0
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ $\in$ is \in, btw $\endgroup$ – glowstonetrees Apr 5 '18 at 13:31
  • $\begingroup$ $I - A$ being invertible does not mean that $A$ is invertible, for example take $A = 0$. $\endgroup$ – Joppy Apr 5 '18 at 13:40
  • $\begingroup$ How can a $p\times q$ matrix be invertible if $p\neq q$? $\endgroup$ – Aweygan Apr 5 '18 at 13:57
1
$\begingroup$

(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 $I_p -K K^*$.

$\endgroup$
2
  • $\begingroup$ could you explain me ur thoughts on i) ->ii) how did you obtain the inverse, (looked for some identites but couldn't find any online). $\endgroup$ – MasterPI Apr 5 '18 at 16:41
  • $\begingroup$ It is a special case of the Woodbury matrix identity $\endgroup$ – Reinhard Meier Apr 5 '18 at 17:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.