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Maximum number of pairwise linearly independent vectors

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2

$begingroup$

Consider vectors $v_1,dots,v_ninmathbbR^d$. My question is: What is the maximum number of such vectors, that are pairwise linearly independent?

Clearly, if we remove the word pairwise the answer is $d$, but it feels the number is larger. Is this known explicitly?

linear-algebra vector-spaces vectors

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asked 8 hours ago

TBTDTBTD

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• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If by pairwise linear independence you mean that one is not multiple of the other, then taking the vectors of norm $1$ is an uncountably infinite set that is pairwise linear independent
  $endgroup$
  – Daniel
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks everyone. Simple, yet nice examples all.
  $endgroup$
  – TBTD
  7 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Consider vectors $v_1,dots,v_ninmathbbR^d$. My question is: What is the maximum number of such vectors, that are pairwise linearly independent?

Clearly, if we remove the word pairwise the answer is $d$, but it feels the number is larger. Is this known explicitly?

linear-algebra vector-spaces vectors

share|cite|improve this question

asked 8 hours ago

TBTDTBTD

2,71655 silver badges1616 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If by pairwise linear independence you mean that one is not multiple of the other, then taking the vectors of norm $1$ is an uncountably infinite set that is pairwise linear independent
  $endgroup$
  – Daniel
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks everyone. Simple, yet nice examples all.
  $endgroup$
  – TBTD
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Consider vectors $v_1,dots,v_ninmathbbR^d$. My question is: What is the maximum number of such vectors, that are pairwise linearly independent?

Clearly, if we remove the word pairwise the answer is $d$, but it feels the number is larger. Is this known explicitly?

linear-algebra vector-spaces vectors

share|cite|improve this question

asked 8 hours ago

TBTDTBTD

2,71655 silver badges1616 bronze badges

$endgroup$

share|cite|improve this question

asked 8 hours ago

TBTDTBTD

2,71655 silver badges1616 bronze badges

share|cite|improve this question

asked 8 hours ago

TBTDTBTD

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asked 8 hours ago

TBTDTBTD

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asked 8 hours ago

TBTDTBTD

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3 Answers
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4

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Consider the case $d=2$. Then you have that the vectors $(1,0)$, $(0,1)$ and $(sqrt2,sqrt2)$ are pairwise linearly independent vectors, for example. Indeed, the set of all vectors of length $1$ (unit circle) consists of pairwise linearly independent vectors (except, of course, those pairs on the same line). Notice that such set is uncountable.

The same argument can be extended to any dimension $d$. Hence, the maximum number is infinity.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

sam wolfesam wolfe

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2

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For e.g. $d=2$ the collection $(1,r)mid rinmathbb R$ is pairwise independent.

Observe that equality $lambda(1,r)+mu(1,r')=(0,0)$ leads to $lambda=mu=0$ if $rneq r'$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

drhabdrhab

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add a comment | 

1

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In dimension $2$ the uncountable set of all vectors $(x,y)$ of length $1$ with $x > 0$ is pairwise independent.

(Added the constraint $x>0$ since the pair $v,-v$ is obviously dependent.)

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Ethan BolkerEthan Bolker

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$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Well, strictly speaking some pairs consist of a vector and its negative but that is a negligible subset of the set of pairs.
  $endgroup$
  – Vincent
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Vincent Yes, of course. Edited.
  $endgroup$
  – Ethan Bolker
  8 hours ago
  

  
  
  
  

add a comment | 

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4

$begingroup$

Consider the case $d=2$. Then you have that the vectors $(1,0)$, $(0,1)$ and $(sqrt2,sqrt2)$ are pairwise linearly independent vectors, for example. Indeed, the set of all vectors of length $1$ (unit circle) consists of pairwise linearly independent vectors (except, of course, those pairs on the same line). Notice that such set is uncountable.

The same argument can be extended to any dimension $d$. Hence, the maximum number is infinity.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

sam wolfesam wolfe

90055 silver badges2525 bronze badges

$endgroup$

add a comment | 

4

$begingroup$

Consider the case $d=2$. Then you have that the vectors $(1,0)$, $(0,1)$ and $(sqrt2,sqrt2)$ are pairwise linearly independent vectors, for example. Indeed, the set of all vectors of length $1$ (unit circle) consists of pairwise linearly independent vectors (except, of course, those pairs on the same line). Notice that such set is uncountable.

The same argument can be extended to any dimension $d$. Hence, the maximum number is infinity.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

sam wolfesam wolfe

90055 silver badges2525 bronze badges

$endgroup$

add a comment | 

$begingroup$

Consider the case $d=2$. Then you have that the vectors $(1,0)$, $(0,1)$ and $(sqrt2,sqrt2)$ are pairwise linearly independent vectors, for example. Indeed, the set of all vectors of length $1$ (unit circle) consists of pairwise linearly independent vectors (except, of course, those pairs on the same line). Notice that such set is uncountable.

The same argument can be extended to any dimension $d$. Hence, the maximum number is infinity.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

sam wolfesam wolfe

90055 silver badges2525 bronze badges

$endgroup$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

sam wolfesam wolfe

90055 silver badges2525 bronze badges

answered 8 hours ago

sam wolfesam wolfe

90055 silver badges2525 bronze badges

answered 8 hours ago

sam wolfesam wolfe

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2

$begingroup$

For e.g. $d=2$ the collection $(1,r)mid rinmathbb R$ is pairwise independent.

Observe that equality $lambda(1,r)+mu(1,r')=(0,0)$ leads to $lambda=mu=0$ if $rneq r'$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

drhabdrhab

112k55 gold badges4949 silver badges140140 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

For e.g. $d=2$ the collection $(1,r)mid rinmathbb R$ is pairwise independent.

Observe that equality $lambda(1,r)+mu(1,r')=(0,0)$ leads to $lambda=mu=0$ if $rneq r'$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

drhabdrhab

112k55 gold badges4949 silver badges140140 bronze badges

$endgroup$

add a comment | 

$begingroup$

For e.g. $d=2$ the collection $(1,r)mid rinmathbb R$ is pairwise independent.

Observe that equality $lambda(1,r)+mu(1,r')=(0,0)$ leads to $lambda=mu=0$ if $rneq r'$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

drhabdrhab

112k55 gold badges4949 silver badges140140 bronze badges

$endgroup$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

drhabdrhab

112k55 gold badges4949 silver badges140140 bronze badges

answered 8 hours ago

drhabdrhab

112k55 gold badges4949 silver badges140140 bronze badges

answered 8 hours ago

drhabdrhab

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1

$begingroup$

In dimension $2$ the uncountable set of all vectors $(x,y)$ of length $1$ with $x > 0$ is pairwise independent.

(Added the constraint $x>0$ since the pair $v,-v$ is obviously dependent.)

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Ethan BolkerEthan Bolker

54.6k55 gold badges6161 silver badges132132 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Well, strictly speaking some pairs consist of a vector and its negative but that is a negligible subset of the set of pairs.
  $endgroup$
  – Vincent
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Vincent Yes, of course. Edited.
  $endgroup$
  – Ethan Bolker
  8 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

In dimension $2$ the uncountable set of all vectors $(x,y)$ of length $1$ with $x > 0$ is pairwise independent.

(Added the constraint $x>0$ since the pair $v,-v$ is obviously dependent.)

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Ethan BolkerEthan Bolker

54.6k55 gold badges6161 silver badges132132 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Well, strictly speaking some pairs consist of a vector and its negative but that is a negligible subset of the set of pairs.
  $endgroup$
  – Vincent
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Vincent Yes, of course. Edited.
  $endgroup$
  – Ethan Bolker
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

In dimension $2$ the uncountable set of all vectors $(x,y)$ of length $1$ with $x > 0$ is pairwise independent.

(Added the constraint $x>0$ since the pair $v,-v$ is obviously dependent.)

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Ethan BolkerEthan Bolker

54.6k55 gold badges6161 silver badges132132 bronze badges

$endgroup$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Ethan BolkerEthan Bolker

54.6k55 gold badges6161 silver badges132132 bronze badges

answered 8 hours ago

Ethan BolkerEthan Bolker

54.6k55 gold badges6161 silver badges132132 bronze badges

answered 8 hours ago

Ethan BolkerEthan Bolker

54.6k55 gold badges6161 silver badges132132 bronze badges