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What does the notation $A^-2$ mean if $A$ is a matrix?

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1

1

$begingroup$

$A^2$ means to multiple the matrix by itself, and $A^-1$ refers to the matrix's inverse. Would $A^-2$ be the square of the inverse or the inverse of the square?

linear-algebra matrices notation exponentiation inverse

share|cite|improve this question

edited Mar 11 at 21:43

Rodrigo de Azevedo

13k41960

asked Mar 11 at 21:32

James RonaldJames Ronald

1807

$endgroup$

add a comment | 

1

1

$begingroup$

$A^2$ means to multiple the matrix by itself, and $A^-1$ refers to the matrix's inverse. Would $A^-2$ be the square of the inverse or the inverse of the square?

linear-algebra matrices notation exponentiation inverse

share|cite|improve this question

edited Mar 11 at 21:43

Rodrigo de Azevedo

13k41960

asked Mar 11 at 21:32

James RonaldJames Ronald

1807

$endgroup$

add a comment | 

$begingroup$

$A^2$ means to multiple the matrix by itself, and $A^-1$ refers to the matrix's inverse. Would $A^-2$ be the square of the inverse or the inverse of the square?

linear-algebra matrices notation exponentiation inverse

share|cite|improve this question

edited Mar 11 at 21:43

Rodrigo de Azevedo

13k41960

asked Mar 11 at 21:32

James RonaldJames Ronald

1807

$endgroup$

share|cite|improve this question

edited Mar 11 at 21:43

Rodrigo de Azevedo

13k41960

asked Mar 11 at 21:32

James RonaldJames Ronald

1807

share|cite|improve this question

edited Mar 11 at 21:43

Rodrigo de Azevedo

13k41960

asked Mar 11 at 21:32

James RonaldJames Ronald

1807

edited Mar 11 at 21:43

Rodrigo de Azevedo

13k41960

edited Mar 11 at 21:43

Rodrigo de Azevedo

13k41960

asked Mar 11 at 21:32

James RonaldJames Ronald

1807

asked Mar 11 at 21:32

James RonaldJames Ronald

1807

3 Answers
3

active

oldest

votes

7

$begingroup$

It means $(A^-1)^2$, so the square of the inverse, which also happens to be the inverse of the square.

share|cite|improve this answer

edited Mar 11 at 21:52

Javi

2,9362829

answered Mar 11 at 21:34

ThinkingThinking

1,22716

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  which also happens to be the inverse of the square!
  $endgroup$
  – Kenny Wong
  Mar 11 at 21:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The expression has a conventional meaning only if $A$ is an invertible matrix (a square matrix of full rank).
  $endgroup$
  – hardmath
  Mar 11 at 21:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you! Also wow @KennyWong I actually tested that and found that it doesn't, but looking back I was incorrect haha. Would you be able to tell me why exactly this rule works? Do exponent laws apply to matrices?
  $endgroup$
  – James Ronald
  Mar 11 at 21:51
  

  
  
  
  

add a comment | 

4

$begingroup$

$$left(A^-1right)^2A^2=left(A^-1A^-1right)(AA) = A^-1left(A^-1Aright)A=A^-1IA=A^-1A=I
$$
which says that $$left(A^-1right)^2=left(A^2right)^-1
$$
This is essentially the same as the proof that $$left(1over xright)^2=1over x^2$$

share|cite|improve this answer

answered Mar 11 at 22:14

saulspatzsaulspatz

17.3k31435

$endgroup$

add a comment | 

1

$begingroup$

It can be shown, via induction, that $(A^-1)^n=(A^n)^-1$, $forall n in mathbbN$. Thus, whenever $A$ is invertible, $A^-n:= (A^-1)^n=(A^n)^-1$.

share|cite|improve this answer

answered Mar 12 at 3:32

Pietro PaparellaPietro Paparella

1,544615

$endgroup$

add a comment | 

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7

$begingroup$

It means $(A^-1)^2$, so the square of the inverse, which also happens to be the inverse of the square.

share|cite|improve this answer

edited Mar 11 at 21:52

Javi

2,9362829

answered Mar 11 at 21:34

ThinkingThinking

1,22716

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  which also happens to be the inverse of the square!
  $endgroup$
  – Kenny Wong
  Mar 11 at 21:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The expression has a conventional meaning only if $A$ is an invertible matrix (a square matrix of full rank).
  $endgroup$
  – hardmath
  Mar 11 at 21:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you! Also wow @KennyWong I actually tested that and found that it doesn't, but looking back I was incorrect haha. Would you be able to tell me why exactly this rule works? Do exponent laws apply to matrices?
  $endgroup$
  – James Ronald
  Mar 11 at 21:51
  

  
  
  
  

add a comment | 

7

$begingroup$

It means $(A^-1)^2$, so the square of the inverse, which also happens to be the inverse of the square.

share|cite|improve this answer

edited Mar 11 at 21:52

Javi

2,9362829

answered Mar 11 at 21:34

ThinkingThinking

1,22716

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  which also happens to be the inverse of the square!
  $endgroup$
  – Kenny Wong
  Mar 11 at 21:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The expression has a conventional meaning only if $A$ is an invertible matrix (a square matrix of full rank).
  $endgroup$
  – hardmath
  Mar 11 at 21:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you! Also wow @KennyWong I actually tested that and found that it doesn't, but looking back I was incorrect haha. Would you be able to tell me why exactly this rule works? Do exponent laws apply to matrices?
  $endgroup$
  – James Ronald
  Mar 11 at 21:51
  

  
  
  
  

add a comment | 

$begingroup$

It means $(A^-1)^2$, so the square of the inverse, which also happens to be the inverse of the square.

share|cite|improve this answer

edited Mar 11 at 21:52

Javi

2,9362829

answered Mar 11 at 21:34

ThinkingThinking

1,22716

$endgroup$

share|cite|improve this answer

edited Mar 11 at 21:52

Javi

2,9362829

answered Mar 11 at 21:34

ThinkingThinking

1,22716

edited Mar 11 at 21:52

Javi

2,9362829

edited Mar 11 at 21:52

Javi

2,9362829

answered Mar 11 at 21:34

ThinkingThinking

1,22716

answered Mar 11 at 21:34

ThinkingThinking

1,22716

4

$begingroup$

$$left(A^-1right)^2A^2=left(A^-1A^-1right)(AA) = A^-1left(A^-1Aright)A=A^-1IA=A^-1A=I
$$
which says that $$left(A^-1right)^2=left(A^2right)^-1
$$
This is essentially the same as the proof that $$left(1over xright)^2=1over x^2$$

share|cite|improve this answer

answered Mar 11 at 22:14

saulspatzsaulspatz

17.3k31435

$endgroup$

add a comment | 

4

$begingroup$

$$left(A^-1right)^2A^2=left(A^-1A^-1right)(AA) = A^-1left(A^-1Aright)A=A^-1IA=A^-1A=I
$$
which says that $$left(A^-1right)^2=left(A^2right)^-1
$$
This is essentially the same as the proof that $$left(1over xright)^2=1over x^2$$

share|cite|improve this answer

answered Mar 11 at 22:14

saulspatzsaulspatz

17.3k31435

$endgroup$

add a comment | 

$begingroup$

$$left(A^-1right)^2A^2=left(A^-1A^-1right)(AA) = A^-1left(A^-1Aright)A=A^-1IA=A^-1A=I
$$
which says that $$left(A^-1right)^2=left(A^2right)^-1
$$
This is essentially the same as the proof that $$left(1over xright)^2=1over x^2$$

share|cite|improve this answer

answered Mar 11 at 22:14

saulspatzsaulspatz

17.3k31435

$endgroup$

share|cite|improve this answer

answered Mar 11 at 22:14

saulspatzsaulspatz

17.3k31435

answered Mar 11 at 22:14

saulspatzsaulspatz

17.3k31435

answered Mar 11 at 22:14

saulspatzsaulspatz

17.3k31435

1

$begingroup$

It can be shown, via induction, that $(A^-1)^n=(A^n)^-1$, $forall n in mathbbN$. Thus, whenever $A$ is invertible, $A^-n:= (A^-1)^n=(A^n)^-1$.

share|cite|improve this answer

answered Mar 12 at 3:32

Pietro PaparellaPietro Paparella

1,544615

$endgroup$

add a comment | 

1

$begingroup$

It can be shown, via induction, that $(A^-1)^n=(A^n)^-1$, $forall n in mathbbN$. Thus, whenever $A$ is invertible, $A^-n:= (A^-1)^n=(A^n)^-1$.

share|cite|improve this answer

answered Mar 12 at 3:32

Pietro PaparellaPietro Paparella

1,544615

$endgroup$

add a comment | 

$begingroup$

It can be shown, via induction, that $(A^-1)^n=(A^n)^-1$, $forall n in mathbbN$. Thus, whenever $A$ is invertible, $A^-n:= (A^-1)^n=(A^n)^-1$.

share|cite|improve this answer

answered Mar 12 at 3:32

Pietro PaparellaPietro Paparella

1,544615

$endgroup$

share|cite|improve this answer

answered Mar 12 at 3:32

Pietro PaparellaPietro Paparella

1,544615

answered Mar 12 at 3:32

Pietro PaparellaPietro Paparella

1,544615

answered Mar 12 at 3:32

Pietro PaparellaPietro Paparella

1,544615