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Is $operatornamerank(AQB)=operatornamerank(AB)$ if $Q$ is non-singular?

For given $ntimes n$ matrix $A$ singular matrix, prove that $operatornamerank(operatornameadjA) leq 1$Prove that matrices have equal rank.Counterexample of Converse of “$operatornamerank (PA) = operatornamerank (A)$ if $P$ is invertible”Rank of orthogonal projection to prep of null-space, right-singular matrix and identity matrix$operatornameRank(A)leq operatornameRank(AB).$If $A$ is of rank $n$ then why is it non-singular?If $A$ is $ktimes l$ and $B$ is $ltimes k$, $lgeq k$ and full-rank, must $AB$ have full-rank?Understanding the low-rank manifoldProve $operatornamerank(A)=operatornamerank(PAQ)$ , when $A$ is a $mtimes n$ matrix and $P$ and $Q$ are invertible.$operatornamerank(A^2)+operatornamerank(B^2)geq2operatornamerank(AB)$ whenever $AB=BA$?

2

$begingroup$

$newcommandrankoperatornamerank$We know that $rank(PA)=rank(AQ)=rank(PAQ)=rank(A)$ where $Ain M_mtimes n(mathbb F), P, Q$ are $mtimes m, ntimes n$ invertible matrices.

mean to say , from the matrix product we can remove the non-singular matrices; rank will not be effected.

After studying this, it came to my mind, then what will happen in the case of $rank(PAQB)$ where $Bin M_ntimes m(mathbb F)$ ?

No idea. The only thing I got is $rank(PAQB)=rank(AQB)$. Can we remove $Q$ as well and write $rank(AQB)=rank(AB)$?

Please help. In case it has been solved earlier, provide me the link.

linear-algebra matrices matrix-rank

share|cite|improve this question

edited Mar 14 at 6:07

Rócherz

2,9863821

asked Jul 23 '15 at 5:08

Anjan3Anjan3

2,298923

$endgroup$

add a comment | 

2

$begingroup$

$newcommandrankoperatornamerank$We know that $rank(PA)=rank(AQ)=rank(PAQ)=rank(A)$ where $Ain M_mtimes n(mathbb F), P, Q$ are $mtimes m, ntimes n$ invertible matrices.

mean to say , from the matrix product we can remove the non-singular matrices; rank will not be effected.

After studying this, it came to my mind, then what will happen in the case of $rank(PAQB)$ where $Bin M_ntimes m(mathbb F)$ ?

No idea. The only thing I got is $rank(PAQB)=rank(AQB)$. Can we remove $Q$ as well and write $rank(AQB)=rank(AB)$?

Please help. In case it has been solved earlier, provide me the link.

linear-algebra matrices matrix-rank

share|cite|improve this question

edited Mar 14 at 6:07

Rócherz

2,9863821

asked Jul 23 '15 at 5:08

Anjan3Anjan3

2,298923

$endgroup$

add a comment | 

$begingroup$

$newcommandrankoperatornamerank$We know that $rank(PA)=rank(AQ)=rank(PAQ)=rank(A)$ where $Ain M_mtimes n(mathbb F), P, Q$ are $mtimes m, ntimes n$ invertible matrices.

mean to say , from the matrix product we can remove the non-singular matrices; rank will not be effected.

After studying this, it came to my mind, then what will happen in the case of $rank(PAQB)$ where $Bin M_ntimes m(mathbb F)$ ?

No idea. The only thing I got is $rank(PAQB)=rank(AQB)$. Can we remove $Q$ as well and write $rank(AQB)=rank(AB)$?

Please help. In case it has been solved earlier, provide me the link.

linear-algebra matrices matrix-rank

share|cite|improve this question

edited Mar 14 at 6:07

Rócherz

2,9863821

asked Jul 23 '15 at 5:08

Anjan3Anjan3

2,298923

$endgroup$

share|cite|improve this question

edited Mar 14 at 6:07

Rócherz

2,9863821

asked Jul 23 '15 at 5:08

Anjan3Anjan3

2,298923

share|cite|improve this question

edited Mar 14 at 6:07

Rócherz

2,9863821

asked Jul 23 '15 at 5:08

Anjan3Anjan3

2,298923

edited Mar 14 at 6:07

Rócherz

2,9863821

edited Mar 14 at 6:07

Rócherz

2,9863821

asked Jul 23 '15 at 5:08

Anjan3Anjan3

2,298923

asked Jul 23 '15 at 5:08

Anjan3Anjan3

2,298923

2 Answers
2

active

oldest

votes

8

$begingroup$

No. Worse, the much weaker property of the product being zero is not preserved by insertion of a nonsingular matrix: If we take $$A = B = pmatrix0&1\0&0, quad Q = pmatrix0&1\1&0,$$ then
$$AQB = A neq 0$$
but $$AB = 0.$$

share|cite|improve this answer

answered Jul 23 '15 at 5:15

TravisTravis

63.5k769150

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thank you. that cleared my doubt.
  $endgroup$
  – Anjan3
  Jul 23 '15 at 5:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome, I'm glad you found it useful.
  $endgroup$
  – Travis
  Jul 23 '15 at 6:04
  

  
  
  
  

add a comment | 

3

$begingroup$

To give an example in the opposite direction of Travis's answer, consider
$$A = B = left[beginarraycc 1 & 0\0 & 0endarrayright],quad Q = left[beginarraycc 0 & 1\1 & 0endarrayright]$$

Then, $AB = A$ has rank $1$, but $AQB = 0$ has rank $0$. Geometrically, $A$ fixes $e_1$, and maps $e_2$ to $0$, and $Q$ interchanged $e_1$ and $e_2$, where $e_1 = left[beginarrayc 1\0endarrayright], e_2 = left[beginarrayc 0\1endarrayright]$ are the standard basis.

share|cite|improve this answer

answered Jul 23 '15 at 5:58

StrantsStrants

5,61521736

$endgroup$

add a comment | 

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8

$begingroup$

No. Worse, the much weaker property of the product being zero is not preserved by insertion of a nonsingular matrix: If we take $$A = B = pmatrix0&1\0&0, quad Q = pmatrix0&1\1&0,$$ then
$$AQB = A neq 0$$
but $$AB = 0.$$

share|cite|improve this answer

answered Jul 23 '15 at 5:15

TravisTravis

63.5k769150

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thank you. that cleared my doubt.
  $endgroup$
  – Anjan3
  Jul 23 '15 at 5:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome, I'm glad you found it useful.
  $endgroup$
  – Travis
  Jul 23 '15 at 6:04
  

  
  
  
  

add a comment | 

8

$begingroup$

No. Worse, the much weaker property of the product being zero is not preserved by insertion of a nonsingular matrix: If we take $$A = B = pmatrix0&1\0&0, quad Q = pmatrix0&1\1&0,$$ then
$$AQB = A neq 0$$
but $$AB = 0.$$

share|cite|improve this answer

answered Jul 23 '15 at 5:15

TravisTravis

63.5k769150

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thank you. that cleared my doubt.
  $endgroup$
  – Anjan3
  Jul 23 '15 at 5:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome, I'm glad you found it useful.
  $endgroup$
  – Travis
  Jul 23 '15 at 6:04
  

  
  
  
  

add a comment | 

$begingroup$

No. Worse, the much weaker property of the product being zero is not preserved by insertion of a nonsingular matrix: If we take $$A = B = pmatrix0&1\0&0, quad Q = pmatrix0&1\1&0,$$ then
$$AQB = A neq 0$$
but $$AB = 0.$$

share|cite|improve this answer

answered Jul 23 '15 at 5:15

TravisTravis

63.5k769150

$endgroup$

share|cite|improve this answer

answered Jul 23 '15 at 5:15

TravisTravis

63.5k769150

answered Jul 23 '15 at 5:15

TravisTravis

63.5k769150

answered Jul 23 '15 at 5:15

TravisTravis

63.5k769150

3

$begingroup$

To give an example in the opposite direction of Travis's answer, consider
$$A = B = left[beginarraycc 1 & 0\0 & 0endarrayright],quad Q = left[beginarraycc 0 & 1\1 & 0endarrayright]$$

Then, $AB = A$ has rank $1$, but $AQB = 0$ has rank $0$. Geometrically, $A$ fixes $e_1$, and maps $e_2$ to $0$, and $Q$ interchanged $e_1$ and $e_2$, where $e_1 = left[beginarrayc 1\0endarrayright], e_2 = left[beginarrayc 0\1endarrayright]$ are the standard basis.

share|cite|improve this answer

answered Jul 23 '15 at 5:58

StrantsStrants

5,61521736

$endgroup$

add a comment | 

3

$begingroup$

To give an example in the opposite direction of Travis's answer, consider
$$A = B = left[beginarraycc 1 & 0\0 & 0endarrayright],quad Q = left[beginarraycc 0 & 1\1 & 0endarrayright]$$

Then, $AB = A$ has rank $1$, but $AQB = 0$ has rank $0$. Geometrically, $A$ fixes $e_1$, and maps $e_2$ to $0$, and $Q$ interchanged $e_1$ and $e_2$, where $e_1 = left[beginarrayc 1\0endarrayright], e_2 = left[beginarrayc 0\1endarrayright]$ are the standard basis.

share|cite|improve this answer

answered Jul 23 '15 at 5:58

StrantsStrants

5,61521736

$endgroup$

add a comment | 

$begingroup$

To give an example in the opposite direction of Travis's answer, consider
$$A = B = left[beginarraycc 1 & 0\0 & 0endarrayright],quad Q = left[beginarraycc 0 & 1\1 & 0endarrayright]$$

Then, $AB = A$ has rank $1$, but $AQB = 0$ has rank $0$. Geometrically, $A$ fixes $e_1$, and maps $e_2$ to $0$, and $Q$ interchanged $e_1$ and $e_2$, where $e_1 = left[beginarrayc 1\0endarrayright], e_2 = left[beginarrayc 0\1endarrayright]$ are the standard basis.

share|cite|improve this answer

answered Jul 23 '15 at 5:58

StrantsStrants

5,61521736

$endgroup$

share|cite|improve this answer

answered Jul 23 '15 at 5:58

StrantsStrants

5,61521736

answered Jul 23 '15 at 5:58

StrantsStrants

5,61521736

answered Jul 23 '15 at 5:58

StrantsStrants

5,61521736