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Determine all $2 times 2$ real matrices $A$ such that $(1) A^2=I$, $(2) A^2=0$

Convert Frobenius canonical form to Jordan canonical formDetermine whether the set $H$ of all matrices form a subspace of $M_2 times 2$Let $A$ be a $3 times 3$ matrix with real entries such that…Determine the matrices that represent the following rotations of $mathbbR^3$Matrices which commute with all the matrices commuting with a given matrixA symmetric real matrix with all diagonal entries unspecified can be completed to be positive semidefiniteCan we show that all $2 times 2$ matrices are sums of matrices with determinant 1?Eigenvalues of a product of matrices, involving Moore-Penrose pseudo inverseHow to show that $M$ is nonsingular (iff $A-BC$ = nonsingular)$beginbmatrixA&B\C&I\endbmatrix$“Perturbed projection” matrix equality

2

$begingroup$

Determine all $2 times 2$ real matrices $A$ such that $(1) A^2=I$, $(2) A^2=0$

I came across this problem recently where I have to determine all the $2times2$ matrices satisfying the aforementioned criteria. The problem is that the equations which I get after multiplication are not so easy to work with and I'm unable to proceed.

If someone could help me with these equations and solve for the various cases, that'd be really helpful.

Answer to part one:
$$A=I$$ $$A=-I$$ $$a_11=a_22=0 , a_12a_21=1 $$$$a_11=-a_22neq0, a_11^2+a_12a_21=1$$

Answer to part two:
$$a_11=a_12=a_22=0$$$$a_11=a_21=a_22=0$$$$a_11=-a_22neq0, a_11^2+a_12a_21=0$$

linear-algebra matrices matrix-equations nilpotence

share|cite|improve this question

edited Mar 17 at 14:18

Rodrigo de Azevedo

13.2k41960

asked Mar 17 at 13:42

s0ulr3aper07s0ulr3aper07

580111

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How are you getting these equations? I can't see what they have to do with the question.
  $endgroup$
  – saulspatz
  Mar 17 at 13:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @saulspatz those are the actual answers given in the textbook. I'm looking for a way to reach them.
  $endgroup$
  – s0ulr3aper07
  Mar 17 at 14:44
  

  
  
  
  

add a comment | 

2

$begingroup$

Determine all $2 times 2$ real matrices $A$ such that $(1) A^2=I$, $(2) A^2=0$

I came across this problem recently where I have to determine all the $2times2$ matrices satisfying the aforementioned criteria. The problem is that the equations which I get after multiplication are not so easy to work with and I'm unable to proceed.

If someone could help me with these equations and solve for the various cases, that'd be really helpful.

Answer to part one:
$$A=I$$ $$A=-I$$ $$a_11=a_22=0 , a_12a_21=1 $$$$a_11=-a_22neq0, a_11^2+a_12a_21=1$$

Answer to part two:
$$a_11=a_12=a_22=0$$$$a_11=a_21=a_22=0$$$$a_11=-a_22neq0, a_11^2+a_12a_21=0$$

linear-algebra matrices matrix-equations nilpotence

share|cite|improve this question

edited Mar 17 at 14:18

Rodrigo de Azevedo

13.2k41960

asked Mar 17 at 13:42

s0ulr3aper07s0ulr3aper07

580111

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How are you getting these equations? I can't see what they have to do with the question.
  $endgroup$
  – saulspatz
  Mar 17 at 13:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @saulspatz those are the actual answers given in the textbook. I'm looking for a way to reach them.
  $endgroup$
  – s0ulr3aper07
  Mar 17 at 14:44
  

  
  
  
  

add a comment | 

$begingroup$

Determine all $2 times 2$ real matrices $A$ such that $(1) A^2=I$, $(2) A^2=0$

I came across this problem recently where I have to determine all the $2times2$ matrices satisfying the aforementioned criteria. The problem is that the equations which I get after multiplication are not so easy to work with and I'm unable to proceed.

If someone could help me with these equations and solve for the various cases, that'd be really helpful.

Answer to part one:
$$A=I$$ $$A=-I$$ $$a_11=a_22=0 , a_12a_21=1 $$$$a_11=-a_22neq0, a_11^2+a_12a_21=1$$

Answer to part two:
$$a_11=a_12=a_22=0$$$$a_11=a_21=a_22=0$$$$a_11=-a_22neq0, a_11^2+a_12a_21=0$$

linear-algebra matrices matrix-equations nilpotence

share|cite|improve this question

edited Mar 17 at 14:18

Rodrigo de Azevedo

13.2k41960

asked Mar 17 at 13:42

s0ulr3aper07s0ulr3aper07

580111

$endgroup$

share|cite|improve this question

edited Mar 17 at 14:18

Rodrigo de Azevedo

13.2k41960

asked Mar 17 at 13:42

s0ulr3aper07s0ulr3aper07

580111

share|cite|improve this question

edited Mar 17 at 14:18

Rodrigo de Azevedo

13.2k41960

asked Mar 17 at 13:42

s0ulr3aper07s0ulr3aper07

580111

edited Mar 17 at 14:18

Rodrigo de Azevedo

13.2k41960

edited Mar 17 at 14:18

Rodrigo de Azevedo

13.2k41960

asked Mar 17 at 13:42

s0ulr3aper07s0ulr3aper07

580111

asked Mar 17 at 13:42

s0ulr3aper07s0ulr3aper07

580111

2 Answers
2

active

oldest

votes

2

$begingroup$

Case $A^2 = I$.

Two possibilities are of course $A = pm I$. If $A ne pm I$, then the characteristic polynomial of $A$ is $x^2 - 1$, which means $A$ has trace $0$ and determinant $-1$. Therefore
beginequation*
beginbmatrix
a & b\
c & -a
endbmatrix,
endequation*
where $a^2 + b c = 1$.

Case $A^2 = 0$.

One possibility is of course $A = 0$. If $A ne 0$, then the characteristic polynomial of $A$ is $x^2$, which means $A$ has trace and determinant $0$. Therefore
beginequation*
beginbmatrix
a & b\
c & -a
endbmatrix,
endequation*
where $a^2 + b c = 0$, and $a, b, c$ are not all zero.

In part one, distinguish now

• the case when $a = pm 1$, and then either $b = 0$ and $b$ is arbitrary, or vice versa;


• the case when $a ne pm 1$, and then $b ne 0$, say, can be chosen arbitrarily, and $c = (1 - a^2)/b ne 0$.

In part two, distinguish now

• the case when $a = 0$, and then either $b = 0$ and $c$ is arbitrary, or vice versa;


• the case when $a ne 0$, and then $b ne 0$, say, can be chosen arbitrarily, and $c = -a^2/bne 0$.

share|cite|improve this answer

edited Mar 18 at 17:06

Widawensen

4,69331446

answered Mar 17 at 14:40

Andreas CarantiAndreas Caranti

56.9k34397

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks a lot! But I have not yet learnt about characteristic polynomial, so if it's not too much to ask for, could you please add some information about it and how you've used it in this context?
  $endgroup$
  – s0ulr3aper07
  Mar 17 at 15:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @s0ulr3aper07 have you been taught about eigenvalues?
  $endgroup$
  – Andreas Caranti
  Mar 17 at 16:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I have learnt about eigenvalues and have come across the characteristic polynomial while doing so. The way you've used it to find trace and determinant is new to me.
  $endgroup$
  – s0ulr3aper07
  Mar 17 at 16:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @s0ulr3aper07, ok, so I believe you just need to know a couple of things. The first one is that the characteristic polynomial $A - x I$ of $2 times 2$ matrix $A$ has the form $x^2 - T x + D$, where $T$ is the trace of $A$, and $D$ is its determinant. It is just a matter of expanding $detleft( beginbmatrixa - x&b\c&d-xendbmatrixright)$, and then...
  $endgroup$
  – Andreas Caranti
  Mar 17 at 17:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The second thing is to note that if $A^2 = I$, and the eigenvalues of $A$ are both $1$, then $A = I$, while if they are both $-1$, then $A = -I$. Just check that if this is not the case, then $A^2 ne I$.
  $endgroup$
  – Andreas Caranti
  Mar 17 at 17:04
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

$$A = beginbmatrix
a & b\
c & d
endbmatrix
Rightarrow A^2 = beginbmatrix
a^2+bc & (a+d)b\
(a+d)c & bc +d^2
endbmatrix$$

In both cases, you have to impose $(a+d)b = (a+d)c = 0$.

1. Case $a+d neq 0$: then $b = c = 0$ and $A^2 = beginbmatrix
   a^2 & 0\
   0 & d^2
   endbmatrix$ and it's easy to continue.


2. Case $a+d = 0$: $A^2 = beginbmatrix
   a^2+bc & 0\
   0 & bc + d^2
   endbmatrix$ and it's easy to continue.

share|cite|improve this answer

answered Mar 17 at 14:02

dcolazindcolazin

3694

$endgroup$

add a comment | 

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2

$begingroup$

Case $A^2 = I$.

Two possibilities are of course $A = pm I$. If $A ne pm I$, then the characteristic polynomial of $A$ is $x^2 - 1$, which means $A$ has trace $0$ and determinant $-1$. Therefore
beginequation*
beginbmatrix
a & b\
c & -a
endbmatrix,
endequation*
where $a^2 + b c = 1$.

Case $A^2 = 0$.

One possibility is of course $A = 0$. If $A ne 0$, then the characteristic polynomial of $A$ is $x^2$, which means $A$ has trace and determinant $0$. Therefore
beginequation*
beginbmatrix
a & b\
c & -a
endbmatrix,
endequation*
where $a^2 + b c = 0$, and $a, b, c$ are not all zero.

In part one, distinguish now

• the case when $a = pm 1$, and then either $b = 0$ and $b$ is arbitrary, or vice versa;


• the case when $a ne pm 1$, and then $b ne 0$, say, can be chosen arbitrarily, and $c = (1 - a^2)/b ne 0$.

In part two, distinguish now

• the case when $a = 0$, and then either $b = 0$ and $c$ is arbitrary, or vice versa;


• the case when $a ne 0$, and then $b ne 0$, say, can be chosen arbitrarily, and $c = -a^2/bne 0$.

share|cite|improve this answer

edited Mar 18 at 17:06

Widawensen

4,69331446

answered Mar 17 at 14:40

Andreas CarantiAndreas Caranti

56.9k34397

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks a lot! But I have not yet learnt about characteristic polynomial, so if it's not too much to ask for, could you please add some information about it and how you've used it in this context?
  $endgroup$
  – s0ulr3aper07
  Mar 17 at 15:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @s0ulr3aper07 have you been taught about eigenvalues?
  $endgroup$
  – Andreas Caranti
  Mar 17 at 16:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I have learnt about eigenvalues and have come across the characteristic polynomial while doing so. The way you've used it to find trace and determinant is new to me.
  $endgroup$
  – s0ulr3aper07
  Mar 17 at 16:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @s0ulr3aper07, ok, so I believe you just need to know a couple of things. The first one is that the characteristic polynomial $A - x I$ of $2 times 2$ matrix $A$ has the form $x^2 - T x + D$, where $T$ is the trace of $A$, and $D$ is its determinant. It is just a matter of expanding $detleft( beginbmatrixa - x&b\c&d-xendbmatrixright)$, and then...
  $endgroup$
  – Andreas Caranti
  Mar 17 at 17:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The second thing is to note that if $A^2 = I$, and the eigenvalues of $A$ are both $1$, then $A = I$, while if they are both $-1$, then $A = -I$. Just check that if this is not the case, then $A^2 ne I$.
  $endgroup$
  – Andreas Caranti
  Mar 17 at 17:04
  

  
  
  
  

 | 
show 1 more comment

2

$begingroup$

Case $A^2 = I$.

Two possibilities are of course $A = pm I$. If $A ne pm I$, then the characteristic polynomial of $A$ is $x^2 - 1$, which means $A$ has trace $0$ and determinant $-1$. Therefore
beginequation*
beginbmatrix
a & b\
c & -a
endbmatrix,
endequation*
where $a^2 + b c = 1$.

Case $A^2 = 0$.

One possibility is of course $A = 0$. If $A ne 0$, then the characteristic polynomial of $A$ is $x^2$, which means $A$ has trace and determinant $0$. Therefore
beginequation*
beginbmatrix
a & b\
c & -a
endbmatrix,
endequation*
where $a^2 + b c = 0$, and $a, b, c$ are not all zero.

In part one, distinguish now

• the case when $a = pm 1$, and then either $b = 0$ and $b$ is arbitrary, or vice versa;


• the case when $a ne pm 1$, and then $b ne 0$, say, can be chosen arbitrarily, and $c = (1 - a^2)/b ne 0$.

In part two, distinguish now

• the case when $a = 0$, and then either $b = 0$ and $c$ is arbitrary, or vice versa;


• the case when $a ne 0$, and then $b ne 0$, say, can be chosen arbitrarily, and $c = -a^2/bne 0$.

share|cite|improve this answer

edited Mar 18 at 17:06

Widawensen

4,69331446

answered Mar 17 at 14:40

Andreas CarantiAndreas Caranti

56.9k34397

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks a lot! But I have not yet learnt about characteristic polynomial, so if it's not too much to ask for, could you please add some information about it and how you've used it in this context?
  $endgroup$
  – s0ulr3aper07
  Mar 17 at 15:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @s0ulr3aper07 have you been taught about eigenvalues?
  $endgroup$
  – Andreas Caranti
  Mar 17 at 16:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I have learnt about eigenvalues and have come across the characteristic polynomial while doing so. The way you've used it to find trace and determinant is new to me.
  $endgroup$
  – s0ulr3aper07
  Mar 17 at 16:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @s0ulr3aper07, ok, so I believe you just need to know a couple of things. The first one is that the characteristic polynomial $A - x I$ of $2 times 2$ matrix $A$ has the form $x^2 - T x + D$, where $T$ is the trace of $A$, and $D$ is its determinant. It is just a matter of expanding $detleft( beginbmatrixa - x&b\c&d-xendbmatrixright)$, and then...
  $endgroup$
  – Andreas Caranti
  Mar 17 at 17:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The second thing is to note that if $A^2 = I$, and the eigenvalues of $A$ are both $1$, then $A = I$, while if they are both $-1$, then $A = -I$. Just check that if this is not the case, then $A^2 ne I$.
  $endgroup$
  – Andreas Caranti
  Mar 17 at 17:04
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

Case $A^2 = I$.

Two possibilities are of course $A = pm I$. If $A ne pm I$, then the characteristic polynomial of $A$ is $x^2 - 1$, which means $A$ has trace $0$ and determinant $-1$. Therefore
beginequation*
beginbmatrix
a & b\
c & -a
endbmatrix,
endequation*
where $a^2 + b c = 1$.

Case $A^2 = 0$.

One possibility is of course $A = 0$. If $A ne 0$, then the characteristic polynomial of $A$ is $x^2$, which means $A$ has trace and determinant $0$. Therefore
beginequation*
beginbmatrix
a & b\
c & -a
endbmatrix,
endequation*
where $a^2 + b c = 0$, and $a, b, c$ are not all zero.

In part one, distinguish now

• the case when $a = pm 1$, and then either $b = 0$ and $b$ is arbitrary, or vice versa;


• the case when $a ne pm 1$, and then $b ne 0$, say, can be chosen arbitrarily, and $c = (1 - a^2)/b ne 0$.

In part two, distinguish now

• the case when $a = 0$, and then either $b = 0$ and $c$ is arbitrary, or vice versa;


• the case when $a ne 0$, and then $b ne 0$, say, can be chosen arbitrarily, and $c = -a^2/bne 0$.

share|cite|improve this answer

edited Mar 18 at 17:06

Widawensen

4,69331446

answered Mar 17 at 14:40

Andreas CarantiAndreas Caranti

56.9k34397

$endgroup$

share|cite|improve this answer

edited Mar 18 at 17:06

Widawensen

4,69331446

answered Mar 17 at 14:40

Andreas CarantiAndreas Caranti

56.9k34397

edited Mar 18 at 17:06

Widawensen

4,69331446

edited Mar 18 at 17:06

Widawensen

4,69331446

answered Mar 17 at 14:40

Andreas CarantiAndreas Caranti

56.9k34397

answered Mar 17 at 14:40

Andreas CarantiAndreas Caranti

56.9k34397

1

$begingroup$

$$A = beginbmatrix
a & b\
c & d
endbmatrix
Rightarrow A^2 = beginbmatrix
a^2+bc & (a+d)b\
(a+d)c & bc +d^2
endbmatrix$$

In both cases, you have to impose $(a+d)b = (a+d)c = 0$.

1. Case $a+d neq 0$: then $b = c = 0$ and $A^2 = beginbmatrix
   a^2 & 0\
   0 & d^2
   endbmatrix$ and it's easy to continue.


2. Case $a+d = 0$: $A^2 = beginbmatrix
   a^2+bc & 0\
   0 & bc + d^2
   endbmatrix$ and it's easy to continue.

share|cite|improve this answer

answered Mar 17 at 14:02

dcolazindcolazin

3694

$endgroup$

add a comment | 

1

$begingroup$

$$A = beginbmatrix
a & b\
c & d
endbmatrix
Rightarrow A^2 = beginbmatrix
a^2+bc & (a+d)b\
(a+d)c & bc +d^2
endbmatrix$$

In both cases, you have to impose $(a+d)b = (a+d)c = 0$.

1. Case $a+d neq 0$: then $b = c = 0$ and $A^2 = beginbmatrix
   a^2 & 0\
   0 & d^2
   endbmatrix$ and it's easy to continue.


2. Case $a+d = 0$: $A^2 = beginbmatrix
   a^2+bc & 0\
   0 & bc + d^2
   endbmatrix$ and it's easy to continue.

share|cite|improve this answer

answered Mar 17 at 14:02

dcolazindcolazin

3694

$endgroup$

add a comment | 

$begingroup$

$$A = beginbmatrix
a & b\
c & d
endbmatrix
Rightarrow A^2 = beginbmatrix
a^2+bc & (a+d)b\
(a+d)c & bc +d^2
endbmatrix$$

In both cases, you have to impose $(a+d)b = (a+d)c = 0$.

1. Case $a+d neq 0$: then $b = c = 0$ and $A^2 = beginbmatrix
   a^2 & 0\
   0 & d^2
   endbmatrix$ and it's easy to continue.


2. Case $a+d = 0$: $A^2 = beginbmatrix
   a^2+bc & 0\
   0 & bc + d^2
   endbmatrix$ and it's easy to continue.

share|cite|improve this answer

answered Mar 17 at 14:02

dcolazindcolazin

3694

$endgroup$

share|cite|improve this answer

answered Mar 17 at 14:02

dcolazindcolazin

3694

answered Mar 17 at 14:02

dcolazindcolazin

3694

answered Mar 17 at 14:02

dcolazindcolazin

3694