Rank of the product of three matrices

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Assume that $AinmathbbR^ntimes nu$, $BinmathbbR^nutimes nu$ and $CinmathbbR^nutimes n$. I'm interested in the rank of the matrix
$$AB^-1C.$$
$A$ and $C$ are full rank ($n$), and of course, $B$ is as well ($nu$). In light of this operation I'm wondering if:

• Is there any explicit condition so that $operatornamerank(M)=n$?


• Can I say something about $operatornamerank(AB^-1C)$ using $operatornamerank(ABC)$ instead? (that would be much easier for my problem).

linear-algebra matrix-rank

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edited Jul 19 at 15:32

user7530

33.4k558109

asked Jul 19 at 15:15

Nico F.

1379

add a comment | 

up vote
3
down vote

favorite

Assume that $AinmathbbR^ntimes nu$, $BinmathbbR^nutimes nu$ and $CinmathbbR^nutimes n$. I'm interested in the rank of the matrix
$$AB^-1C.$$
$A$ and $C$ are full rank ($n$), and of course, $B$ is as well ($nu$). In light of this operation I'm wondering if:

• Is there any explicit condition so that $operatornamerank(M)=n$?


• Can I say something about $operatornamerank(AB^-1C)$ using $operatornamerank(ABC)$ instead? (that would be much easier for my problem).

linear-algebra matrix-rank

share|cite|improve this question

edited Jul 19 at 15:32

user7530

33.4k558109

asked Jul 19 at 15:15

Nico F.

1379

add a comment | 

up vote
3
down vote

favorite

up vote
3
down vote

favorite

Assume that $AinmathbbR^ntimes nu$, $BinmathbbR^nutimes nu$ and $CinmathbbR^nutimes n$. I'm interested in the rank of the matrix
$$AB^-1C.$$
$A$ and $C$ are full rank ($n$), and of course, $B$ is as well ($nu$). In light of this operation I'm wondering if:

• Is there any explicit condition so that $operatornamerank(M)=n$?


• Can I say something about $operatornamerank(AB^-1C)$ using $operatornamerank(ABC)$ instead? (that would be much easier for my problem).

linear-algebra matrix-rank

share|cite|improve this question

edited Jul 19 at 15:32

user7530

33.4k558109

asked Jul 19 at 15:15

Nico F.

1379

Assume that $AinmathbbR^ntimes nu$, $BinmathbbR^nutimes nu$ and $CinmathbbR^nutimes n$. I'm interested in the rank of the matrix
$$AB^-1C.$$
$A$ and $C$ are full rank ($n$), and of course, $B$ is as well ($nu$). In light of this operation I'm wondering if:

• Is there any explicit condition so that $operatornamerank(M)=n$?


• Can I say something about $operatornamerank(AB^-1C)$ using $operatornamerank(ABC)$ instead? (that would be much easier for my problem).

linear-algebra matrix-rank

share|cite|improve this question

edited Jul 19 at 15:32

user7530

33.4k558109

asked Jul 19 at 15:15

Nico F.

1379

share|cite|improve this question

share|cite|improve this question

edited Jul 19 at 15:32

user7530

33.4k558109

edited Jul 19 at 15:32

user7530

33.4k558109

edited Jul 19 at 15:32

user7530

33.4k558109

33.4k558109

asked Jul 19 at 15:15

Nico F.

1379

asked Jul 19 at 15:15

Nico F.

1379

asked Jul 19 at 15:15

Nico F.

1379

1379

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
2
down vote



accepted

Let me answer you second question first: unfortunately no, the conditions are not equivalent. Take
$$A=beginbmatrix0 \ 1 \ 0endbmatrix, B = beginbmatrix0 & 1 & 0\0 & 0 & 1\ 1 & 0 & 0endbmatrix, C = beginbmatrix1 \ 0 \ 0endbmatrix.$$
Then $A^TBC = 0$ but $A^T B^-1C = 1.$ Here the matrix $B$ was carefully crafted to rotate the image of $C$ into the kernel of $A^T$, but not vice-versa.

For the first question, I know of no simple condition (simpler than performing the multiplication and probing the product's rank). You can take the SVD of $A$ and $C$ and try to reason how B maps the different subspaces to each other, but at that point you're already better off taking the determinant...

share|cite|improve this answer

answered Jul 19 at 15:47

user7530

33.4k558109

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks a million for the answer. Extremely clear example! I'll follow your advice with the SVD... It will be a little messy since $A$,$B$ and $C$ are the result of several Kronecker products/sums. Thanks again!
  – Nico F.
  Jul 20 at 9:28
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Let me answer you second question first: unfortunately no, the conditions are not equivalent. Take
$$A=beginbmatrix0 \ 1 \ 0endbmatrix, B = beginbmatrix0 & 1 & 0\0 & 0 & 1\ 1 & 0 & 0endbmatrix, C = beginbmatrix1 \ 0 \ 0endbmatrix.$$
Then $A^TBC = 0$ but $A^T B^-1C = 1.$ Here the matrix $B$ was carefully crafted to rotate the image of $C$ into the kernel of $A^T$, but not vice-versa.

For the first question, I know of no simple condition (simpler than performing the multiplication and probing the product's rank). You can take the SVD of $A$ and $C$ and try to reason how B maps the different subspaces to each other, but at that point you're already better off taking the determinant...

share|cite|improve this answer

answered Jul 19 at 15:47

user7530

33.4k558109

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks a million for the answer. Extremely clear example! I'll follow your advice with the SVD... It will be a little messy since $A$,$B$ and $C$ are the result of several Kronecker products/sums. Thanks again!
  – Nico F.
  Jul 20 at 9:28
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

Let me answer you second question first: unfortunately no, the conditions are not equivalent. Take
$$A=beginbmatrix0 \ 1 \ 0endbmatrix, B = beginbmatrix0 & 1 & 0\0 & 0 & 1\ 1 & 0 & 0endbmatrix, C = beginbmatrix1 \ 0 \ 0endbmatrix.$$
Then $A^TBC = 0$ but $A^T B^-1C = 1.$ Here the matrix $B$ was carefully crafted to rotate the image of $C$ into the kernel of $A^T$, but not vice-versa.

For the first question, I know of no simple condition (simpler than performing the multiplication and probing the product's rank). You can take the SVD of $A$ and $C$ and try to reason how B maps the different subspaces to each other, but at that point you're already better off taking the determinant...

share|cite|improve this answer

answered Jul 19 at 15:47

user7530

33.4k558109

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks a million for the answer. Extremely clear example! I'll follow your advice with the SVD... It will be a little messy since $A$,$B$ and $C$ are the result of several Kronecker products/sums. Thanks again!
  – Nico F.
  Jul 20 at 9:28
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Let me answer you second question first: unfortunately no, the conditions are not equivalent. Take
$$A=beginbmatrix0 \ 1 \ 0endbmatrix, B = beginbmatrix0 & 1 & 0\0 & 0 & 1\ 1 & 0 & 0endbmatrix, C = beginbmatrix1 \ 0 \ 0endbmatrix.$$
Then $A^TBC = 0$ but $A^T B^-1C = 1.$ Here the matrix $B$ was carefully crafted to rotate the image of $C$ into the kernel of $A^T$, but not vice-versa.

For the first question, I know of no simple condition (simpler than performing the multiplication and probing the product's rank). You can take the SVD of $A$ and $C$ and try to reason how B maps the different subspaces to each other, but at that point you're already better off taking the determinant...

share|cite|improve this answer

answered Jul 19 at 15:47

user7530

33.4k558109

Let me answer you second question first: unfortunately no, the conditions are not equivalent. Take
$$A=beginbmatrix0 \ 1 \ 0endbmatrix, B = beginbmatrix0 & 1 & 0\0 & 0 & 1\ 1 & 0 & 0endbmatrix, C = beginbmatrix1 \ 0 \ 0endbmatrix.$$
Then $A^TBC = 0$ but $A^T B^-1C = 1.$ Here the matrix $B$ was carefully crafted to rotate the image of $C$ into the kernel of $A^T$, but not vice-versa.

For the first question, I know of no simple condition (simpler than performing the multiplication and probing the product's rank). You can take the SVD of $A$ and $C$ and try to reason how B maps the different subspaces to each other, but at that point you're already better off taking the determinant...

share|cite|improve this answer

answered Jul 19 at 15:47

user7530

33.4k558109

share|cite|improve this answer

share|cite|improve this answer

answered Jul 19 at 15:47

user7530

33.4k558109

answered Jul 19 at 15:47

user7530

33.4k558109

answered Jul 19 at 15:47

user7530

33.4k558109

33.4k558109

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks a million for the answer. Extremely clear example! I'll follow your advice with the SVD... It will be a little messy since $A$,$B$ and $C$ are the result of several Kronecker products/sums. Thanks again!
  – Nico F.
  Jul 20 at 9:28
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks a million for the answer. Extremely clear example! I'll follow your advice with the SVD... It will be a little messy since $A$,$B$ and $C$ are the result of several Kronecker products/sums. Thanks again!
  – Nico F.
  Jul 20 at 9:28
  

  
  
  
  

Thanks a million for the answer. Extremely clear example! I'll follow your advice with the SVD... It will be a little messy since $A$,$B$ and $C$ are the result of several Kronecker products/sums. Thanks again!
– Nico F.
Jul 20 at 9:28

Thanks a million for the answer. Extremely clear example! I'll follow your advice with the SVD... It will be a little messy since $A$,$B$ and $C$ are the result of several Kronecker products/sums. Thanks again!
– Nico F.
Jul 20 at 9:28

add a comment | 

 

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