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Generic vector of matrix of rank one [closed]

What is generic rank?form of symmetric matrix of rank onerank of a diagonal matrix + rank-one perturbationIf $AB=I$, A is full rank, but not square matrix, shall we have $BA=I$?Rank-one update of singular matrixEfficient low rank matrix-vector multiplicationRank One decreaseCan a non-square matrix have a full rank?What is a generic matrixRank of a matrix of L.T. which is not one one

-4

$begingroup$

What is the generic vector for a non square matrix of rank one ?

linear-algebra matrices eigenvalues-eigenvectors

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edited Mar 17 at 21:08

Mathimatica

asked Mar 17 at 20:51

MathimaticaMathimatica

926

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closed as off-topic by Morgan Rodgers, mrtaurho, YiFan, jmerry, Eric Wofsey Mar 17 at 21:38

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Morgan Rodgers, mrtaurho, YiFan

If this question can be reworded to fit the rules in the help center, please edit the question.

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  $begingroup$
  Are you sure you want $n neq m$?
  $endgroup$
  – Klaus
  Mar 17 at 20:53
  

  
  
  
  


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  Yes, because if the matrix is square of rank one it is obvious that this vector exist and we can compute it explicitly, but i need to find such vector for no square matrix of rank one.
  $endgroup$
  – Mathimatica
  Mar 17 at 20:56
  
  

  
  
  
  


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  @Mathimatica No, it can't be $;nneq m;$ and also $;Mv=alpha v;$ ...
  $endgroup$
  – DonAntonio
  Mar 17 at 20:57
  

  
  
  
  


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  $Mv$ and $alpha v$ have different dimensions, so I don't undestand your question.
  $endgroup$
  – Klaus
  Mar 17 at 20:58
  

  
  
  
  


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  $begingroup$
  That was not a helpful edit just now. "Generic vector" is not standard terminology in mathematical English, and you removed the portions of the question that explained it - even though that explanation is inconsistent with the question as it is now. Also, "the" generic vector? Do you want one of them, or all of them? It is not at all clear what you're asking for.
  $endgroup$
  – jmerry
  Mar 17 at 21:13
  

  
  
  
  

add a comment | 

-4

$begingroup$

What is the generic vector for a non square matrix of rank one ?

linear-algebra matrices eigenvalues-eigenvectors

share|cite|improve this question

edited Mar 17 at 21:08

Mathimatica

asked Mar 17 at 20:51

MathimaticaMathimatica

926

$endgroup$

closed as off-topic by Morgan Rodgers, mrtaurho, YiFan, jmerry, Eric Wofsey Mar 17 at 21:38

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Morgan Rodgers, mrtaurho, YiFan

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Are you sure you want $n neq m$?
  $endgroup$
  – Klaus
  Mar 17 at 20:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, because if the matrix is square of rank one it is obvious that this vector exist and we can compute it explicitly, but i need to find such vector for no square matrix of rank one.
  $endgroup$
  – Mathimatica
  Mar 17 at 20:56
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Mathimatica No, it can't be $;nneq m;$ and also $;Mv=alpha v;$ ...
  $endgroup$
  – DonAntonio
  Mar 17 at 20:57
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  $Mv$ and $alpha v$ have different dimensions, so I don't undestand your question.
  $endgroup$
  – Klaus
  Mar 17 at 20:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That was not a helpful edit just now. "Generic vector" is not standard terminology in mathematical English, and you removed the portions of the question that explained it - even though that explanation is inconsistent with the question as it is now. Also, "the" generic vector? Do you want one of them, or all of them? It is not at all clear what you're asking for.
  $endgroup$
  – jmerry
  Mar 17 at 21:13
  

  
  
  
  

add a comment | 

$begingroup$

What is the generic vector for a non square matrix of rank one ?

linear-algebra matrices eigenvalues-eigenvectors

share|cite|improve this question

edited Mar 17 at 21:08

Mathimatica

asked Mar 17 at 20:51

MathimaticaMathimatica

926

$endgroup$

share|cite|improve this question

edited Mar 17 at 21:08

Mathimatica

asked Mar 17 at 20:51

MathimaticaMathimatica

926

share|cite|improve this question

edited Mar 17 at 21:08

Mathimatica

asked Mar 17 at 20:51

MathimaticaMathimatica

926

asked Mar 17 at 20:51

MathimaticaMathimatica

926

asked Mar 17 at 20:51

MathimaticaMathimatica

926

1 Answer
1

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oldest

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1

$begingroup$

First, in $;Mv=alpha v;$ , the left side has order $;mtimes 1;$ (or if you prefer, it is a column vector with $;m;$ entries), whereas the right side is $;ntimes 1;$ (column vector with $,n;$ entries) , so for the equality to be true if must be that $;n=m;$ and thus $;M;$ is a square matrix of order $;ntimes n;$ .

And now: assuming $;nge2;$ , since $;M;$ has rank $;1<n;$ we get that $;ker Mneq 0;$ and thus there exists $;vec vneq vec 0;$ s.t. $;Mvec v=vec 0=0cdot vec v;$

share|cite|improve this answer

answered Mar 17 at 21:01

DonAntonioDonAntonio

180k1494233

$endgroup$

add a comment | 

1

$begingroup$

First, in $;Mv=alpha v;$ , the left side has order $;mtimes 1;$ (or if you prefer, it is a column vector with $;m;$ entries), whereas the right side is $;ntimes 1;$ (column vector with $,n;$ entries) , so for the equality to be true if must be that $;n=m;$ and thus $;M;$ is a square matrix of order $;ntimes n;$ .

And now: assuming $;nge2;$ , since $;M;$ has rank $;1<n;$ we get that $;ker Mneq 0;$ and thus there exists $;vec vneq vec 0;$ s.t. $;Mvec v=vec 0=0cdot vec v;$

share|cite|improve this answer

answered Mar 17 at 21:01

DonAntonioDonAntonio

180k1494233

$endgroup$

add a comment | 

1

$begingroup$

First, in $;Mv=alpha v;$ , the left side has order $;mtimes 1;$ (or if you prefer, it is a column vector with $;m;$ entries), whereas the right side is $;ntimes 1;$ (column vector with $,n;$ entries) , so for the equality to be true if must be that $;n=m;$ and thus $;M;$ is a square matrix of order $;ntimes n;$ .

And now: assuming $;nge2;$ , since $;M;$ has rank $;1<n;$ we get that $;ker Mneq 0;$ and thus there exists $;vec vneq vec 0;$ s.t. $;Mvec v=vec 0=0cdot vec v;$

share|cite|improve this answer

answered Mar 17 at 21:01

DonAntonioDonAntonio

180k1494233

$endgroup$

add a comment | 

$begingroup$

First, in $;Mv=alpha v;$ , the left side has order $;mtimes 1;$ (or if you prefer, it is a column vector with $;m;$ entries), whereas the right side is $;ntimes 1;$ (column vector with $,n;$ entries) , so for the equality to be true if must be that $;n=m;$ and thus $;M;$ is a square matrix of order $;ntimes n;$ .

And now: assuming $;nge2;$ , since $;M;$ has rank $;1<n;$ we get that $;ker Mneq 0;$ and thus there exists $;vec vneq vec 0;$ s.t. $;Mvec v=vec 0=0cdot vec v;$

share|cite|improve this answer

answered Mar 17 at 21:01

DonAntonioDonAntonio

180k1494233

$endgroup$

share|cite|improve this answer

answered Mar 17 at 21:01

DonAntonioDonAntonio

180k1494233

answered Mar 17 at 21:01

DonAntonioDonAntonio

180k1494233

answered Mar 17 at 21:01

DonAntonioDonAntonio

180k1494233