for i) to be invertible we need to have that ${ KK }^{ * }$ is invertible. If ${ KK }^{ * }$ is invertible we must have that $K$ and ${K }^{ * }$ are also invertible so trivially ${ K }^{ * }K$ is invertible since its also a product of invertible matrices. And for the third part its now obvious that the given Matrix has a full rank beause of the invertiblity of $K$ and ${K }^{ * }$. Is this reasoning correct?

And a further question if I'd like to get the inverse of the last matrix how do I proceed, I get stuck finding the concrete inverse of i) which I need for the next step in $$\left.\begin{pmatrix}{ I }_{ p }&K\\ { 0 }&{ I }_{ p }-{ KK }^{ * }\end{pmatrix}\middle|\begin{pmatrix}{ K }^{ * }&0\\ 0 & I \end{pmatrix}\right..$$

Appreicate any help to understand this problem

• $\in$ is \in, btw – glowstonetrees Apr 5 '18 at 13:31
• $I - A$ being invertible does not mean that $A$ is invertible, for example take $A = 0$. – Joppy Apr 5 '18 at 13:40
• How can a $p\times q$ matrix be invertible if $p\neq q$? – Aweygan Apr 5 '18 at 13:57

(i) => (ii) $I_q + K^*(I_p-KK^*)^{-1} K$ is an inverse matrix of $I_q -K^*K$.
(ii) => (iii) Since $$\begin{bmatrix} I_p & K\\ K^*& I_q \end{bmatrix} \begin{bmatrix} I_p & -K\\ 0& I_q \end{bmatrix} = \begin{bmatrix} I_p& 0\\ K^*& I_q -K^* K \end{bmatrix}$$ and the last matrix is invertible by assumption (ii).
(iii) => (i) We have $$\begin{bmatrix} I_p & K\\ K^*& I_q \end{bmatrix} \begin{bmatrix} I_p & 0\\ -K^*& I_q \end{bmatrix} = \begin{bmatrix} I_p-KK^*& K\\ 0& I_q \end{bmatrix}$$ By the assumption (iii) then the last matrix is invertible, hence so is $I_p -K K^*$.