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Determinant of block matrix with commuting blocks

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4

2

$begingroup$

I know that given a $2N times 2N$ block matrix with $N times N$ blocks like

$$mathbfS = beginpmatrix
A & B\
C & D
endpmatrix$$

we can calculate $$det(mathbfS)=det(AD-BD^-1CD)$$ and so clearly if $D$ and $C$ commute this reduces to $det(AD-BC)$, which is a very nice property.

My question is, for a general $nNtimes nN$ matrix with $Ntimes N$ blocks where all of the blocks commute with each other, can we find the determinant in a similar way? That is, by first finding the determinant treating the blocks like scalars and then taking the determinant of the resulting $Ntimes N$ matrix.

Thanks in advance!

matrices determinant block-matrices

share|cite|improve this question

edited 15 hours ago

Rodrigo de Azevedo

13.1k41960

asked May 24 '15 at 5:53

GordonGordon

353

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $D$ is not invertible then what about this identity $det(mathbfS)=det(AD-BD^-1CD)$
  $endgroup$
  – Sry
  Jun 20 '15 at 6:54
  

  
  
  
  

add a comment | 

4

2

$begingroup$

I know that given a $2N times 2N$ block matrix with $N times N$ blocks like

$$mathbfS = beginpmatrix
A & B\
C & D
endpmatrix$$

we can calculate $$det(mathbfS)=det(AD-BD^-1CD)$$ and so clearly if $D$ and $C$ commute this reduces to $det(AD-BC)$, which is a very nice property.

My question is, for a general $nNtimes nN$ matrix with $Ntimes N$ blocks where all of the blocks commute with each other, can we find the determinant in a similar way? That is, by first finding the determinant treating the blocks like scalars and then taking the determinant of the resulting $Ntimes N$ matrix.

Thanks in advance!

matrices determinant block-matrices

share|cite|improve this question

edited 15 hours ago

Rodrigo de Azevedo

13.1k41960

asked May 24 '15 at 5:53

GordonGordon

353

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $D$ is not invertible then what about this identity $det(mathbfS)=det(AD-BD^-1CD)$
  $endgroup$
  – Sry
  Jun 20 '15 at 6:54
  

  
  
  
  

add a comment | 

$begingroup$

I know that given a $2N times 2N$ block matrix with $N times N$ blocks like

$$mathbfS = beginpmatrix
A & B\
C & D
endpmatrix$$

we can calculate $$det(mathbfS)=det(AD-BD^-1CD)$$ and so clearly if $D$ and $C$ commute this reduces to $det(AD-BC)$, which is a very nice property.

My question is, for a general $nNtimes nN$ matrix with $Ntimes N$ blocks where all of the blocks commute with each other, can we find the determinant in a similar way? That is, by first finding the determinant treating the blocks like scalars and then taking the determinant of the resulting $Ntimes N$ matrix.

Thanks in advance!

matrices determinant block-matrices

share|cite|improve this question

edited 15 hours ago

Rodrigo de Azevedo

13.1k41960

asked May 24 '15 at 5:53

GordonGordon

353

$endgroup$

share|cite|improve this question

edited 15 hours ago

Rodrigo de Azevedo

13.1k41960

asked May 24 '15 at 5:53

GordonGordon

353

share|cite|improve this question

edited 15 hours ago

Rodrigo de Azevedo

13.1k41960

asked May 24 '15 at 5:53

GordonGordon

353

edited 15 hours ago

Rodrigo de Azevedo

13.1k41960

edited 15 hours ago

Rodrigo de Azevedo

13.1k41960

asked May 24 '15 at 5:53

GordonGordon

353

asked May 24 '15 at 5:53

GordonGordon

353