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I am trying to understand how to compute the inverse of a square block matrix defined as follows

$$\begin{bmatrix}2{\bf I}&-{\bf X}\\{\bf X}'&{\bf 0}\end{bmatrix}$$

where ${\bf I}$ is a $T\times T$ identity matrix, ${\bf X}$ is a $T\times K$ matrix of scalars and ${\bf 0}$ is a $K\times K$ null matrix.

I actually have no idea about that, so any hint is welcome.

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  • $\begingroup$ Does ${\bf X}'$ denote the transpose of ${\bf X}$? $\endgroup$ – Luckyluck63 May 1 '16 at 8:35
  • $\begingroup$ Yes, sorry I should have written that $\endgroup$ – PhDing May 1 '16 at 9:18
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In general, we have $$\left[\begin{array}{cc}{\bf V} & {\bf W} \\ {\bf X} & {\bf Y}\end{array}\right]^{-1} = \left[\begin{array}{cc}{\bf V}^{-1} + {\bf V}^{-1}{\bf W}\left({\bf Y}-{\bf X}{\bf V}^{-1}{\bf W}\right)^{-1}{\bf X}{\bf V}^{-1}& -{\bf V}^{-1}{\bf W}\left({\bf Y}-{\bf X}{\bf V}^{-1}{\bf W}\right)^{-1} \\ -\left({\bf Y}-{\bf X}{\bf V}^{-1}{\bf W}\right)^{-1}{\bf X}{\bf V}^{-1} & \left({\bf Y}-{\bf X}{\bf V}^{-1}{\bf W}\right)^{-1}\end{array}\right].$$ This can be derived through Woodbury matrix identity or in general Schur complement. Now you can plug in your matrices to get the appropriate inverse.

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    $\begingroup$ Thanks to the remark from @JeanMarie the above formula holds if ${\bf X}'{\bf W}$ is invertible. $\endgroup$ – Luckyluck63 May 1 '16 at 10:28
  • $\begingroup$ Thanks for the editing, I appreciate : I will pay attention no to "slent" the name of matrices... $\endgroup$ – Jean Marie May 1 '16 at 10:34
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Let

$${\bf M}=\begin{bmatrix}2{\bf I}&-{\bf X}\\{\bf X}'&{\bf 0}\end{bmatrix}$$

The formula given by @Luckyluck63 works if and only if ${\bf X}'{\bf X}$ is invertible, and this is the necessary and sufficient condition under which ${\bf M}$ is invertible.

This condition can be re-written in the following way:

The format $T \times K$ of ${\bf X}$ must be with $T \geq K$ (i.e., square or "portrait" shape) AND ${\bf X}$ must be full rank (i.e., rank$({\bf X})=K$).

In all other cases, even if ${\bf X}$ is full rank, ${\bf M}$ is not invertible, for the following reason: as $T < K$, ${\bf X}$ has a "landscape" format; thus its columns are necessarily linked, or said otherwise, there exists a certain non trivial linear combination of its columns equal to zero. Thus, the last $K$ columns of ${\bf M}$ are linked by the same linear combination. Therefore $det({\bf M})=0$, establishing that ${\bf M}$ is not invertible.

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  • $\begingroup$ I am pretty sure $M$ is invertible, since I am working in Econometrics, under the strong Gauss-Markov hypotheses, among which there is the fact that $X'X$ is non singular $\endgroup$ – PhDing May 1 '16 at 10:05
  • $\begingroup$ It means that necessarily, you have $X$ in the portrait format $T \geq K$. As this hypothesis did not appear in your question, I have addressed the general issue... $\endgroup$ – Jean Marie May 1 '16 at 10:14
  • $\begingroup$ And I thank you for that! $\endgroup$ – PhDing May 1 '16 at 10:20

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