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Real Inner Product Space, Hermitian Operator $T = S^{n}$ for n odd

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1

1

$begingroup$

Let V be a finite dimensional inner product space over $mathbb{R}$, and $T: Vrightarrow V$ hermitian.

Suppose n is an odd positive integer.

Want to show:
$exists S:Vrightarrow V $ such that $T = S^{n}$

Here, I know that T is Hermitian if $T = T^{*}$ its complex conjugate and so, $T = T^{t}$ because it is a real operator.

I am having trouble with this question because I don't know where the criteria that n is odd will come in. Will the reasoning be similar to this? https://math.stackexchange.com/a/89715/651806

linear-algebra inner-product-space

share|cite|improve this question

asked Mar 18 at 20:09

Kevin AllenKevin Allen

154

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Because $x mapsto sqrt[n]{x}$ is bijective. And $T$ can be diagonalised...
  $endgroup$
  – copper.hat
  Mar 18 at 20:26
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If all you have is a real scalar product you don't have a notion of hermitian. You mean $T$ is symmetric? The latter means that $T$ can be diagonalized by a orthogonal transformation and the eigenvalues are real. Then you need to find the $n$-th root of said numbers. If $n$ is odd there's always a real solution.
  $endgroup$
  – lcv
  Mar 18 at 20:28
  

  
  
  
  

add a comment | 

1

1

$begingroup$

Let V be a finite dimensional inner product space over $mathbb{R}$, and $T: Vrightarrow V$ hermitian.

Suppose n is an odd positive integer.

Want to show:
$exists S:Vrightarrow V $ such that $T = S^{n}$

Here, I know that T is Hermitian if $T = T^{*}$ its complex conjugate and so, $T = T^{t}$ because it is a real operator.

I am having trouble with this question because I don't know where the criteria that n is odd will come in. Will the reasoning be similar to this? https://math.stackexchange.com/a/89715/651806

linear-algebra inner-product-space

share|cite|improve this question

asked Mar 18 at 20:09

Kevin AllenKevin Allen

154

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Because $x mapsto sqrt[n]{x}$ is bijective. And $T$ can be diagonalised...
  $endgroup$
  – copper.hat
  Mar 18 at 20:26
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If all you have is a real scalar product you don't have a notion of hermitian. You mean $T$ is symmetric? The latter means that $T$ can be diagonalized by a orthogonal transformation and the eigenvalues are real. Then you need to find the $n$-th root of said numbers. If $n$ is odd there's always a real solution.
  $endgroup$
  – lcv
  Mar 18 at 20:28
  

  
  
  
  

add a comment | 

$begingroup$

Let V be a finite dimensional inner product space over $mathbb{R}$, and $T: Vrightarrow V$ hermitian.

Suppose n is an odd positive integer.

Want to show:
$exists S:Vrightarrow V $ such that $T = S^{n}$

Here, I know that T is Hermitian if $T = T^{*}$ its complex conjugate and so, $T = T^{t}$ because it is a real operator.

I am having trouble with this question because I don't know where the criteria that n is odd will come in. Will the reasoning be similar to this? https://math.stackexchange.com/a/89715/651806

linear-algebra inner-product-space

share|cite|improve this question

asked Mar 18 at 20:09

Kevin AllenKevin Allen

154

$endgroup$

share|cite|improve this question

asked Mar 18 at 20:09

Kevin AllenKevin Allen

154

share|cite|improve this question

asked Mar 18 at 20:09

Kevin AllenKevin Allen

154

asked Mar 18 at 20:09

Kevin AllenKevin Allen

154

asked Mar 18 at 20:09

Kevin AllenKevin Allen

154

1 Answer
1

active

oldest

votes

1

$begingroup$

On a finite-dimensional real inner product space, the notions of hermitian and symmetric for operators coincide; that is,

$T^dagger = T^t = T; tag 1$

since $T$ is symmetric, it may be diagonalized by some orthogonal operator

$O:V to V, tag 2$

$OO^t = O^tO = I, tag 3$

$OTO^t = D = text{diag} (t_1, t_2, ldots, t_m), ; t_i in Bbb R, ; 1 le i le m, tag 4$

where

$m = dim_{Bbb R}V; tag 5$

since each $t_i in Bbb R$, for odd $n in Bbb N$ there exists

$rho_i in Bbb R, ; rho_i^n =t_i; tag 6$

we observe that this assertion fails for even $n$, since negative reals do not have even roots; we set

$R = text{diag} (rho_1, rho_2, ldots, rho_m); tag 7$

then

$R^n = text{diag} (rho_1^n, rho_2^n, ldots, rho_m^n) =
text{diag} (t_1, t_2, ldots, t_m) = D; tag 8$

it follows that

$T = O^tDO = O^tR^nO = (O^tRO)^n, tag 9$

where we have used the general property that matrix conjugation preserves products:

$O^tAOO^TBO = O^tABO, tag{10}$

in affirming (9). We close by simply setting

$S = O^tRO. tag{11}$

share|cite|improve this answer

answered Mar 18 at 20:46

Robert LewisRobert Lewis

48.6k23167

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, and the diagonalisability property follows from the Spectral Theorem, right?
  $endgroup$
  – Kevin Allen
  Mar 18 at 22:41
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @KevinAllen: yes, more or less; see en.wikipedia.org/wiki/Spectral_theorem.
  $endgroup$
  – Robert Lewis
  Mar 19 at 16:40
  

  
  
  
  

add a comment | 

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1

$begingroup$

On a finite-dimensional real inner product space, the notions of hermitian and symmetric for operators coincide; that is,

$T^dagger = T^t = T; tag 1$

since $T$ is symmetric, it may be diagonalized by some orthogonal operator

$O:V to V, tag 2$

$OO^t = O^tO = I, tag 3$

$OTO^t = D = text{diag} (t_1, t_2, ldots, t_m), ; t_i in Bbb R, ; 1 le i le m, tag 4$

where

$m = dim_{Bbb R}V; tag 5$

since each $t_i in Bbb R$, for odd $n in Bbb N$ there exists

$rho_i in Bbb R, ; rho_i^n =t_i; tag 6$

we observe that this assertion fails for even $n$, since negative reals do not have even roots; we set

$R = text{diag} (rho_1, rho_2, ldots, rho_m); tag 7$

then

$R^n = text{diag} (rho_1^n, rho_2^n, ldots, rho_m^n) =
text{diag} (t_1, t_2, ldots, t_m) = D; tag 8$

it follows that

$T = O^tDO = O^tR^nO = (O^tRO)^n, tag 9$

where we have used the general property that matrix conjugation preserves products:

$O^tAOO^TBO = O^tABO, tag{10}$

in affirming (9). We close by simply setting

$S = O^tRO. tag{11}$

share|cite|improve this answer

answered Mar 18 at 20:46

Robert LewisRobert Lewis

48.6k23167

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, and the diagonalisability property follows from the Spectral Theorem, right?
  $endgroup$
  – Kevin Allen
  Mar 18 at 22:41
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @KevinAllen: yes, more or less; see en.wikipedia.org/wiki/Spectral_theorem.
  $endgroup$
  – Robert Lewis
  Mar 19 at 16:40
  

  
  
  
  

add a comment | 

1

$begingroup$

On a finite-dimensional real inner product space, the notions of hermitian and symmetric for operators coincide; that is,

$T^dagger = T^t = T; tag 1$

since $T$ is symmetric, it may be diagonalized by some orthogonal operator

$O:V to V, tag 2$

$OO^t = O^tO = I, tag 3$

$OTO^t = D = text{diag} (t_1, t_2, ldots, t_m), ; t_i in Bbb R, ; 1 le i le m, tag 4$

where

$m = dim_{Bbb R}V; tag 5$

since each $t_i in Bbb R$, for odd $n in Bbb N$ there exists

$rho_i in Bbb R, ; rho_i^n =t_i; tag 6$

we observe that this assertion fails for even $n$, since negative reals do not have even roots; we set

$R = text{diag} (rho_1, rho_2, ldots, rho_m); tag 7$

then

$R^n = text{diag} (rho_1^n, rho_2^n, ldots, rho_m^n) =
text{diag} (t_1, t_2, ldots, t_m) = D; tag 8$

it follows that

$T = O^tDO = O^tR^nO = (O^tRO)^n, tag 9$

where we have used the general property that matrix conjugation preserves products:

$O^tAOO^TBO = O^tABO, tag{10}$

in affirming (9). We close by simply setting

$S = O^tRO. tag{11}$

share|cite|improve this answer

answered Mar 18 at 20:46

Robert LewisRobert Lewis

48.6k23167

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, and the diagonalisability property follows from the Spectral Theorem, right?
  $endgroup$
  – Kevin Allen
  Mar 18 at 22:41
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @KevinAllen: yes, more or less; see en.wikipedia.org/wiki/Spectral_theorem.
  $endgroup$
  – Robert Lewis
  Mar 19 at 16:40
  

  
  
  
  

add a comment | 

$begingroup$

On a finite-dimensional real inner product space, the notions of hermitian and symmetric for operators coincide; that is,

$T^dagger = T^t = T; tag 1$

since $T$ is symmetric, it may be diagonalized by some orthogonal operator

$O:V to V, tag 2$

$OO^t = O^tO = I, tag 3$

$OTO^t = D = text{diag} (t_1, t_2, ldots, t_m), ; t_i in Bbb R, ; 1 le i le m, tag 4$

where

$m = dim_{Bbb R}V; tag 5$

since each $t_i in Bbb R$, for odd $n in Bbb N$ there exists

$rho_i in Bbb R, ; rho_i^n =t_i; tag 6$

we observe that this assertion fails for even $n$, since negative reals do not have even roots; we set

$R = text{diag} (rho_1, rho_2, ldots, rho_m); tag 7$

then

$R^n = text{diag} (rho_1^n, rho_2^n, ldots, rho_m^n) =
text{diag} (t_1, t_2, ldots, t_m) = D; tag 8$

it follows that

$T = O^tDO = O^tR^nO = (O^tRO)^n, tag 9$

where we have used the general property that matrix conjugation preserves products:

$O^tAOO^TBO = O^tABO, tag{10}$

in affirming (9). We close by simply setting

$S = O^tRO. tag{11}$

share|cite|improve this answer

answered Mar 18 at 20:46

Robert LewisRobert Lewis

48.6k23167

$endgroup$

share|cite|improve this answer

answered Mar 18 at 20:46

Robert LewisRobert Lewis

48.6k23167

answered Mar 18 at 20:46

Robert LewisRobert Lewis

48.6k23167

answered Mar 18 at 20:46

Robert LewisRobert Lewis

48.6k23167