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Inverse of a square block matrix

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1

$begingroup$

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

$$beginbmatrix2bf I&-bf X\bf X'&bf 0endbmatrix$$

where $bf I$ is a $Ttimes T$ identity matrix, $bf X$ is a $Ttimes K$ matrix of scalars and $bf 0$ is a $Ktimes K$ null matrix.

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

linear-algebra matrices inverse block-matrices

share|cite|improve this question

edited 18 hours ago

Rodrigo de Azevedo

13.1k41960

asked May 1 '16 at 7:37

PhDingPhDing

266316

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

1

$begingroup$

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

$$beginbmatrix2bf I&-bf X\bf X'&bf 0endbmatrix$$

where $bf I$ is a $Ttimes T$ identity matrix, $bf X$ is a $Ttimes K$ matrix of scalars and $bf 0$ is a $Ktimes K$ null matrix.

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

linear-algebra matrices inverse block-matrices

share|cite|improve this question

edited 18 hours ago

Rodrigo de Azevedo

13.1k41960

asked May 1 '16 at 7:37

PhDingPhDing

266316

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

$begingroup$

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

$$beginbmatrix2bf I&-bf X\bf X'&bf 0endbmatrix$$

where $bf I$ is a $Ttimes T$ identity matrix, $bf X$ is a $Ttimes K$ matrix of scalars and $bf 0$ is a $Ktimes K$ null matrix.

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

linear-algebra matrices inverse block-matrices

share|cite|improve this question

edited 18 hours ago

Rodrigo de Azevedo

13.1k41960

asked May 1 '16 at 7:37

PhDingPhDing

266316

$endgroup$

share|cite|improve this question

edited 18 hours ago

Rodrigo de Azevedo

13.1k41960

asked May 1 '16 at 7:37

PhDingPhDing

266316

share|cite|improve this question

edited 18 hours ago

Rodrigo de Azevedo

13.1k41960

asked May 1 '16 at 7:37

PhDingPhDing

266316

edited 18 hours ago

Rodrigo de Azevedo

13.1k41960

edited 18 hours ago

Rodrigo de Azevedo

13.1k41960

asked May 1 '16 at 7:37

PhDingPhDing

266316

asked May 1 '16 at 7:37

PhDingPhDing

266316

2 Answers
2

active

oldest

votes

2

$begingroup$

In general, we have
$$left[beginarrayccbf V & bf W \ bf X & bf Yendarrayright]^-1 = left[beginarrayccbf V^-1 + bf V^-1bf Wleft(bf Y-bf Xbf V^-1bf Wright)^-1bf Xbf V^-1& -bf V^-1bf Wleft(bf Y-bf Xbf V^-1bf Wright)^-1 \ -left(bf Y-bf Xbf V^-1bf Wright)^-1bf Xbf V^-1 & left(bf Y-bf Xbf V^-1bf Wright)^-1endarrayright].$$
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.

share|cite|improve this answer

edited May 1 '16 at 8:53

answered May 1 '16 at 8:48

Luckyluck63Luckyluck63

434211

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

2

$begingroup$

Let

$$bf M=beginbmatrix2bf I&-bf X\bf X'&bf 0endbmatrix$$

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.

share|cite|improve this answer

edited May 1 '16 at 10:31

Luckyluck63

434211

answered May 1 '16 at 10:00

Jean MarieJean Marie

30.5k42154

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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2

$begingroup$

In general, we have
$$left[beginarrayccbf V & bf W \ bf X & bf Yendarrayright]^-1 = left[beginarrayccbf V^-1 + bf V^-1bf Wleft(bf Y-bf Xbf V^-1bf Wright)^-1bf Xbf V^-1& -bf V^-1bf Wleft(bf Y-bf Xbf V^-1bf Wright)^-1 \ -left(bf Y-bf Xbf V^-1bf Wright)^-1bf Xbf V^-1 & left(bf Y-bf Xbf V^-1bf Wright)^-1endarrayright].$$
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.

share|cite|improve this answer

edited May 1 '16 at 8:53

answered May 1 '16 at 8:48

Luckyluck63Luckyluck63

434211

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

2

$begingroup$

In general, we have
$$left[beginarrayccbf V & bf W \ bf X & bf Yendarrayright]^-1 = left[beginarrayccbf V^-1 + bf V^-1bf Wleft(bf Y-bf Xbf V^-1bf Wright)^-1bf Xbf V^-1& -bf V^-1bf Wleft(bf Y-bf Xbf V^-1bf Wright)^-1 \ -left(bf Y-bf Xbf V^-1bf Wright)^-1bf Xbf V^-1 & left(bf Y-bf Xbf V^-1bf Wright)^-1endarrayright].$$
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.

share|cite|improve this answer

edited May 1 '16 at 8:53

answered May 1 '16 at 8:48

Luckyluck63Luckyluck63

434211

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

$begingroup$

In general, we have
$$left[beginarrayccbf V & bf W \ bf X & bf Yendarrayright]^-1 = left[beginarrayccbf V^-1 + bf V^-1bf Wleft(bf Y-bf Xbf V^-1bf Wright)^-1bf Xbf V^-1& -bf V^-1bf Wleft(bf Y-bf Xbf V^-1bf Wright)^-1 \ -left(bf Y-bf Xbf V^-1bf Wright)^-1bf Xbf V^-1 & left(bf Y-bf Xbf V^-1bf Wright)^-1endarrayright].$$
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.

share|cite|improve this answer

edited May 1 '16 at 8:53

answered May 1 '16 at 8:48

Luckyluck63Luckyluck63

434211

$endgroup$

share|cite|improve this answer

edited May 1 '16 at 8:53

answered May 1 '16 at 8:48

Luckyluck63Luckyluck63

434211

answered May 1 '16 at 8:48

Luckyluck63Luckyluck63

434211

answered May 1 '16 at 8:48

Luckyluck63Luckyluck63

434211

2

$begingroup$

Let

$$bf M=beginbmatrix2bf I&-bf X\bf X'&bf 0endbmatrix$$

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.

share|cite|improve this answer

edited May 1 '16 at 10:31

Luckyluck63

434211

answered May 1 '16 at 10:00

Jean MarieJean Marie

30.5k42154

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

2

$begingroup$

Let

$$bf M=beginbmatrix2bf I&-bf X\bf X'&bf 0endbmatrix$$

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.

share|cite|improve this answer

edited May 1 '16 at 10:31

Luckyluck63

434211

answered May 1 '16 at 10:00

Jean MarieJean Marie

30.5k42154

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

$begingroup$

Let

$$bf M=beginbmatrix2bf I&-bf X\bf X'&bf 0endbmatrix$$

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.

share|cite|improve this answer

edited May 1 '16 at 10:31

Luckyluck63

434211

answered May 1 '16 at 10:00

Jean MarieJean Marie

30.5k42154

$endgroup$

share|cite|improve this answer

edited May 1 '16 at 10:31

Luckyluck63

434211

answered May 1 '16 at 10:00

Jean MarieJean Marie

30.5k42154

edited May 1 '16 at 10:31

Luckyluck63

434211

edited May 1 '16 at 10:31

Luckyluck63

434211

answered May 1 '16 at 10:00

Jean MarieJean Marie

30.5k42154

answered May 1 '16 at 10:00

Jean MarieJean Marie

30.5k42154