Fdtxjr

Did I make a mistake by ccing email to boss to others?

Typing CO_2 easily

Telemetry for feature health

Ways of geometrical multiplication

Why the "ls" command is showing the permissions of files in a FAT32 partition?

Why can't the Brexit deadlock in the UK parliament be solved with a plurality vote?

Are Captain Marvel's powers affected by Thanos breaking the Tesseract and claiming the stone?

What is the smallest number n> 5 so that 5 ^ n ends with "3125"?

When and why was runway 07/25 at Kai Tak removed?

Isometric embedding of a genus g surface

Can I cause damage to electrical appliances by unplugging them when they are turned on?

Echo with obfuscation

What is the meaning of the following sentence?

How do I prevent inappropriate ads from appearing in my game?

How to reduce predictors the right way for a logistic regression model

The Digit Triangles

What the heck is gets(stdin) on site coderbyte?

Does Doodling or Improvising on the Piano Have Any Benefits?

"Oh no!" in Latin

Sigmoid with a slope but no asymptotes?

What is the meaning of "You've never met a graph you didn't like?"

What is this high flying aircraft over Pennsylvania?

Would a primitive species be able to learn English from reading books alone?

Anime with legendary swords made from talismans and a man who could change them with a shattered body

If $operatornamerank (AB) = operatornamerank (BA)$ for any $B$, then is $A$ invertible?

How to prove and interpret $operatornamerank(AB) leq operatornamemin(operatornamerank(A), operatornamerank(B))$?Show that $operatornamerank(A+B) leq operatornamerank(A) + operatornamerank(B)$Show that $operatornamerank(A) = operatornamerank(B)$$rm rank(BA)=rm rank(B)$ if $A in mathbbR^n times n$ is invertible?Prove $operatornamerank (BA) = operatornamerank (A) = operatornamerank (AC)$, for $B$ and $C$ invertible matricesFor nonzeros $A,B,Cin M_n(mathbbR)$, $ABC=0$. Show $operatornamerank(A)+operatornamerank(B)+operatornamerank(C)le 2n$Counterexample of Converse of “$operatornamerank (PA) = operatornamerank (A)$ if $P$ is invertible”Proof for if $A$ is invertible then $AB$ is invertibleShow that $operatornamerank(A)=operatornamerank(B)$Does $operatornamerank (A^2) = operatornamerank (A)$ for any matrix $Ain operatornameMat_n times n$?

5

0

$begingroup$

Let $A$ and $B$ be two (nonzero) real square matrices and suppose that $operatornamerank (AB) = operatornamerank (BA)$ for any $B$. Can one prove that $A$ is invertible? (The converse is true and a simple linear algebra question but I'm stuck on this one.)

linear-algebra matrices inverse matrix-rank

share|cite|improve this question

edited Mar 14 at 5:14

Rócherz

2,9863821

asked Aug 28 '16 at 15:23

BlaiseBlaise

384

$endgroup$

add a comment | 

5

0

$begingroup$

Let $A$ and $B$ be two (nonzero) real square matrices and suppose that $operatornamerank (AB) = operatornamerank (BA)$ for any $B$. Can one prove that $A$ is invertible? (The converse is true and a simple linear algebra question but I'm stuck on this one.)

linear-algebra matrices inverse matrix-rank

share|cite|improve this question

edited Mar 14 at 5:14

Rócherz

2,9863821

asked Aug 28 '16 at 15:23

BlaiseBlaise

384

$endgroup$

add a comment | 

$begingroup$

Let $A$ and $B$ be two (nonzero) real square matrices and suppose that $operatornamerank (AB) = operatornamerank (BA)$ for any $B$. Can one prove that $A$ is invertible? (The converse is true and a simple linear algebra question but I'm stuck on this one.)

linear-algebra matrices inverse matrix-rank

share|cite|improve this question

edited Mar 14 at 5:14

Rócherz

2,9863821

asked Aug 28 '16 at 15:23

BlaiseBlaise

384

$endgroup$

share|cite|improve this question

edited Mar 14 at 5:14

Rócherz

2,9863821

asked Aug 28 '16 at 15:23

BlaiseBlaise

384

share|cite|improve this question

edited Mar 14 at 5:14

Rócherz

2,9863821

asked Aug 28 '16 at 15:23

BlaiseBlaise

384

edited Mar 14 at 5:14

Rócherz

2,9863821

edited Mar 14 at 5:14

Rócherz

2,9863821

asked Aug 28 '16 at 15:23

BlaiseBlaise

384

asked Aug 28 '16 at 15:23

BlaiseBlaise

384

2 Answers
2

active

oldest

votes

8

$begingroup$

If $A$ is not invertible, there is a nonzero vector $u$ in the null space of $A$. Also, since you assume $A neq 0$, there is a nonzero vector $v$ in the column space of $A$. Use these observations to construct a rank one matrix $B$ such that $AB = 0$ and $BA neq 0$.

share|cite|improve this answer

answered Aug 28 '16 at 15:30

Mike FMike F

12.5k23583

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Technically we should consider the row space of $A$ (the column space of $A^T$)
  $endgroup$
  – Omnomnomnom
  Aug 28 '16 at 15:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Omnomnomnom: I don't think it's obligatory to talk about the row space of $A$, I was just thinking of taking $B = u v^t$ where $u in mathrmnull(A)$ and $v in mathrmcol(A)$ are nonzero, but I agree one could also argue slightly differently using the row space of $A$.
  $endgroup$
  – Mike F
  Aug 28 '16 at 17:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is not generally true that $v^TAneq 0$ (Or, for that matter, that $Av=0$) when $v$ is in col(A).
  $endgroup$
  – Omnomnomnom
  Aug 29 '16 at 2:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That should be true! If $v in mathrmcol(A)$ is nonzero, then take some $w$ with $Aw=v$. Note that $v^T A w = v^T v = |v|^2 neq 0$, so $v^TA neq 0$ too.
  $endgroup$
  – Mike F
  Aug 29 '16 at 3:05
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ah, so you're right. I didn't know, and I had a different solution in mind, but that all works.
  $endgroup$
  – Omnomnomnom
  Aug 29 '16 at 3:37
  

  
  
  
  

add a comment | 

1

$begingroup$

Let's construct the $B$ matrix along the lines given by Mike F. In the image of $A$ we have a nonzero vector $v$ which has a nonzero antecedent $w$ so that $v=A w$.
In components we define $B_ij=u_i w_j$. This matrix is nonzero. Left multiplication by $A$ gives 0 because $u$ is in the kernel of $A$. If we do the right-multiplication by $A$ we find a matrix which is $u_i v_j$ which is nonzero.
Strictly speaking there is a transposition when dealing with the right stuff.
Great ! thanks.

share|cite|improve this answer

answered Aug 28 '16 at 16:38

BlaiseBlaise

384

$endgroup$

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1906390%2fif-operatornamerank-ab-operatornamerank-ba-for-any-b-then-is-a%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

8

$begingroup$

If $A$ is not invertible, there is a nonzero vector $u$ in the null space of $A$. Also, since you assume $A neq 0$, there is a nonzero vector $v$ in the column space of $A$. Use these observations to construct a rank one matrix $B$ such that $AB = 0$ and $BA neq 0$.

share|cite|improve this answer

answered Aug 28 '16 at 15:30

Mike FMike F

12.5k23583

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Technically we should consider the row space of $A$ (the column space of $A^T$)
  $endgroup$
  – Omnomnomnom
  Aug 28 '16 at 15:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Omnomnomnom: I don't think it's obligatory to talk about the row space of $A$, I was just thinking of taking $B = u v^t$ where $u in mathrmnull(A)$ and $v in mathrmcol(A)$ are nonzero, but I agree one could also argue slightly differently using the row space of $A$.
  $endgroup$
  – Mike F
  Aug 28 '16 at 17:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is not generally true that $v^TAneq 0$ (Or, for that matter, that $Av=0$) when $v$ is in col(A).
  $endgroup$
  – Omnomnomnom
  Aug 29 '16 at 2:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That should be true! If $v in mathrmcol(A)$ is nonzero, then take some $w$ with $Aw=v$. Note that $v^T A w = v^T v = |v|^2 neq 0$, so $v^TA neq 0$ too.
  $endgroup$
  – Mike F
  Aug 29 '16 at 3:05
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ah, so you're right. I didn't know, and I had a different solution in mind, but that all works.
  $endgroup$
  – Omnomnomnom
  Aug 29 '16 at 3:37
  

  
  
  
  

add a comment | 

8

$begingroup$

If $A$ is not invertible, there is a nonzero vector $u$ in the null space of $A$. Also, since you assume $A neq 0$, there is a nonzero vector $v$ in the column space of $A$. Use these observations to construct a rank one matrix $B$ such that $AB = 0$ and $BA neq 0$.

share|cite|improve this answer

answered Aug 28 '16 at 15:30

Mike FMike F

12.5k23583

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Technically we should consider the row space of $A$ (the column space of $A^T$)
  $endgroup$
  – Omnomnomnom
  Aug 28 '16 at 15:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Omnomnomnom: I don't think it's obligatory to talk about the row space of $A$, I was just thinking of taking $B = u v^t$ where $u in mathrmnull(A)$ and $v in mathrmcol(A)$ are nonzero, but I agree one could also argue slightly differently using the row space of $A$.
  $endgroup$
  – Mike F
  Aug 28 '16 at 17:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is not generally true that $v^TAneq 0$ (Or, for that matter, that $Av=0$) when $v$ is in col(A).
  $endgroup$
  – Omnomnomnom
  Aug 29 '16 at 2:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That should be true! If $v in mathrmcol(A)$ is nonzero, then take some $w$ with $Aw=v$. Note that $v^T A w = v^T v = |v|^2 neq 0$, so $v^TA neq 0$ too.
  $endgroup$
  – Mike F
  Aug 29 '16 at 3:05
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ah, so you're right. I didn't know, and I had a different solution in mind, but that all works.
  $endgroup$
  – Omnomnomnom
  Aug 29 '16 at 3:37
  

  
  
  
  

add a comment | 

$begingroup$

If $A$ is not invertible, there is a nonzero vector $u$ in the null space of $A$. Also, since you assume $A neq 0$, there is a nonzero vector $v$ in the column space of $A$. Use these observations to construct a rank one matrix $B$ such that $AB = 0$ and $BA neq 0$.

share|cite|improve this answer

answered Aug 28 '16 at 15:30

Mike FMike F

12.5k23583

$endgroup$

share|cite|improve this answer

answered Aug 28 '16 at 15:30

Mike FMike F

12.5k23583

answered Aug 28 '16 at 15:30

Mike FMike F

12.5k23583

answered Aug 28 '16 at 15:30

Mike FMike F

12.5k23583

1

$begingroup$

Let's construct the $B$ matrix along the lines given by Mike F. In the image of $A$ we have a nonzero vector $v$ which has a nonzero antecedent $w$ so that $v=A w$.
In components we define $B_ij=u_i w_j$. This matrix is nonzero. Left multiplication by $A$ gives 0 because $u$ is in the kernel of $A$. If we do the right-multiplication by $A$ we find a matrix which is $u_i v_j$ which is nonzero.
Strictly speaking there is a transposition when dealing with the right stuff.
Great ! thanks.

share|cite|improve this answer

answered Aug 28 '16 at 16:38

BlaiseBlaise

384

$endgroup$

add a comment | 

1

$begingroup$

Let's construct the $B$ matrix along the lines given by Mike F. In the image of $A$ we have a nonzero vector $v$ which has a nonzero antecedent $w$ so that $v=A w$.
In components we define $B_ij=u_i w_j$. This matrix is nonzero. Left multiplication by $A$ gives 0 because $u$ is in the kernel of $A$. If we do the right-multiplication by $A$ we find a matrix which is $u_i v_j$ which is nonzero.
Strictly speaking there is a transposition when dealing with the right stuff.
Great ! thanks.

share|cite|improve this answer

answered Aug 28 '16 at 16:38

BlaiseBlaise

384

$endgroup$

add a comment | 

$begingroup$

Let's construct the $B$ matrix along the lines given by Mike F. In the image of $A$ we have a nonzero vector $v$ which has a nonzero antecedent $w$ so that $v=A w$.
In components we define $B_ij=u_i w_j$. This matrix is nonzero. Left multiplication by $A$ gives 0 because $u$ is in the kernel of $A$. If we do the right-multiplication by $A$ we find a matrix which is $u_i v_j$ which is nonzero.
Strictly speaking there is a transposition when dealing with the right stuff.
Great ! thanks.

share|cite|improve this answer

answered Aug 28 '16 at 16:38

BlaiseBlaise

384

$endgroup$

share|cite|improve this answer

answered Aug 28 '16 at 16:38

BlaiseBlaise

384

answered Aug 28 '16 at 16:38

BlaiseBlaise

384

answered Aug 28 '16 at 16:38

BlaiseBlaise

384