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Prove $operatornamerank (BA) = operatornamerank (A) = operatornamerank (AC)$, for $B$ and $C$ invertible matrices

Show that $operatornamerank(A+B) leq operatornamerank(A) + operatornamerank(B)$Show that $operatornamerank(A) = operatornamerank(B)$Prove that $operatornamerank(A) + operatornamerank(B) ge operatornamerank(A + B)$Rank of the sum of two matricesHow to prove a property of ranks: $operatornamerank(AB)= operatornamerank(B)- dim(operatornameIm B cap ker A)$How to prove $operatornamerank(AB) = operatornamerankA$?How to prove $operatornamerank(XGY^T) = operatornamerank(G)$?Show that $operatornamerank(A)=operatornamerank(B)$Prove $operatornamerank(A)=operatornamerank(PAQ)$ , when $A$ is a $mtimes n$ matrix and $P$ and $Q$ are invertible.$DeclareMathOperatorrankrankrank(A) = rank(B)$, Prove there exist $U, V$ invertible matrices such that: $A = UBV$

2

2

$begingroup$

This is my question:

Let $Ain F^m times n$, $Bin F^m times m$ and $Cin F^n times n$ be invertible matrices. Prove that $$operatornamerank (BA) = operatornamerank (A) = operatornamerank (AC).$$

How to proceed with this question? A complete answer will help my proper understanding. :)

linear-algebra linear-transformations matrix-rank

share|cite|improve this question

edited Mar 14 at 5:09

Rócherz

2,9863821

asked Apr 30 '15 at 13:39

Half-Blood princeHalf-Blood prince

5561521

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Let's start with this: what definition of rank are you most comfortable with?
  $endgroup$
  – Omnomnomnom
  Apr 30 '15 at 15:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  dim(R(T))=Rank(T) where R(T) is range of transformation T
  $endgroup$
  – Half-Blood prince
  Apr 30 '15 at 17:31
  

  
  
  
  

add a comment | 

2

2

$begingroup$

This is my question:

Let $Ain F^m times n$, $Bin F^m times m$ and $Cin F^n times n$ be invertible matrices. Prove that $$operatornamerank (BA) = operatornamerank (A) = operatornamerank (AC).$$

How to proceed with this question? A complete answer will help my proper understanding. :)

linear-algebra linear-transformations matrix-rank

share|cite|improve this question

edited Mar 14 at 5:09

Rócherz

2,9863821

asked Apr 30 '15 at 13:39

Half-Blood princeHalf-Blood prince

5561521

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Let's start with this: what definition of rank are you most comfortable with?
  $endgroup$
  – Omnomnomnom
  Apr 30 '15 at 15:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  dim(R(T))=Rank(T) where R(T) is range of transformation T
  $endgroup$
  – Half-Blood prince
  Apr 30 '15 at 17:31
  

  
  
  
  

add a comment | 

$begingroup$

This is my question:

Let $Ain F^m times n$, $Bin F^m times m$ and $Cin F^n times n$ be invertible matrices. Prove that $$operatornamerank (BA) = operatornamerank (A) = operatornamerank (AC).$$

How to proceed with this question? A complete answer will help my proper understanding. :)

linear-algebra linear-transformations matrix-rank

share|cite|improve this question

edited Mar 14 at 5:09

Rócherz

2,9863821

asked Apr 30 '15 at 13:39

Half-Blood princeHalf-Blood prince

5561521

$endgroup$

share|cite|improve this question

edited Mar 14 at 5:09

Rócherz

2,9863821

asked Apr 30 '15 at 13:39

Half-Blood princeHalf-Blood prince

5561521

share|cite|improve this question

edited Mar 14 at 5:09

Rócherz

2,9863821

asked Apr 30 '15 at 13:39

Half-Blood princeHalf-Blood prince

5561521

edited Mar 14 at 5:09

Rócherz

2,9863821

edited Mar 14 at 5:09

Rócherz

2,9863821

asked Apr 30 '15 at 13:39

Half-Blood princeHalf-Blood prince

5561521

asked Apr 30 '15 at 13:39

Half-Blood princeHalf-Blood prince

5561521

2 Answers
2

active

oldest

votes

2

$begingroup$

The key is to note that because $A$ and $C$ are invertible, they are isomorphisms of vector spaces. Thus, if $U$ is a subspace of $V$, we have $dim(B(U)) = dim(U)$.

For the first part: note that $B(Bbb R^n) = R(B)$ is a subspace of $Bbb R^m$. It follows that
$$
dim(R(AB)) = dim(AB(Bbb R^n))
= dim(A[B(Bbb R^n)])
= dim(A[R(B)]) = dim(R(B))
$$
we conclude that $AB$ has the same rank as $B$.

For the second: note that $dim(R(C)) = n$, from which it follows that $R(C) = Bbb R^n$. So, we have
$$
R(BC) = BC(Bbb R^n) = B[C(Bbb R^n)]
= B[Bbb R^n] = R(B)
$$
it follows that $BC$ has the same rank as $B$.

share|cite|improve this answer

edited May 1 '15 at 11:28

answered Apr 30 '15 at 19:31

OmnomnomnomOmnomnomnom

129k792185

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how did you write $dim(A[R(B)])=dim(R(B))$ ?
  $endgroup$
  – Half-Blood prince
  May 1 '15 at 7:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $A$ is an a isomorphism
  $endgroup$
  – Omnomnomnom
  May 1 '15 at 11:29
  

  
  
  
  

add a comment | 

0

$begingroup$

To show that $operatornamerank (A) = operatornamerank (AC)$, column space of $AC$ is a subspace of column space of $A$. Now, see that $A = (AC)C^-1$, which says that column space of $A$ is a subspace of column space of $AC$. For the first part you may use row rank of $A$ equals the column rank of $A$.

share|cite|improve this answer

edited Mar 14 at 5:11

Rócherz

2,9863821

answered Apr 30 '15 at 15:37

SAUVIKSAUVIK

162111

$endgroup$

add a comment | 

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2

$begingroup$

The key is to note that because $A$ and $C$ are invertible, they are isomorphisms of vector spaces. Thus, if $U$ is a subspace of $V$, we have $dim(B(U)) = dim(U)$.

For the first part: note that $B(Bbb R^n) = R(B)$ is a subspace of $Bbb R^m$. It follows that
$$
dim(R(AB)) = dim(AB(Bbb R^n))
= dim(A[B(Bbb R^n)])
= dim(A[R(B)]) = dim(R(B))
$$
we conclude that $AB$ has the same rank as $B$.

For the second: note that $dim(R(C)) = n$, from which it follows that $R(C) = Bbb R^n$. So, we have
$$
R(BC) = BC(Bbb R^n) = B[C(Bbb R^n)]
= B[Bbb R^n] = R(B)
$$
it follows that $BC$ has the same rank as $B$.

share|cite|improve this answer

edited May 1 '15 at 11:28

answered Apr 30 '15 at 19:31

OmnomnomnomOmnomnomnom

129k792185

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how did you write $dim(A[R(B)])=dim(R(B))$ ?
  $endgroup$
  – Half-Blood prince
  May 1 '15 at 7:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $A$ is an a isomorphism
  $endgroup$
  – Omnomnomnom
  May 1 '15 at 11:29
  

  
  
  
  

add a comment | 

2

$begingroup$

The key is to note that because $A$ and $C$ are invertible, they are isomorphisms of vector spaces. Thus, if $U$ is a subspace of $V$, we have $dim(B(U)) = dim(U)$.

For the first part: note that $B(Bbb R^n) = R(B)$ is a subspace of $Bbb R^m$. It follows that
$$
dim(R(AB)) = dim(AB(Bbb R^n))
= dim(A[B(Bbb R^n)])
= dim(A[R(B)]) = dim(R(B))
$$
we conclude that $AB$ has the same rank as $B$.

For the second: note that $dim(R(C)) = n$, from which it follows that $R(C) = Bbb R^n$. So, we have
$$
R(BC) = BC(Bbb R^n) = B[C(Bbb R^n)]
= B[Bbb R^n] = R(B)
$$
it follows that $BC$ has the same rank as $B$.

share|cite|improve this answer

edited May 1 '15 at 11:28

answered Apr 30 '15 at 19:31

OmnomnomnomOmnomnomnom

129k792185

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how did you write $dim(A[R(B)])=dim(R(B))$ ?
  $endgroup$
  – Half-Blood prince
  May 1 '15 at 7:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $A$ is an a isomorphism
  $endgroup$
  – Omnomnomnom
  May 1 '15 at 11:29
  

  
  
  
  

add a comment | 

$begingroup$

The key is to note that because $A$ and $C$ are invertible, they are isomorphisms of vector spaces. Thus, if $U$ is a subspace of $V$, we have $dim(B(U)) = dim(U)$.

For the first part: note that $B(Bbb R^n) = R(B)$ is a subspace of $Bbb R^m$. It follows that
$$
dim(R(AB)) = dim(AB(Bbb R^n))
= dim(A[B(Bbb R^n)])
= dim(A[R(B)]) = dim(R(B))
$$
we conclude that $AB$ has the same rank as $B$.

For the second: note that $dim(R(C)) = n$, from which it follows that $R(C) = Bbb R^n$. So, we have
$$
R(BC) = BC(Bbb R^n) = B[C(Bbb R^n)]
= B[Bbb R^n] = R(B)
$$
it follows that $BC$ has the same rank as $B$.

share|cite|improve this answer

edited May 1 '15 at 11:28

answered Apr 30 '15 at 19:31

OmnomnomnomOmnomnomnom

129k792185

$endgroup$

share|cite|improve this answer

edited May 1 '15 at 11:28

answered Apr 30 '15 at 19:31

OmnomnomnomOmnomnomnom

129k792185

answered Apr 30 '15 at 19:31

OmnomnomnomOmnomnomnom

129k792185

answered Apr 30 '15 at 19:31

OmnomnomnomOmnomnomnom

129k792185

0

$begingroup$

To show that $operatornamerank (A) = operatornamerank (AC)$, column space of $AC$ is a subspace of column space of $A$. Now, see that $A = (AC)C^-1$, which says that column space of $A$ is a subspace of column space of $AC$. For the first part you may use row rank of $A$ equals the column rank of $A$.

share|cite|improve this answer

edited Mar 14 at 5:11

Rócherz

2,9863821

answered Apr 30 '15 at 15:37

SAUVIKSAUVIK

162111

$endgroup$

add a comment | 

0

$begingroup$

To show that $operatornamerank (A) = operatornamerank (AC)$, column space of $AC$ is a subspace of column space of $A$. Now, see that $A = (AC)C^-1$, which says that column space of $A$ is a subspace of column space of $AC$. For the first part you may use row rank of $A$ equals the column rank of $A$.

share|cite|improve this answer

edited Mar 14 at 5:11

Rócherz

2,9863821

answered Apr 30 '15 at 15:37

SAUVIKSAUVIK

162111

$endgroup$

add a comment | 

$begingroup$

To show that $operatornamerank (A) = operatornamerank (AC)$, column space of $AC$ is a subspace of column space of $A$. Now, see that $A = (AC)C^-1$, which says that column space of $A$ is a subspace of column space of $AC$. For the first part you may use row rank of $A$ equals the column rank of $A$.

share|cite|improve this answer

edited Mar 14 at 5:11

Rócherz

2,9863821

answered Apr 30 '15 at 15:37

SAUVIKSAUVIK

162111

$endgroup$

share|cite|improve this answer

edited Mar 14 at 5:11

Rócherz

2,9863821

answered Apr 30 '15 at 15:37

SAUVIKSAUVIK

162111

edited Mar 14 at 5:11

Rócherz

2,9863821

edited Mar 14 at 5:11

Rócherz

2,9863821

answered Apr 30 '15 at 15:37

SAUVIKSAUVIK

162111

answered Apr 30 '15 at 15:37

SAUVIKSAUVIK

162111