No. of distinct matrices satisfying a certain equation [duplicate]

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This question already has an answer here:

• 
  Are there infinitely many $Ain mathbbC^2 times 2$ satisfying $A^3 = A$?
  
  3 answers
  
  

[TIFR-GS 2011, Part B, problem no. 1] Let $A$ be a $2×2$ matrix with complex entries. The no. of $2×2$ matrices A with complex entries satisfying the equation $A^3=A$ is infinite. They asked if the statement is true or false.

From the given equation, I've found that the eigenvalues of $A$ can be $0$, $1$, and $-1$. Now I took a trial matrix and tried to satisfy the equation, but it doesn't help in any way. Any help would be appreciated.

linear-algebra eigenvalues-eigenvectors

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edited 10 hours ago

asked 12 hours ago

Anik Bhowmick

849

marked as duplicate by Community♦ 9 hours ago

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The identity doesn't have trace $1$ or determinant $0$.
  – Arnaud Mortier
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay, identity matrix just have the eigenvalue $1$. Thanks for correcting my route.
  – Anik Bhowmick
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Plus, there is the possible eigenvalue $-1$.
  – Arnaud Mortier
  12 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, sorry, I computer it wrong. $0$, $1$, $-1$ are the roots of characteristic equation.
  – Anik Bhowmick
  12 hours ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

This question already has an answer here:

• 
  Are there infinitely many $Ain mathbbC^2 times 2$ satisfying $A^3 = A$?
  
  3 answers
  
  

[TIFR-GS 2011, Part B, problem no. 1] Let $A$ be a $2×2$ matrix with complex entries. The no. of $2×2$ matrices A with complex entries satisfying the equation $A^3=A$ is infinite. They asked if the statement is true or false.

From the given equation, I've found that the eigenvalues of $A$ can be $0$, $1$, and $-1$. Now I took a trial matrix and tried to satisfy the equation, but it doesn't help in any way. Any help would be appreciated.

linear-algebra eigenvalues-eigenvectors

share|cite|improve this question

edited 10 hours ago

asked 12 hours ago

Anik Bhowmick

849

marked as duplicate by Community♦ 9 hours ago

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The identity doesn't have trace $1$ or determinant $0$.
  – Arnaud Mortier
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay, identity matrix just have the eigenvalue $1$. Thanks for correcting my route.
  – Anik Bhowmick
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Plus, there is the possible eigenvalue $-1$.
  – Arnaud Mortier
  12 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, sorry, I computer it wrong. $0$, $1$, $-1$ are the roots of characteristic equation.
  – Anik Bhowmick
  12 hours ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

This question already has an answer here:

• 
  Are there infinitely many $Ain mathbbC^2 times 2$ satisfying $A^3 = A$?
  
  3 answers
  
  

[TIFR-GS 2011, Part B, problem no. 1] Let $A$ be a $2×2$ matrix with complex entries. The no. of $2×2$ matrices A with complex entries satisfying the equation $A^3=A$ is infinite. They asked if the statement is true or false.

From the given equation, I've found that the eigenvalues of $A$ can be $0$, $1$, and $-1$. Now I took a trial matrix and tried to satisfy the equation, but it doesn't help in any way. Any help would be appreciated.

linear-algebra eigenvalues-eigenvectors

share|cite|improve this question

edited 10 hours ago

asked 12 hours ago

Anik Bhowmick

849

This question already has an answer here:

• 
  Are there infinitely many $Ain mathbbC^2 times 2$ satisfying $A^3 = A$?
  
  3 answers
  
  

[TIFR-GS 2011, Part B, problem no. 1] Let $A$ be a $2×2$ matrix with complex entries. The no. of $2×2$ matrices A with complex entries satisfying the equation $A^3=A$ is infinite. They asked if the statement is true or false.

From the given equation, I've found that the eigenvalues of $A$ can be $0$, $1$, and $-1$. Now I took a trial matrix and tried to satisfy the equation, but it doesn't help in any way. Any help would be appreciated.

This question already has an answer here:

• 
  Are there infinitely many $Ain mathbbC^2 times 2$ satisfying $A^3 = A$?
  
  3 answers
  
  

linear-algebra eigenvalues-eigenvectors

share|cite|improve this question

edited 10 hours ago

asked 12 hours ago

Anik Bhowmick

849

share|cite|improve this question

share|cite|improve this question

edited 10 hours ago

edited 10 hours ago

edited 10 hours ago

asked 12 hours ago

Anik Bhowmick

849

asked 12 hours ago

Anik Bhowmick

849

asked 12 hours ago

Anik Bhowmick

849

849

marked as duplicate by Community♦ 9 hours ago

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Community♦ 9 hours ago

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The identity doesn't have trace $1$ or determinant $0$.
  – Arnaud Mortier
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay, identity matrix just have the eigenvalue $1$. Thanks for correcting my route.
  – Anik Bhowmick
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Plus, there is the possible eigenvalue $-1$.
  – Arnaud Mortier
  12 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, sorry, I computer it wrong. $0$, $1$, $-1$ are the roots of characteristic equation.
  – Anik Bhowmick
  12 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The identity doesn't have trace $1$ or determinant $0$.
  – Arnaud Mortier
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay, identity matrix just have the eigenvalue $1$. Thanks for correcting my route.
  – Anik Bhowmick
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Plus, there is the possible eigenvalue $-1$.
  – Arnaud Mortier
  12 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, sorry, I computer it wrong. $0$, $1$, $-1$ are the roots of characteristic equation.
  – Anik Bhowmick
  12 hours ago
  

  
  
  
  

The identity doesn't have trace $1$ or determinant $0$.
– Arnaud Mortier
12 hours ago

The identity doesn't have trace $1$ or determinant $0$.
– Arnaud Mortier
12 hours ago

Okay, identity matrix just have the eigenvalue $1$. Thanks for correcting my route.
– Anik Bhowmick
12 hours ago

Okay, identity matrix just have the eigenvalue $1$. Thanks for correcting my route.
– Anik Bhowmick
12 hours ago

Plus, there is the possible eigenvalue $-1$.
– Arnaud Mortier
12 hours ago

Plus, there is the possible eigenvalue $-1$.
– Arnaud Mortier
12 hours ago

Yes, sorry, I computer it wrong. $0$, $1$, $-1$ are the roots of characteristic equation.
– Anik Bhowmick
12 hours ago

Yes, sorry, I computer it wrong. $0$, $1$, $-1$ are the roots of characteristic equation.
– Anik Bhowmick
12 hours ago

add a comment | 

2 Answers
2

active

oldest

votes

up vote
4
down vote

For any basis $x_1,x_2$ of $mathbbC^2$, any linear transformation $A:mathbbC^2tomathbbC^2$ with eigenvectors $x_1$ and $x_2$ and eigenvalues $-1, 0,$ or $1$ satisfies the equation.

share|cite|improve this answer

answered 12 hours ago

alphacapture

1,682319

• 
  
  
  

  
  

  
  
  
  
  
  

  
  (+1) Nice point of view.
  – Arnaud Mortier
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Got it !! Since any matrix is some sort of linear transformation, it'll happen for all L.T having eigenvalues $0$, $1$, $-1$ !!
  – Anik Bhowmick
  12 hours ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

Since $X^3-X$ has only roots with multiplicity $1$, such a matrix $A$ has to be diagonalizable. So you are looking at the set $$leftA=Pleft(matrixa&0\0&bright)P^-1mid Ptext is invertible and a,bin-1,0,1right$$
The three subsets corresponding to $a=b$ have one element each, now what happens when $aneq b$?

For instance, $$leftleft(matrix1&0\z&1right)left(matrix1&0\0&-1right)left(matrix1&0\-z&1right)mid zinBbb Cright$$
is infinite.

share|cite|improve this answer

answered 12 hours ago

Arnaud Mortier

17.6k21757

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote

For any basis $x_1,x_2$ of $mathbbC^2$, any linear transformation $A:mathbbC^2tomathbbC^2$ with eigenvectors $x_1$ and $x_2$ and eigenvalues $-1, 0,$ or $1$ satisfies the equation.

share|cite|improve this answer

answered 12 hours ago

alphacapture

1,682319

• 
  
  
  

  
  

  
  
  
  
  
  

  
  (+1) Nice point of view.
  – Arnaud Mortier
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Got it !! Since any matrix is some sort of linear transformation, it'll happen for all L.T having eigenvalues $0$, $1$, $-1$ !!
  – Anik Bhowmick
  12 hours ago
  

  
  
  
  

add a comment | 

up vote
4
down vote

For any basis $x_1,x_2$ of $mathbbC^2$, any linear transformation $A:mathbbC^2tomathbbC^2$ with eigenvectors $x_1$ and $x_2$ and eigenvalues $-1, 0,$ or $1$ satisfies the equation.

share|cite|improve this answer

answered 12 hours ago

alphacapture

1,682319

• 
  
  
  

  
  

  
  
  
  
  
  

  
  (+1) Nice point of view.
  – Arnaud Mortier
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Got it !! Since any matrix is some sort of linear transformation, it'll happen for all L.T having eigenvalues $0$, $1$, $-1$ !!
  – Anik Bhowmick
  12 hours ago
  

  
  
  
  

add a comment | 

up vote
4
down vote

up vote
4
down vote

For any basis $x_1,x_2$ of $mathbbC^2$, any linear transformation $A:mathbbC^2tomathbbC^2$ with eigenvectors $x_1$ and $x_2$ and eigenvalues $-1, 0,$ or $1$ satisfies the equation.

share|cite|improve this answer

answered 12 hours ago

alphacapture

1,682319

For any basis $x_1,x_2$ of $mathbbC^2$, any linear transformation $A:mathbbC^2tomathbbC^2$ with eigenvectors $x_1$ and $x_2$ and eigenvalues $-1, 0,$ or $1$ satisfies the equation.

share|cite|improve this answer

answered 12 hours ago

alphacapture

1,682319

share|cite|improve this answer

share|cite|improve this answer

answered 12 hours ago

alphacapture

1,682319

answered 12 hours ago

alphacapture

1,682319

answered 12 hours ago

alphacapture

1,682319

1,682319

• 
  
  
  

  
  

  
  
  
  
  
  

  
  (+1) Nice point of view.
  – Arnaud Mortier
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Got it !! Since any matrix is some sort of linear transformation, it'll happen for all L.T having eigenvalues $0$, $1$, $-1$ !!
  – Anik Bhowmick
  12 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  (+1) Nice point of view.
  – Arnaud Mortier
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Got it !! Since any matrix is some sort of linear transformation, it'll happen for all L.T having eigenvalues $0$, $1$, $-1$ !!
  – Anik Bhowmick
  12 hours ago
  

  
  
  
  

(+1) Nice point of view.
– Arnaud Mortier
12 hours ago

(+1) Nice point of view.
– Arnaud Mortier
12 hours ago

Got it !! Since any matrix is some sort of linear transformation, it'll happen for all L.T having eigenvalues $0$, $1$, $-1$ !!
– Anik Bhowmick
12 hours ago

Got it !! Since any matrix is some sort of linear transformation, it'll happen for all L.T having eigenvalues $0$, $1$, $-1$ !!
– Anik Bhowmick
12 hours ago

add a comment | 

up vote
2
down vote

Since $X^3-X$ has only roots with multiplicity $1$, such a matrix $A$ has to be diagonalizable. So you are looking at the set $$leftA=Pleft(matrixa&0\0&bright)P^-1mid Ptext is invertible and a,bin-1,0,1right$$
The three subsets corresponding to $a=b$ have one element each, now what happens when $aneq b$?

For instance, $$leftleft(matrix1&0\z&1right)left(matrix1&0\0&-1right)left(matrix1&0\-z&1right)mid zinBbb Cright$$
is infinite.

share|cite|improve this answer

answered 12 hours ago

Arnaud Mortier

17.6k21757

add a comment | 

up vote
2
down vote

Since $X^3-X$ has only roots with multiplicity $1$, such a matrix $A$ has to be diagonalizable. So you are looking at the set $$leftA=Pleft(matrixa&0\0&bright)P^-1mid Ptext is invertible and a,bin-1,0,1right$$
The three subsets corresponding to $a=b$ have one element each, now what happens when $aneq b$?

For instance, $$leftleft(matrix1&0\z&1right)left(matrix1&0\0&-1right)left(matrix1&0\-z&1right)mid zinBbb Cright$$
is infinite.

share|cite|improve this answer

answered 12 hours ago

Arnaud Mortier

17.6k21757

add a comment | 

up vote
2
down vote

up vote
2
down vote

Since $X^3-X$ has only roots with multiplicity $1$, such a matrix $A$ has to be diagonalizable. So you are looking at the set $$leftA=Pleft(matrixa&0\0&bright)P^-1mid Ptext is invertible and a,bin-1,0,1right$$
The three subsets corresponding to $a=b$ have one element each, now what happens when $aneq b$?

For instance, $$leftleft(matrix1&0\z&1right)left(matrix1&0\0&-1right)left(matrix1&0\-z&1right)mid zinBbb Cright$$
is infinite.

share|cite|improve this answer

answered 12 hours ago

Arnaud Mortier

17.6k21757

Since $X^3-X$ has only roots with multiplicity $1$, such a matrix $A$ has to be diagonalizable. So you are looking at the set $$leftA=Pleft(matrixa&0\0&bright)P^-1mid Ptext is invertible and a,bin-1,0,1right$$
The three subsets corresponding to $a=b$ have one element each, now what happens when $aneq b$?

For instance, $$leftleft(matrix1&0\z&1right)left(matrix1&0\0&-1right)left(matrix1&0\-z&1right)mid zinBbb Cright$$
is infinite.

share|cite|improve this answer

answered 12 hours ago

Arnaud Mortier

17.6k21757

share|cite|improve this answer

share|cite|improve this answer

answered 12 hours ago

Arnaud Mortier

17.6k21757

answered 12 hours ago

Arnaud Mortier

17.6k21757

answered 12 hours ago

Arnaud Mortier

17.6k21757

17.6k21757

add a comment | 

add a comment |