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quadratic equations on 2 by 2 matrices

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3

$begingroup$

For real non-zero $2times 2$ matrices, can we say:

For any $A,B$, there are at most two matrices $X$ such that $XX + AX + B =0$

Is there a way to see this without going in the direction of writing the open form result of $XX + AX + B$ and solve each element is equal to zero, etc.?

linear-algebra matrices quadratic-forms

share|cite|improve this question

edited 8 hours ago

independentvariable

asked 8 hours ago

independentvariableindependentvariable

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$endgroup$

add a comment
 | 

3

$begingroup$

For real non-zero $2times 2$ matrices, can we say:

For any $A,B$, there are at most two matrices $X$ such that $XX + AX + B =0$

Is there a way to see this without going in the direction of writing the open form result of $XX + AX + B$ and solve each element is equal to zero, etc.?

linear-algebra matrices quadratic-forms

share|cite|improve this question

edited 8 hours ago

independentvariable

asked 8 hours ago

independentvariableindependentvariable

36311 silver badge1111 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

For real non-zero $2times 2$ matrices, can we say:

For any $A,B$, there are at most two matrices $X$ such that $XX + AX + B =0$

Is there a way to see this without going in the direction of writing the open form result of $XX + AX + B$ and solve each element is equal to zero, etc.?

linear-algebra matrices quadratic-forms

share|cite|improve this question

edited 8 hours ago

independentvariable

asked 8 hours ago

independentvariableindependentvariable

36311 silver badge1111 bronze badges

$endgroup$

share|cite|improve this question

edited 8 hours ago

independentvariable

asked 8 hours ago

independentvariableindependentvariable

36311 silver badge1111 bronze badges

share|cite|improve this question

edited 8 hours ago

independentvariable

asked 8 hours ago

independentvariableindependentvariable

36311 silver badge1111 bronze badges

asked 8 hours ago

independentvariableindependentvariable

36311 silver badge1111 bronze badges

asked 8 hours ago

independentvariableindependentvariable

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2 Answers
2

active

oldest

votes

4

$begingroup$

No. E.g. over $mathbb R$, we have $pmatrix1&t\ 0&-1^2=I$ for any $t$. It also follows that
$$
pmatrix1&t\ 0&-1^2+pmatrix0&0\ 0&1pmatrix1&t\ 0&-1+pmatrix-1&0\ 0&0=0
$$
for any $t$.

Edit. Another counterexample (inspired by Dietrich Burde's answer). Let $u,v$ be any two vectors such that $v^Tu=0$. Then $X=uv^T$ is a solution to the equation $X^2=0$. In particular, if the underlying field is $GF(2)$ (the field consisting of only two elements $0$ and $1$, with $1+1=0$), the equation $Y^2=0$ for a $2times2$ matrix $Y$ has exactly four solutions
$$
Yinleftpmatrix0&0\ 0&0, pmatrix0&1\ 0&0, pmatrix0&0\ 1&0, pmatrix1&1\ 1&1right.
$$
And so does the equation $X^2=I$ over $GF(2)$: just put $X=Y+I$.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

user1551user1551

81.1k66 gold badges7474 silver badges137137 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how did you come up with this so fast? may I ask your technique?
  $endgroup$
  – independentvariable
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @independentvariable It may not be hard for someone to figure out a counterexample, but I just happen to know that $X^2=I$ has infinitely many solutions over a field of characteristic zero.
  $endgroup$
  – user1551
  7 hours ago
  

  
  
  
  

add a comment
 | 

3

$begingroup$

No, take for example $A=0$ and
$$
B=beginpmatrix 0 & 1 cr 0 & 0 endpmatrix.
$$
Then $X^2=-B$ has no solution, since $X$ is nilpotent, so that $X^2=0$ but $Bneq 0$.

share|cite|improve this answer

answered 8 hours ago

Dietrich BurdeDietrich Burde

87.9k66 gold badges5050 silver badges111111 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The matrices are non zero though.
  $endgroup$
  – independentvariable
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It doesn't matter really. You can adapt this example with nonzero $A$. Quadratic matrix solutions need not have any solution.
  $endgroup$
  – Dietrich Burde
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But the question is about showing it can have more than 2, isnt it?
  $endgroup$
  – independentvariable
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, sorry, I thought of "at least". So quadratic matrix equations can have no solution, or a number of solutions including infinity. It comes from the fact that the matrix algebra is not commutative.
  $endgroup$
  – Dietrich Burde
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Still, good for knowledge :)
  $endgroup$
  – independentvariable
  7 hours ago
  

  
  
  
  

add a comment
 | 

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4

$begingroup$

No. E.g. over $mathbb R$, we have $pmatrix1&t\ 0&-1^2=I$ for any $t$. It also follows that
$$
pmatrix1&t\ 0&-1^2+pmatrix0&0\ 0&1pmatrix1&t\ 0&-1+pmatrix-1&0\ 0&0=0
$$
for any $t$.

Edit. Another counterexample (inspired by Dietrich Burde's answer). Let $u,v$ be any two vectors such that $v^Tu=0$. Then $X=uv^T$ is a solution to the equation $X^2=0$. In particular, if the underlying field is $GF(2)$ (the field consisting of only two elements $0$ and $1$, with $1+1=0$), the equation $Y^2=0$ for a $2times2$ matrix $Y$ has exactly four solutions
$$
Yinleftpmatrix0&0\ 0&0, pmatrix0&1\ 0&0, pmatrix0&0\ 1&0, pmatrix1&1\ 1&1right.
$$
And so does the equation $X^2=I$ over $GF(2)$: just put $X=Y+I$.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

user1551user1551

81.1k66 gold badges7474 silver badges137137 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how did you come up with this so fast? may I ask your technique?
  $endgroup$
  – independentvariable
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @independentvariable It may not be hard for someone to figure out a counterexample, but I just happen to know that $X^2=I$ has infinitely many solutions over a field of characteristic zero.
  $endgroup$
  – user1551
  7 hours ago
  

  
  
  
  

add a comment
 | 

4

$begingroup$

No. E.g. over $mathbb R$, we have $pmatrix1&t\ 0&-1^2=I$ for any $t$. It also follows that
$$
pmatrix1&t\ 0&-1^2+pmatrix0&0\ 0&1pmatrix1&t\ 0&-1+pmatrix-1&0\ 0&0=0
$$
for any $t$.

Edit. Another counterexample (inspired by Dietrich Burde's answer). Let $u,v$ be any two vectors such that $v^Tu=0$. Then $X=uv^T$ is a solution to the equation $X^2=0$. In particular, if the underlying field is $GF(2)$ (the field consisting of only two elements $0$ and $1$, with $1+1=0$), the equation $Y^2=0$ for a $2times2$ matrix $Y$ has exactly four solutions
$$
Yinleftpmatrix0&0\ 0&0, pmatrix0&1\ 0&0, pmatrix0&0\ 1&0, pmatrix1&1\ 1&1right.
$$
And so does the equation $X^2=I$ over $GF(2)$: just put $X=Y+I$.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

user1551user1551

81.1k66 gold badges7474 silver badges137137 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how did you come up with this so fast? may I ask your technique?
  $endgroup$
  – independentvariable
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @independentvariable It may not be hard for someone to figure out a counterexample, but I just happen to know that $X^2=I$ has infinitely many solutions over a field of characteristic zero.
  $endgroup$
  – user1551
  7 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

No. E.g. over $mathbb R$, we have $pmatrix1&t\ 0&-1^2=I$ for any $t$. It also follows that
$$
pmatrix1&t\ 0&-1^2+pmatrix0&0\ 0&1pmatrix1&t\ 0&-1+pmatrix-1&0\ 0&0=0
$$
for any $t$.

Edit. Another counterexample (inspired by Dietrich Burde's answer). Let $u,v$ be any two vectors such that $v^Tu=0$. Then $X=uv^T$ is a solution to the equation $X^2=0$. In particular, if the underlying field is $GF(2)$ (the field consisting of only two elements $0$ and $1$, with $1+1=0$), the equation $Y^2=0$ for a $2times2$ matrix $Y$ has exactly four solutions
$$
Yinleftpmatrix0&0\ 0&0, pmatrix0&1\ 0&0, pmatrix0&0\ 1&0, pmatrix1&1\ 1&1right.
$$
And so does the equation $X^2=I$ over $GF(2)$: just put $X=Y+I$.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

user1551user1551

81.1k66 gold badges7474 silver badges137137 bronze badges

$endgroup$

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

user1551user1551

81.1k66 gold badges7474 silver badges137137 bronze badges

answered 8 hours ago

user1551user1551

81.1k66 gold badges7474 silver badges137137 bronze badges

answered 8 hours ago

user1551user1551

81.1k66 gold badges7474 silver badges137137 bronze badges

3

$begingroup$

No, take for example $A=0$ and
$$
B=beginpmatrix 0 & 1 cr 0 & 0 endpmatrix.
$$
Then $X^2=-B$ has no solution, since $X$ is nilpotent, so that $X^2=0$ but $Bneq 0$.

share|cite|improve this answer

answered 8 hours ago

Dietrich BurdeDietrich Burde

87.9k66 gold badges5050 silver badges111111 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The matrices are non zero though.
  $endgroup$
  – independentvariable
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It doesn't matter really. You can adapt this example with nonzero $A$. Quadratic matrix solutions need not have any solution.
  $endgroup$
  – Dietrich Burde
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But the question is about showing it can have more than 2, isnt it?
  $endgroup$
  – independentvariable
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, sorry, I thought of "at least". So quadratic matrix equations can have no solution, or a number of solutions including infinity. It comes from the fact that the matrix algebra is not commutative.
  $endgroup$
  – Dietrich Burde
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Still, good for knowledge :)
  $endgroup$
  – independentvariable
  7 hours ago
  

  
  
  
  

add a comment
 | 

3

$begingroup$

No, take for example $A=0$ and
$$
B=beginpmatrix 0 & 1 cr 0 & 0 endpmatrix.
$$
Then $X^2=-B$ has no solution, since $X$ is nilpotent, so that $X^2=0$ but $Bneq 0$.

share|cite|improve this answer

answered 8 hours ago

Dietrich BurdeDietrich Burde

87.9k66 gold badges5050 silver badges111111 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The matrices are non zero though.
  $endgroup$
  – independentvariable
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It doesn't matter really. You can adapt this example with nonzero $A$. Quadratic matrix solutions need not have any solution.
  $endgroup$
  – Dietrich Burde
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But the question is about showing it can have more than 2, isnt it?
  $endgroup$
  – independentvariable
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, sorry, I thought of "at least". So quadratic matrix equations can have no solution, or a number of solutions including infinity. It comes from the fact that the matrix algebra is not commutative.
  $endgroup$
  – Dietrich Burde
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Still, good for knowledge :)
  $endgroup$
  – independentvariable
  7 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

No, take for example $A=0$ and
$$
B=beginpmatrix 0 & 1 cr 0 & 0 endpmatrix.
$$
Then $X^2=-B$ has no solution, since $X$ is nilpotent, so that $X^2=0$ but $Bneq 0$.

share|cite|improve this answer

answered 8 hours ago

Dietrich BurdeDietrich Burde

87.9k66 gold badges5050 silver badges111111 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Dietrich BurdeDietrich Burde

87.9k66 gold badges5050 silver badges111111 bronze badges

answered 8 hours ago

Dietrich BurdeDietrich Burde

87.9k66 gold badges5050 silver badges111111 bronze badges

answered 8 hours ago

Dietrich BurdeDietrich Burde

87.9k66 gold badges5050 silver badges111111 bronze badges