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How to prove the following rank equality?

Equality case in the Frobenius rank inequalityEqualities of rank inequalitiesMatrix rank inequalityRank of a block lower triangular matrixProving a matrix rank equalityLet $A$ and $B$ be two square Hermitian matrices such that $AB = 0 = BA$. Prove that $mboxrank (A + B) = mboxrank(A) + mboxrank (B)$.Rank of a matrix sum: equality caseOn the equality of the rank-sum inequalityRank of projection matrix $P$$a_k = rank(A^k+1) - rank(A^k)$ is increasing

1

$begingroup$

Let $A,Bin M_n$. Show that $$mboxrank,(A-ABA)=mboxrank,(A)+mboxrank,(I_n-BA)-n$$

I have showed one part by using Sylvester rank inequality. How to prove the equality case?

linear-algebra matrices matrix-rank

share|cite|improve this question

edited Mar 17 at 12:57

Rodrigo de Azevedo

13.2k41960

asked Mar 17 at 12:42

J.DoeJ.Doe

341

$endgroup$

add a comment | 

1

$begingroup$

Let $A,Bin M_n$. Show that $$mboxrank,(A-ABA)=mboxrank,(A)+mboxrank,(I_n-BA)-n$$

I have showed one part by using Sylvester rank inequality. How to prove the equality case?

linear-algebra matrices matrix-rank

share|cite|improve this question

edited Mar 17 at 12:57

Rodrigo de Azevedo

13.2k41960

asked Mar 17 at 12:42

J.DoeJ.Doe

341

$endgroup$

add a comment | 

$begingroup$

Let $A,Bin M_n$. Show that $$mboxrank,(A-ABA)=mboxrank,(A)+mboxrank,(I_n-BA)-n$$

I have showed one part by using Sylvester rank inequality. How to prove the equality case?

linear-algebra matrices matrix-rank

share|cite|improve this question

edited Mar 17 at 12:57

Rodrigo de Azevedo

13.2k41960

asked Mar 17 at 12:42

J.DoeJ.Doe

341

$endgroup$

share|cite|improve this question

edited Mar 17 at 12:57

Rodrigo de Azevedo

13.2k41960

asked Mar 17 at 12:42

J.DoeJ.Doe

341

share|cite|improve this question

edited Mar 17 at 12:57

Rodrigo de Azevedo

13.2k41960

asked Mar 17 at 12:42

J.DoeJ.Doe

341

edited Mar 17 at 12:57

Rodrigo de Azevedo

13.2k41960

edited Mar 17 at 12:57

Rodrigo de Azevedo

13.2k41960

asked Mar 17 at 12:42

J.DoeJ.Doe

341

asked Mar 17 at 12:42

J.DoeJ.Doe

341

1 Answer
1

active

oldest

votes

1

$begingroup$

Fix a basis of $mathbb F^n$ (where $mathbb F$ is the field of entries of $A$ and $B$) so that $A$ and $B$ can be considered as linear transformations. Now, since $A - ABA = A(I - BA)$, we have
$requireAMScdDeclareMathOperatorimimDeclareMathOperatorrankrankDeclareMathOperatornullnull$

beginCD
mathbb F^n @>I-BA>> im(I - BA) @>A>> im(A - ABA)\
endCD

so that by the rank-nullity theorem applied on the second map, $A$,

$$dim im(I - BA) = dim im(A - ABA) + null A$$

where $null A$ is the nullity of $A$ in $im(I - BA)$. Thus,

beginequation
rank(I - BA) = rank(A - ABA) + null A. tag1labelrk-null
endequation

We shall show that the nullity of $A$ in $im(I - BA)$ is the nullity of $A$ in $mathbb F^n$ itself.

Let $x in ker_mathbb F^N A$. Then $Ax = 0 implies BAx = 0$, so we may write $x = x - BAx = (I - BA)x$, which shows that $ker_mathbb F^n A subseteq im(I - BA)$. Thus, $ker_im(I - BA) A = ker_mathbb F^n cap im(I - BA) = ker_mathbb F^n A$. Thus, $null A = n - rank A$ in $im(I - BA)$.

Then the result follows from eqrefrk-null.

share|cite|improve this answer

answered Mar 17 at 13:53

M. VinayM. Vinay

7,11322135

$endgroup$

add a comment | 

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1

$begingroup$

Fix a basis of $mathbb F^n$ (where $mathbb F$ is the field of entries of $A$ and $B$) so that $A$ and $B$ can be considered as linear transformations. Now, since $A - ABA = A(I - BA)$, we have
$requireAMScdDeclareMathOperatorimimDeclareMathOperatorrankrankDeclareMathOperatornullnull$

beginCD
mathbb F^n @>I-BA>> im(I - BA) @>A>> im(A - ABA)\
endCD

so that by the rank-nullity theorem applied on the second map, $A$,

$$dim im(I - BA) = dim im(A - ABA) + null A$$

where $null A$ is the nullity of $A$ in $im(I - BA)$. Thus,

beginequation
rank(I - BA) = rank(A - ABA) + null A. tag1labelrk-null
endequation

We shall show that the nullity of $A$ in $im(I - BA)$ is the nullity of $A$ in $mathbb F^n$ itself.

Let $x in ker_mathbb F^N A$. Then $Ax = 0 implies BAx = 0$, so we may write $x = x - BAx = (I - BA)x$, which shows that $ker_mathbb F^n A subseteq im(I - BA)$. Thus, $ker_im(I - BA) A = ker_mathbb F^n cap im(I - BA) = ker_mathbb F^n A$. Thus, $null A = n - rank A$ in $im(I - BA)$.

Then the result follows from eqrefrk-null.

share|cite|improve this answer

answered Mar 17 at 13:53

M. VinayM. Vinay

7,11322135

$endgroup$

add a comment | 

1

$begingroup$

Fix a basis of $mathbb F^n$ (where $mathbb F$ is the field of entries of $A$ and $B$) so that $A$ and $B$ can be considered as linear transformations. Now, since $A - ABA = A(I - BA)$, we have
$requireAMScdDeclareMathOperatorimimDeclareMathOperatorrankrankDeclareMathOperatornullnull$

beginCD
mathbb F^n @>I-BA>> im(I - BA) @>A>> im(A - ABA)\
endCD

so that by the rank-nullity theorem applied on the second map, $A$,

$$dim im(I - BA) = dim im(A - ABA) + null A$$

where $null A$ is the nullity of $A$ in $im(I - BA)$. Thus,

beginequation
rank(I - BA) = rank(A - ABA) + null A. tag1labelrk-null
endequation

We shall show that the nullity of $A$ in $im(I - BA)$ is the nullity of $A$ in $mathbb F^n$ itself.

Let $x in ker_mathbb F^N A$. Then $Ax = 0 implies BAx = 0$, so we may write $x = x - BAx = (I - BA)x$, which shows that $ker_mathbb F^n A subseteq im(I - BA)$. Thus, $ker_im(I - BA) A = ker_mathbb F^n cap im(I - BA) = ker_mathbb F^n A$. Thus, $null A = n - rank A$ in $im(I - BA)$.

Then the result follows from eqrefrk-null.

share|cite|improve this answer

answered Mar 17 at 13:53

M. VinayM. Vinay

7,11322135

$endgroup$

add a comment | 

$begingroup$

Fix a basis of $mathbb F^n$ (where $mathbb F$ is the field of entries of $A$ and $B$) so that $A$ and $B$ can be considered as linear transformations. Now, since $A - ABA = A(I - BA)$, we have
$requireAMScdDeclareMathOperatorimimDeclareMathOperatorrankrankDeclareMathOperatornullnull$

beginCD
mathbb F^n @>I-BA>> im(I - BA) @>A>> im(A - ABA)\
endCD

so that by the rank-nullity theorem applied on the second map, $A$,

$$dim im(I - BA) = dim im(A - ABA) + null A$$

where $null A$ is the nullity of $A$ in $im(I - BA)$. Thus,

beginequation
rank(I - BA) = rank(A - ABA) + null A. tag1labelrk-null
endequation

We shall show that the nullity of $A$ in $im(I - BA)$ is the nullity of $A$ in $mathbb F^n$ itself.

Let $x in ker_mathbb F^N A$. Then $Ax = 0 implies BAx = 0$, so we may write $x = x - BAx = (I - BA)x$, which shows that $ker_mathbb F^n A subseteq im(I - BA)$. Thus, $ker_im(I - BA) A = ker_mathbb F^n cap im(I - BA) = ker_mathbb F^n A$. Thus, $null A = n - rank A$ in $im(I - BA)$.

Then the result follows from eqrefrk-null.

share|cite|improve this answer

answered Mar 17 at 13:53

M. VinayM. Vinay

7,11322135

$endgroup$

share|cite|improve this answer

answered Mar 17 at 13:53

M. VinayM. Vinay

7,11322135

answered Mar 17 at 13:53

M. VinayM. Vinay

7,11322135

answered Mar 17 at 13:53

M. VinayM. Vinay

7,11322135