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If $T$ is an invertible linear transformation and $vecv$ is an eigenvector of $T$, then $vecv$ is an eigenvector of $T^-1$

1

$begingroup$

I saw there is a proof for invertible matrices, but I don't know how to put this mathematically for a transformation.
How do I prove an invertible linear transformation has the same eigenvectors as its inverse?

linear-algebra eigenvalues-eigenvectors transformation

share|cite|improve this question

edited Apr 13 at 12:08

YuiTo Cheng

2,65641037

asked Apr 13 at 7:18

DGRDGR

163

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can't you just fix bases and consider your linear transformation as a matrix?
  $endgroup$
  – dcolazin
  Apr 13 at 7:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yup...I suppose that's what should have been done in the first place. I have posted an algebraic solution.
  $endgroup$
  – DGR
  Apr 13 at 7:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wouldn’t the proof be exactly the same as the one for matrices?
  $endgroup$
  – amd
  Apr 13 at 18:42
  

  
  
  
  

add a comment | 

1

$begingroup$

I saw there is a proof for invertible matrices, but I don't know how to put this mathematically for a transformation.
How do I prove an invertible linear transformation has the same eigenvectors as its inverse?

linear-algebra eigenvalues-eigenvectors transformation

share|cite|improve this question

edited Apr 13 at 12:08

YuiTo Cheng

2,65641037

asked Apr 13 at 7:18

DGRDGR

163

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can't you just fix bases and consider your linear transformation as a matrix?
  $endgroup$
  – dcolazin
  Apr 13 at 7:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yup...I suppose that's what should have been done in the first place. I have posted an algebraic solution.
  $endgroup$
  – DGR
  Apr 13 at 7:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wouldn’t the proof be exactly the same as the one for matrices?
  $endgroup$
  – amd
  Apr 13 at 18:42
  

  
  
  
  

add a comment | 

$begingroup$

I saw there is a proof for invertible matrices, but I don't know how to put this mathematically for a transformation.
How do I prove an invertible linear transformation has the same eigenvectors as its inverse?

linear-algebra eigenvalues-eigenvectors transformation

share|cite|improve this question

edited Apr 13 at 12:08

YuiTo Cheng

2,65641037

asked Apr 13 at 7:18

DGRDGR

163

$endgroup$

share|cite|improve this question

edited Apr 13 at 12:08

YuiTo Cheng

2,65641037

asked Apr 13 at 7:18

DGRDGR

163

share|cite|improve this question

edited Apr 13 at 12:08

YuiTo Cheng

2,65641037

asked Apr 13 at 7:18

DGRDGR

163

edited Apr 13 at 12:08

YuiTo Cheng

2,65641037

edited Apr 13 at 12:08

YuiTo Cheng

2,65641037

asked Apr 13 at 7:18

DGRDGR

163

asked Apr 13 at 7:18

DGRDGR

163

2 Answers
2

active

oldest

votes

4

$begingroup$

Observe that for an invertible matrix $A$ with eigenvector $mathbf v$ and corresponding eigenvalue $lambdaneq 0$, you have that
beginalign*
mathbf v &= I mathbf v\&=A^-1A mathbf v\&=lambda A^-1 mathbf v
endalign*
Hence $A^-1 mathbf v = lambda^-1mathbf v$

share|cite|improve this answer

answered Apr 13 at 7:27

P. QuintonP. Quinton

2,0001214

$endgroup$

add a comment | 

1

$begingroup$

OK, so Iv'e found a solution!

beginalign*\
Tv &= lambda v vert *T^-1 \
T^-1Tv &= T^-1lambda v \
v &= lambda T^-1v vert *lambda^-1 & text(Invertible transformation, $lambdaneq 0$) \
lambda^-1v &= T^-1v
endalign*

For $T^-1$, eigenvalue $lambda^-1$, eigenvector is $v$ (same as eigenvector of $T$).

Hope this helps!

share|cite|improve this answer

edited Apr 13 at 7:56

Widawensen

4,81531447

answered Apr 13 at 7:25

DGRDGR

163

$endgroup$

add a comment | 

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4

$begingroup$

Observe that for an invertible matrix $A$ with eigenvector $mathbf v$ and corresponding eigenvalue $lambdaneq 0$, you have that
beginalign*
mathbf v &= I mathbf v\&=A^-1A mathbf v\&=lambda A^-1 mathbf v
endalign*
Hence $A^-1 mathbf v = lambda^-1mathbf v$

share|cite|improve this answer

answered Apr 13 at 7:27

P. QuintonP. Quinton

2,0001214

$endgroup$

add a comment | 

4

$begingroup$

Observe that for an invertible matrix $A$ with eigenvector $mathbf v$ and corresponding eigenvalue $lambdaneq 0$, you have that
beginalign*
mathbf v &= I mathbf v\&=A^-1A mathbf v\&=lambda A^-1 mathbf v
endalign*
Hence $A^-1 mathbf v = lambda^-1mathbf v$

share|cite|improve this answer

answered Apr 13 at 7:27

P. QuintonP. Quinton

2,0001214

$endgroup$

add a comment | 

$begingroup$

Observe that for an invertible matrix $A$ with eigenvector $mathbf v$ and corresponding eigenvalue $lambdaneq 0$, you have that
beginalign*
mathbf v &= I mathbf v\&=A^-1A mathbf v\&=lambda A^-1 mathbf v
endalign*
Hence $A^-1 mathbf v = lambda^-1mathbf v$

share|cite|improve this answer

answered Apr 13 at 7:27

P. QuintonP. Quinton

2,0001214

$endgroup$

share|cite|improve this answer

answered Apr 13 at 7:27

P. QuintonP. Quinton

2,0001214

answered Apr 13 at 7:27

P. QuintonP. Quinton

2,0001214

answered Apr 13 at 7:27

P. QuintonP. Quinton

2,0001214

1

$begingroup$

OK, so Iv'e found a solution!

beginalign*\
Tv &= lambda v vert *T^-1 \
T^-1Tv &= T^-1lambda v \
v &= lambda T^-1v vert *lambda^-1 & text(Invertible transformation, $lambdaneq 0$) \
lambda^-1v &= T^-1v
endalign*

For $T^-1$, eigenvalue $lambda^-1$, eigenvector is $v$ (same as eigenvector of $T$).

Hope this helps!

share|cite|improve this answer

edited Apr 13 at 7:56

Widawensen

4,81531447

answered Apr 13 at 7:25

DGRDGR

163

$endgroup$

add a comment | 

1

$begingroup$

OK, so Iv'e found a solution!

beginalign*\
Tv &= lambda v vert *T^-1 \
T^-1Tv &= T^-1lambda v \
v &= lambda T^-1v vert *lambda^-1 & text(Invertible transformation, $lambdaneq 0$) \
lambda^-1v &= T^-1v
endalign*

For $T^-1$, eigenvalue $lambda^-1$, eigenvector is $v$ (same as eigenvector of $T$).

Hope this helps!

share|cite|improve this answer

edited Apr 13 at 7:56

Widawensen

4,81531447

answered Apr 13 at 7:25

DGRDGR

163

$endgroup$

add a comment | 

$begingroup$

OK, so Iv'e found a solution!

beginalign*\
Tv &= lambda v vert *T^-1 \
T^-1Tv &= T^-1lambda v \
v &= lambda T^-1v vert *lambda^-1 & text(Invertible transformation, $lambdaneq 0$) \
lambda^-1v &= T^-1v
endalign*

For $T^-1$, eigenvalue $lambda^-1$, eigenvector is $v$ (same as eigenvector of $T$).

Hope this helps!

share|cite|improve this answer

edited Apr 13 at 7:56

Widawensen

4,81531447

answered Apr 13 at 7:25

DGRDGR

163

$endgroup$

share|cite|improve this answer

edited Apr 13 at 7:56

Widawensen

4,81531447

answered Apr 13 at 7:25

DGRDGR

163

edited Apr 13 at 7:56

Widawensen

4,81531447

edited Apr 13 at 7:56

Widawensen

4,81531447

answered Apr 13 at 7:25

DGRDGR

163

answered Apr 13 at 7:25

DGRDGR

163