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$A^2 + A + I_n = 0 implies$ matrix $A$ is invertible

Invertible matrices over a commutative ring and their determinantsIf $A^2+2A+I_n=O_n$ then $A$ is invertibleUsing determinants, show that for all $n times n$ symmetric matrices $A$, the matrix $A^2 + I_n$ is invertible.Prove that A + B is invertible iff $I_n$ + $A^-1$B is invertible (matrices)Product or sum of invertible matrix give an invertible matrix?Nullspace and column space of invertible matrixProve that the matrix $I_n - A$ is invertibleHelp with proof or counterexample: $A^3=0 implies I_n+A$ is invertibleInvertible matrix and spanInvertible matrix implies square matrix

1

$begingroup$

I need to proof that if
$A^2 + A + I_n = 0$ then matrix $A$ is invertible.

I can see why $A^2 + A$ is invertible, but can't find a way to proof it on $A$.

linear-algebra matrices

share|cite|improve this question

edited Mar 11 at 10:02

Andrews

1,2691421

asked Mar 11 at 9:22

Ron AmirRon Amir

82

New contributor

Ron Amir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The usual terminology is "A is invertible" (and not reversable).
  $endgroup$
  – P Vanchinathan
  Mar 11 at 9:29
  

  
  
  
  

add a comment | 

1

$begingroup$

I need to proof that if
$A^2 + A + I_n = 0$ then matrix $A$ is invertible.

I can see why $A^2 + A$ is invertible, but can't find a way to proof it on $A$.

linear-algebra matrices

share|cite|improve this question

edited Mar 11 at 10:02

Andrews

1,2691421

asked Mar 11 at 9:22

Ron AmirRon Amir

82

New contributor

Ron Amir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The usual terminology is "A is invertible" (and not reversable).
  $endgroup$
  – P Vanchinathan
  Mar 11 at 9:29
  

  
  
  
  

add a comment | 

$begingroup$

I need to proof that if
$A^2 + A + I_n = 0$ then matrix $A$ is invertible.

I can see why $A^2 + A$ is invertible, but can't find a way to proof it on $A$.

linear-algebra matrices

share|cite|improve this question

edited Mar 11 at 10:02

Andrews

1,2691421

asked Mar 11 at 9:22

Ron AmirRon Amir

82

New contributor

Ron Amir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

share|cite|improve this question

edited Mar 11 at 10:02

Andrews

1,2691421

asked Mar 11 at 9:22

Ron AmirRon Amir

82

New contributor

Ron Amir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|cite|improve this question

edited Mar 11 at 10:02

Andrews

1,2691421

asked Mar 11 at 9:22

Ron AmirRon Amir

82

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Ron Amir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited Mar 11 at 10:02

Andrews

1,2691421

edited Mar 11 at 10:02

Andrews

1,2691421

asked Mar 11 at 9:22

Ron AmirRon Amir

82

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asked Mar 11 at 9:22

Ron AmirRon Amir

82

4 Answers
4

active

oldest

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2

$begingroup$

Let $Ax=0$, then $0=(A^2+A+I_n)x=I_nx=x$, hence $ker(A)=0$.

Conclusion ?

share|cite|improve this answer

answered Mar 11 at 9:26

FredFred

48.3k1849

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
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  I would add general conclusion that if only $p(A)=0$ with non-zero part $I_n$ then $A$ must be invertible.
  $endgroup$
  – Widawensen
  Mar 11 at 10:48
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, if $ p(x)=x^n+a_n-1x^n-1+...+a_1x$ and $p(A)+I_n=0$, then $A$ is invertible.
  $endgroup$
  – Fred
  Mar 11 at 10:58
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why the downvote? ?
  $endgroup$
  – Fred
  Mar 11 at 19:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I upvoted both answers, they seem to be very good. I'm also very interested what weaknesses someone noticed in the reasoning about kernel ..
  $endgroup$
  – Widawensen
  2 days ago
  
  

  
  
  
  

add a comment | 

4

$begingroup$

$I_n=-A(A+I_n)$, hence $1= det(I_n)= det(-A)det(A+I_n)=(-1)^n det(A) det(A+I_n).$

This gives $ det(A) ne 0$.

share|cite|improve this answer

answered Mar 11 at 9:37

FredFred

48.3k1849

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  And more explicitly, $A^-1=-A-I_n$.
  $endgroup$
  – Bernard
  Mar 11 at 9:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dietrich: $A=-I_n$ doe not satisfy the equation $A^2 + A + I_n = 0.$
  $endgroup$
  – Fred
  Mar 11 at 9:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Fred Good point!
  $endgroup$
  – Dietrich Burde
  Mar 11 at 9:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This solution is definitely also worth of upvoting. Even more simple shows that $A$ has the inverse $ I=A(-A-I)$
  $endgroup$
  – Widawensen
  Mar 11 at 13:05
  
  

  
  
  
  

add a comment | 

1

$begingroup$

The inverse is $-(A+I)$, since by direct calculation

$$-(A+I)A=-A^2-A=I.$$

share|cite|improve this answer

edited Mar 11 at 20:27

Yves Daoust

130k676229

answered Mar 11 at 18:23

Beavis TrumpBeavis Trump

111

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Beavis Trump is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is the best answer here. I have allowed myself to reformat. Undo if you want.
  $endgroup$
  – Yves Daoust
  Mar 11 at 20:25
  
  

  
  
  
  

add a comment | 

1

$begingroup$

To show that any square matrix A is invertible, you need to show that there exists a square matrix $A^-1$ (called the inverse of A) such that $AA^-1 = A^-1A = I_n $. That's the definition of a matrix being invertible.

Now, we can use the equation $A^2 + A + I_n = O_n$ to find a potential candidate for the inverse $A^-1$ such that the equation $AA^-1 = A^-1A = I_n $ holds. Let us solve for $I_n$:

$A^2 + A + I_n = O_n$ ⇒ $I_n = -A^2 - A$ ⇒ $I_n = A(-A - I_n)$

Here, we have our candidate inverse matrix, which is just $A^-1 = (-A - I_n)$. Now, you must show that it satisfies the equation $AA^-1 = A^-1A = I_n$.

Let $A^-1 = -A - I_n$. Then $AA^-1 = A(-A - I_n) = -A^2 - AI_n = -A^2 - A = -(A^2 + A) = I_n$

Similarly, $A^-1A = (-A - I_n)A = -A^2 - I_nA = -A^2 - A = I_n$.

By the definition, A is invertible.

share|cite|improve this answer

answered Mar 11 at 20:33

TimTim

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2

$begingroup$

Let $Ax=0$, then $0=(A^2+A+I_n)x=I_nx=x$, hence $ker(A)=0$.

Conclusion ?

share|cite|improve this answer

answered Mar 11 at 9:26

FredFred

48.3k1849

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I would add general conclusion that if only $p(A)=0$ with non-zero part $I_n$ then $A$ must be invertible.
  $endgroup$
  – Widawensen
  Mar 11 at 10:48
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, if $ p(x)=x^n+a_n-1x^n-1+...+a_1x$ and $p(A)+I_n=0$, then $A$ is invertible.
  $endgroup$
  – Fred
  Mar 11 at 10:58
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why the downvote? ?
  $endgroup$
  – Fred
  Mar 11 at 19:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I upvoted both answers, they seem to be very good. I'm also very interested what weaknesses someone noticed in the reasoning about kernel ..
  $endgroup$
  – Widawensen
  2 days ago
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Let $Ax=0$, then $0=(A^2+A+I_n)x=I_nx=x$, hence $ker(A)=0$.

Conclusion ?

share|cite|improve this answer

answered Mar 11 at 9:26

FredFred

48.3k1849

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I would add general conclusion that if only $p(A)=0$ with non-zero part $I_n$ then $A$ must be invertible.
  $endgroup$
  – Widawensen
  Mar 11 at 10:48
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, if $ p(x)=x^n+a_n-1x^n-1+...+a_1x$ and $p(A)+I_n=0$, then $A$ is invertible.
  $endgroup$
  – Fred
  Mar 11 at 10:58
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why the downvote? ?
  $endgroup$
  – Fred
  Mar 11 at 19:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I upvoted both answers, they seem to be very good. I'm also very interested what weaknesses someone noticed in the reasoning about kernel ..
  $endgroup$
  – Widawensen
  2 days ago
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $Ax=0$, then $0=(A^2+A+I_n)x=I_nx=x$, hence $ker(A)=0$.

Conclusion ?

share|cite|improve this answer

answered Mar 11 at 9:26

FredFred

48.3k1849

$endgroup$

share|cite|improve this answer

answered Mar 11 at 9:26

FredFred

48.3k1849

answered Mar 11 at 9:26

FredFred

48.3k1849

answered Mar 11 at 9:26

FredFred

48.3k1849

4

$begingroup$

$I_n=-A(A+I_n)$, hence $1= det(I_n)= det(-A)det(A+I_n)=(-1)^n det(A) det(A+I_n).$

This gives $ det(A) ne 0$.

share|cite|improve this answer

answered Mar 11 at 9:37

FredFred

48.3k1849

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  And more explicitly, $A^-1=-A-I_n$.
  $endgroup$
  – Bernard
  Mar 11 at 9:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dietrich: $A=-I_n$ doe not satisfy the equation $A^2 + A + I_n = 0.$
  $endgroup$
  – Fred
  Mar 11 at 9:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Fred Good point!
  $endgroup$
  – Dietrich Burde
  Mar 11 at 9:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This solution is definitely also worth of upvoting. Even more simple shows that $A$ has the inverse $ I=A(-A-I)$
  $endgroup$
  – Widawensen
  Mar 11 at 13:05
  
  

  
  
  
  

add a comment | 

4

$begingroup$

$I_n=-A(A+I_n)$, hence $1= det(I_n)= det(-A)det(A+I_n)=(-1)^n det(A) det(A+I_n).$

This gives $ det(A) ne 0$.

share|cite|improve this answer

answered Mar 11 at 9:37

FredFred

48.3k1849

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  And more explicitly, $A^-1=-A-I_n$.
  $endgroup$
  – Bernard
  Mar 11 at 9:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dietrich: $A=-I_n$ doe not satisfy the equation $A^2 + A + I_n = 0.$
  $endgroup$
  – Fred
  Mar 11 at 9:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Fred Good point!
  $endgroup$
  – Dietrich Burde
  Mar 11 at 9:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This solution is definitely also worth of upvoting. Even more simple shows that $A$ has the inverse $ I=A(-A-I)$
  $endgroup$
  – Widawensen
  Mar 11 at 13:05
  
  

  
  
  
  

add a comment | 

$begingroup$

$I_n=-A(A+I_n)$, hence $1= det(I_n)= det(-A)det(A+I_n)=(-1)^n det(A) det(A+I_n).$

This gives $ det(A) ne 0$.

share|cite|improve this answer

answered Mar 11 at 9:37

FredFred

48.3k1849

$endgroup$

share|cite|improve this answer

answered Mar 11 at 9:37

FredFred

48.3k1849

answered Mar 11 at 9:37

FredFred

48.3k1849

answered Mar 11 at 9:37

FredFred

48.3k1849

1

$begingroup$

The inverse is $-(A+I)$, since by direct calculation

$$-(A+I)A=-A^2-A=I.$$

share|cite|improve this answer

edited Mar 11 at 20:27

Yves Daoust

130k676229

answered Mar 11 at 18:23

Beavis TrumpBeavis Trump

111

New contributor

Beavis Trump is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is the best answer here. I have allowed myself to reformat. Undo if you want.
  $endgroup$
  – Yves Daoust
  Mar 11 at 20:25
  
  

  
  
  
  

add a comment | 

1

$begingroup$

The inverse is $-(A+I)$, since by direct calculation

$$-(A+I)A=-A^2-A=I.$$

share|cite|improve this answer

edited Mar 11 at 20:27

Yves Daoust

130k676229

answered Mar 11 at 18:23

Beavis TrumpBeavis Trump

111

New contributor

Beavis Trump is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is the best answer here. I have allowed myself to reformat. Undo if you want.
  $endgroup$
  – Yves Daoust
  Mar 11 at 20:25
  
  

  
  
  
  

add a comment | 

$begingroup$

The inverse is $-(A+I)$, since by direct calculation

$$-(A+I)A=-A^2-A=I.$$

share|cite|improve this answer

edited Mar 11 at 20:27

Yves Daoust

130k676229

answered Mar 11 at 18:23

Beavis TrumpBeavis Trump

111

New contributor

Beavis Trump is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

share|cite|improve this answer

edited Mar 11 at 20:27

Yves Daoust

130k676229

answered Mar 11 at 18:23

Beavis TrumpBeavis Trump

111

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edited Mar 11 at 20:27

Yves Daoust

130k676229

edited Mar 11 at 20:27

Yves Daoust

130k676229

answered Mar 11 at 18:23

Beavis TrumpBeavis Trump

111

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answered Mar 11 at 18:23

Beavis TrumpBeavis Trump

111

1

$begingroup$

To show that any square matrix A is invertible, you need to show that there exists a square matrix $A^-1$ (called the inverse of A) such that $AA^-1 = A^-1A = I_n $. That's the definition of a matrix being invertible.

Now, we can use the equation $A^2 + A + I_n = O_n$ to find a potential candidate for the inverse $A^-1$ such that the equation $AA^-1 = A^-1A = I_n $ holds. Let us solve for $I_n$:

$A^2 + A + I_n = O_n$ ⇒ $I_n = -A^2 - A$ ⇒ $I_n = A(-A - I_n)$

Here, we have our candidate inverse matrix, which is just $A^-1 = (-A - I_n)$. Now, you must show that it satisfies the equation $AA^-1 = A^-1A = I_n$.

Let $A^-1 = -A - I_n$. Then $AA^-1 = A(-A - I_n) = -A^2 - AI_n = -A^2 - A = -(A^2 + A) = I_n$

Similarly, $A^-1A = (-A - I_n)A = -A^2 - I_nA = -A^2 - A = I_n$.

By the definition, A is invertible.

share|cite|improve this answer

answered Mar 11 at 20:33

TimTim

465

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Tim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

add a comment | 

1

$begingroup$

To show that any square matrix A is invertible, you need to show that there exists a square matrix $A^-1$ (called the inverse of A) such that $AA^-1 = A^-1A = I_n $. That's the definition of a matrix being invertible.

Now, we can use the equation $A^2 + A + I_n = O_n$ to find a potential candidate for the inverse $A^-1$ such that the equation $AA^-1 = A^-1A = I_n $ holds. Let us solve for $I_n$:

$A^2 + A + I_n = O_n$ ⇒ $I_n = -A^2 - A$ ⇒ $I_n = A(-A - I_n)$

Here, we have our candidate inverse matrix, which is just $A^-1 = (-A - I_n)$. Now, you must show that it satisfies the equation $AA^-1 = A^-1A = I_n$.

Let $A^-1 = -A - I_n$. Then $AA^-1 = A(-A - I_n) = -A^2 - AI_n = -A^2 - A = -(A^2 + A) = I_n$

Similarly, $A^-1A = (-A - I_n)A = -A^2 - I_nA = -A^2 - A = I_n$.

By the definition, A is invertible.

share|cite|improve this answer

answered Mar 11 at 20:33

TimTim

465

New contributor

Tim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

To show that any square matrix A is invertible, you need to show that there exists a square matrix $A^-1$ (called the inverse of A) such that $AA^-1 = A^-1A = I_n $. That's the definition of a matrix being invertible.

Now, we can use the equation $A^2 + A + I_n = O_n$ to find a potential candidate for the inverse $A^-1$ such that the equation $AA^-1 = A^-1A = I_n $ holds. Let us solve for $I_n$:

$A^2 + A + I_n = O_n$ ⇒ $I_n = -A^2 - A$ ⇒ $I_n = A(-A - I_n)$

Here, we have our candidate inverse matrix, which is just $A^-1 = (-A - I_n)$. Now, you must show that it satisfies the equation $AA^-1 = A^-1A = I_n$.

Let $A^-1 = -A - I_n$. Then $AA^-1 = A(-A - I_n) = -A^2 - AI_n = -A^2 - A = -(A^2 + A) = I_n$

Similarly, $A^-1A = (-A - I_n)A = -A^2 - I_nA = -A^2 - A = I_n$.

By the definition, A is invertible.

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answered Mar 11 at 20:33

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answered Mar 11 at 20:33

TimTim

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Tim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Mar 11 at 20:33

TimTim

465

New contributor

Tim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Mar 11 at 20:33

TimTim

465