Rank of Vandermond form of four matrices

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Consider four matrices A, B, C, D, all with size $2n times n$. Combining any two of them will get a full rank matrix. For example, [A B] (2n * 2n) is full rank.

Could we say the following matrix (4n*4n)

$beginbmatrix
A&B&C&D\
A&2B&3C&4D
endbmatrix$
always has a full rank?

linear-algebra

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edited Aug 4 at 1:13

asked Aug 3 at 23:21

spacehopper

92

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Consider four matrices A, B, C, D, all with size $2n times n$. Combining any two of them will get a full rank matrix. For example, [A B] (2n * 2n) is full rank.

Could we say the following matrix (4n*4n)

$beginbmatrix
A&B&C&D\
A&2B&3C&4D
endbmatrix$
always has a full rank?

linear-algebra

share|cite|improve this question

edited Aug 4 at 1:13

asked Aug 3 at 23:21

spacehopper

92

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Consider four matrices A, B, C, D, all with size $2n times n$. Combining any two of them will get a full rank matrix. For example, [A B] (2n * 2n) is full rank.

Could we say the following matrix (4n*4n)

$beginbmatrix
A&B&C&D\
A&2B&3C&4D
endbmatrix$
always has a full rank?

linear-algebra

share|cite|improve this question

edited Aug 4 at 1:13

asked Aug 3 at 23:21

spacehopper

92

Consider four matrices A, B, C, D, all with size $2n times n$. Combining any two of them will get a full rank matrix. For example, [A B] (2n * 2n) is full rank.

Could we say the following matrix (4n*4n)

$beginbmatrix
A&B&C&D\
A&2B&3C&4D
endbmatrix$
always has a full rank?

linear-algebra

share|cite|improve this question

edited Aug 4 at 1:13

asked Aug 3 at 23:21

spacehopper

92

share|cite|improve this question

share|cite|improve this question

edited Aug 4 at 1:13

edited Aug 4 at 1:13

edited Aug 4 at 1:13

asked Aug 3 at 23:21

spacehopper

92

asked Aug 3 at 23:21

spacehopper

92

asked Aug 3 at 23:21

spacehopper

92

92

add a comment | 

add a comment | 

1 Answer
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Here is a counterexample:
$$
beginpmatrix4 & 1 & 0 & 1\1 & 0 & 1 & 1\4 & 2 & 0 & 4\1 & 0 & 3 & 4endpmatrixbeginpmatrix1\-6\-3\2endpmatrix = beginpmatrix0\0\0\0endpmatrix.
$$

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answered Aug 4 at 2:25

amsmath

1,545113

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wow awesome. I am curious how you find this counterexample. If we add one more condition "all entries in A B C D are 0 or 1", do you think it will make this statement be true?
  – spacehopper
  Aug 4 at 2:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  How could you have then that each 2-combi is full rank (especially when $n=1$)?
  – amsmath
  Aug 4 at 3:18
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote

Here is a counterexample:
$$
beginpmatrix4 & 1 & 0 & 1\1 & 0 & 1 & 1\4 & 2 & 0 & 4\1 & 0 & 3 & 4endpmatrixbeginpmatrix1\-6\-3\2endpmatrix = beginpmatrix0\0\0\0endpmatrix.
$$

share|cite|improve this answer

answered Aug 4 at 2:25

amsmath

1,545113

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wow awesome. I am curious how you find this counterexample. If we add one more condition "all entries in A B C D are 0 or 1", do you think it will make this statement be true?
  – spacehopper
  Aug 4 at 2:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  How could you have then that each 2-combi is full rank (especially when $n=1$)?
  – amsmath
  Aug 4 at 3:18
  

  
  
  
  

add a comment | 

up vote
0
down vote

Here is a counterexample:
$$
beginpmatrix4 & 1 & 0 & 1\1 & 0 & 1 & 1\4 & 2 & 0 & 4\1 & 0 & 3 & 4endpmatrixbeginpmatrix1\-6\-3\2endpmatrix = beginpmatrix0\0\0\0endpmatrix.
$$

share|cite|improve this answer

answered Aug 4 at 2:25

amsmath

1,545113

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wow awesome. I am curious how you find this counterexample. If we add one more condition "all entries in A B C D are 0 or 1", do you think it will make this statement be true?
  – spacehopper
  Aug 4 at 2:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  How could you have then that each 2-combi is full rank (especially when $n=1$)?
  – amsmath
  Aug 4 at 3:18
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

Here is a counterexample:
$$
beginpmatrix4 & 1 & 0 & 1\1 & 0 & 1 & 1\4 & 2 & 0 & 4\1 & 0 & 3 & 4endpmatrixbeginpmatrix1\-6\-3\2endpmatrix = beginpmatrix0\0\0\0endpmatrix.
$$

share|cite|improve this answer

answered Aug 4 at 2:25

amsmath

1,545113

Here is a counterexample:
$$
beginpmatrix4 & 1 & 0 & 1\1 & 0 & 1 & 1\4 & 2 & 0 & 4\1 & 0 & 3 & 4endpmatrixbeginpmatrix1\-6\-3\2endpmatrix = beginpmatrix0\0\0\0endpmatrix.
$$

share|cite|improve this answer

answered Aug 4 at 2:25

amsmath

1,545113

share|cite|improve this answer

share|cite|improve this answer

answered Aug 4 at 2:25

amsmath

1,545113

answered Aug 4 at 2:25

amsmath

1,545113

answered Aug 4 at 2:25

amsmath

1,545113

1,545113

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wow awesome. I am curious how you find this counterexample. If we add one more condition "all entries in A B C D are 0 or 1", do you think it will make this statement be true?
  – spacehopper
  Aug 4 at 2:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  How could you have then that each 2-combi is full rank (especially when $n=1$)?
  – amsmath
  Aug 4 at 3:18
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wow awesome. I am curious how you find this counterexample. If we add one more condition "all entries in A B C D are 0 or 1", do you think it will make this statement be true?
  – spacehopper
  Aug 4 at 2:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  How could you have then that each 2-combi is full rank (especially when $n=1$)?
  – amsmath
  Aug 4 at 3:18
  

  
  
  
  

Wow awesome. I am curious how you find this counterexample. If we add one more condition "all entries in A B C D are 0 or 1", do you think it will make this statement be true?
– spacehopper
Aug 4 at 2:40

Wow awesome. I am curious how you find this counterexample. If we add one more condition "all entries in A B C D are 0 or 1", do you think it will make this statement be true?
– spacehopper
Aug 4 at 2:40

How could you have then that each 2-combi is full rank (especially when $n=1$)?
– amsmath
Aug 4 at 3:18

How could you have then that each 2-combi is full rank (especially when $n=1$)?
– amsmath
Aug 4 at 3:18

add a comment | 

 

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