Rank is less than $frac n2$

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Let $A in mathbbR^n times n$ such that $A^2 =0$. Prove that $mboxrank(A) leq frac n2$.

With Cayley-Hamilton, the characteristic polynomial is $chi_A=X^2$. I also know $dim A = dim(Im(A)) + dim(Kernel(A))$ so $n = dim(Im(A)) + dim(Kernel(A))$.

linear-algebra matrices matrix-rank nilpotence

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edited Jul 25 at 15:38

Rodrigo de Azevedo

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asked Jul 25 at 15:34

Marc

1005

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  "With Cayley–Hamilton the Characteristic polynomial is $chi_A = X^2$." No, the characteristic polynomial has degree $n$. The minimal polynomial is either $X=0$ or $X^2=0$.
  – Chappers
  Jul 25 at 15:36
  

  
  
  
  

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up vote
0
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favorite

Let $A in mathbbR^n times n$ such that $A^2 =0$. Prove that $mboxrank(A) leq frac n2$.

With Cayley-Hamilton, the characteristic polynomial is $chi_A=X^2$. I also know $dim A = dim(Im(A)) + dim(Kernel(A))$ so $n = dim(Im(A)) + dim(Kernel(A))$.

linear-algebra matrices matrix-rank nilpotence

share|cite|improve this question

edited Jul 25 at 15:38

Rodrigo de Azevedo

12.6k41751

asked Jul 25 at 15:34

Marc

1005

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  "With Cayley–Hamilton the Characteristic polynomial is $chi_A = X^2$." No, the characteristic polynomial has degree $n$. The minimal polynomial is either $X=0$ or $X^2=0$.
  – Chappers
  Jul 25 at 15:36
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Let $A in mathbbR^n times n$ such that $A^2 =0$. Prove that $mboxrank(A) leq frac n2$.

With Cayley-Hamilton, the characteristic polynomial is $chi_A=X^2$. I also know $dim A = dim(Im(A)) + dim(Kernel(A))$ so $n = dim(Im(A)) + dim(Kernel(A))$.

linear-algebra matrices matrix-rank nilpotence

share|cite|improve this question

edited Jul 25 at 15:38

Rodrigo de Azevedo

12.6k41751

asked Jul 25 at 15:34

Marc

1005

Let $A in mathbbR^n times n$ such that $A^2 =0$. Prove that $mboxrank(A) leq frac n2$.

With Cayley-Hamilton, the characteristic polynomial is $chi_A=X^2$. I also know $dim A = dim(Im(A)) + dim(Kernel(A))$ so $n = dim(Im(A)) + dim(Kernel(A))$.

linear-algebra matrices matrix-rank nilpotence

share|cite|improve this question

edited Jul 25 at 15:38

Rodrigo de Azevedo

12.6k41751

asked Jul 25 at 15:34

Marc

1005

share|cite|improve this question

share|cite|improve this question

edited Jul 25 at 15:38

Rodrigo de Azevedo

12.6k41751

edited Jul 25 at 15:38

Rodrigo de Azevedo

12.6k41751

edited Jul 25 at 15:38

Rodrigo de Azevedo

12.6k41751

12.6k41751

asked Jul 25 at 15:34

Marc

1005

asked Jul 25 at 15:34

Marc

1005

asked Jul 25 at 15:34

Marc

1005

1005

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  "With Cayley–Hamilton the Characteristic polynomial is $chi_A = X^2$." No, the characteristic polynomial has degree $n$. The minimal polynomial is either $X=0$ or $X^2=0$.
  – Chappers
  Jul 25 at 15:36
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  "With Cayley–Hamilton the Characteristic polynomial is $chi_A = X^2$." No, the characteristic polynomial has degree $n$. The minimal polynomial is either $X=0$ or $X^2=0$.
  – Chappers
  Jul 25 at 15:36
  

  
  
  
  

1

1

"With Cayley–Hamilton the Characteristic polynomial is $chi_A = X^2$." No, the characteristic polynomial has degree $n$. The minimal polynomial is either $X=0$ or $X^2=0$.
– Chappers
Jul 25 at 15:36

"With Cayley–Hamilton the Characteristic polynomial is $chi_A = X^2$." No, the characteristic polynomial has degree $n$. The minimal polynomial is either $X=0$ or $X^2=0$.
– Chappers
Jul 25 at 15:36

add a comment | 

2 Answers
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accepted

Hint:

$Im(A) subset ker A$ and see what will happen!

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answered Jul 25 at 15:36

Chinnapparaj R

1,559315

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Hint: Characteristic poly is $(x^2)^n/2$, as the order of the matrix $A$ is $n$.

$x^2$ is an anihilating polynomial. If $A$ is non-zero then $x^2$ is minimal polynomial of $A$.

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answered Jul 25 at 15:38

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote



accepted

Hint:

$Im(A) subset ker A$ and see what will happen!

share|cite|improve this answer

answered Jul 25 at 15:36

Chinnapparaj R

1,559315

add a comment | 

up vote
4
down vote



accepted

Hint:

$Im(A) subset ker A$ and see what will happen!

share|cite|improve this answer

answered Jul 25 at 15:36

Chinnapparaj R

1,559315

add a comment | 

up vote
4
down vote



accepted

up vote
4
down vote



accepted

Hint:

$Im(A) subset ker A$ and see what will happen!

share|cite|improve this answer

answered Jul 25 at 15:36

Chinnapparaj R

1,559315

Hint:

$Im(A) subset ker A$ and see what will happen!

share|cite|improve this answer

answered Jul 25 at 15:36

Chinnapparaj R

1,559315

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 15:36

Chinnapparaj R

1,559315

answered Jul 25 at 15:36

Chinnapparaj R

1,559315

answered Jul 25 at 15:36

Chinnapparaj R

1,559315

1,559315

add a comment | 

add a comment | 

up vote
0
down vote

Hint: Characteristic poly is $(x^2)^n/2$, as the order of the matrix $A$ is $n$.

$x^2$ is an anihilating polynomial. If $A$ is non-zero then $x^2$ is minimal polynomial of $A$.

share|cite|improve this answer

answered Jul 25 at 15:38

Empty

7,73742154

add a comment | 

up vote
0
down vote

Hint: Characteristic poly is $(x^2)^n/2$, as the order of the matrix $A$ is $n$.

$x^2$ is an anihilating polynomial. If $A$ is non-zero then $x^2$ is minimal polynomial of $A$.

share|cite|improve this answer

answered Jul 25 at 15:38

Empty

7,73742154

add a comment | 

up vote
0
down vote

up vote
0
down vote

Hint: Characteristic poly is $(x^2)^n/2$, as the order of the matrix $A$ is $n$.

$x^2$ is an anihilating polynomial. If $A$ is non-zero then $x^2$ is minimal polynomial of $A$.

share|cite|improve this answer

answered Jul 25 at 15:38

Empty

7,73742154

Hint: Characteristic poly is $(x^2)^n/2$, as the order of the matrix $A$ is $n$.

$x^2$ is an anihilating polynomial. If $A$ is non-zero then $x^2$ is minimal polynomial of $A$.

share|cite|improve this answer

answered Jul 25 at 15:38

Empty

7,73742154

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 15:38

Empty

7,73742154

answered Jul 25 at 15:38

Empty

7,73742154

answered Jul 25 at 15:38

Empty

7,73742154

7,73742154

add a comment | 

add a comment | 

 

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