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Real Inner Product Space, Hermitian Operator $T = S^n$ for n odd
The Next CEO of Stack OverflowMatrices for which $mathbfA^-1=-mathbfA$Prove non-zero eigenvalues of skew-Hermitian operator are pure imaginaryOrthogonal Operator Infinite Dimensional Inner Product SpaceOperator norm of symmetric Matrix in Hilbert Space with Hermitian Inner Productinner product and hermitian scalar productA conjugate LT in an inner product spaceInverse of hermitian operatorComplex vs. Real Inner-Product SpacesUnitary operator on inner product spaceNormal matrix over real inner product space with real eigenvalues is HermitianIs it possible to recognize when an endomorphism of a finite dimensional vector space is unitary for some choice of inner product?
$begingroup$
Let V be a finite dimensional inner product space over $mathbbR$, 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
asked Mar 18 at 20:09
Kevin AllenKevin Allen
154
$endgroup$
add a comment |
$begingroup$
Let V be a finite dimensional inner product space over $mathbbR$, 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
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 $mathbbR$, 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
asked Mar 18 at 20:09
Kevin AllenKevin Allen
154
$endgroup$
Let V be a finite dimensional inner product space over $mathbbR$, 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
linear-algebra inner-product-space
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
asked Mar 18 at 20:09
Kevin AllenKevin Allen
154
asked Mar 18 at 20:09
Kevin AllenKevin Allen
154
154
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
$begingroup$
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
1
1
$begingroup$
Because $x mapsto sqrt[n]x$ is bijective. And $T$ can be diagonalised...
$endgroup$
– copper.hat
Mar 18 at 20:26
$begingroup$
Because $x mapsto sqrt[n]x$ is bijective. And $T$ can be diagonalised...
$endgroup$
– copper.hat
Mar 18 at 20:26
1
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
$begingroup$
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 Answer
1
active
oldest
votes
$begingroup$
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
$O:V to V, tag 2$
$OO^t = O^tO = I, tag 3$
$OTO^t = D = textdiag (t_1, t_2, ldots, t_m), ; t_i in Bbb R, ; 1 le i le m, tag 4$
where
$m = dim_Bbb RV; tag 5$
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
$R = textdiag (rho_1, rho_2, ldots, rho_m); tag 7$
then
$R^n = textdiag (rho_1^n, rho_2^n, ldots, rho_m^n) =
textdiag (t_1, t_2, ldots, t_m) = D; tag 8$
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:
$O^tAOO^TBO = O^tABO, tag10$
in affirming (9). We close by simply setting
$S = O^tRO. tag11$
answered Mar 18 at 20:46
Robert LewisRobert Lewis
48.5k23167
$endgroup$
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 Answer
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oldest
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votes
active
oldest
votes
$begingroup$
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
$O:V to V, tag 2$
$OO^t = O^tO = I, tag 3$
$OTO^t = D = textdiag (t_1, t_2, ldots, t_m), ; t_i in Bbb R, ; 1 le i le m, tag 4$
where
$m = dim_Bbb RV; tag 5$
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
$R = textdiag (rho_1, rho_2, ldots, rho_m); tag 7$
then
$R^n = textdiag (rho_1^n, rho_2^n, ldots, rho_m^n) =
textdiag (t_1, t_2, ldots, t_m) = D; tag 8$
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:
$O^tAOO^TBO = O^tABO, tag10$
in affirming (9). We close by simply setting
$S = O^tRO. tag11$
answered Mar 18 at 20:46
Robert LewisRobert Lewis
48.5k23167
$endgroup$
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
$O:V to V, tag 2$
$OO^t = O^tO = I, tag 3$
$OTO^t = D = textdiag (t_1, t_2, ldots, t_m), ; t_i in Bbb R, ; 1 le i le m, tag 4$
where
$m = dim_Bbb RV; tag 5$
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
$R = textdiag (rho_1, rho_2, ldots, rho_m); tag 7$
then
$R^n = textdiag (rho_1^n, rho_2^n, ldots, rho_m^n) =
textdiag (t_1, t_2, ldots, t_m) = D; tag 8$
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:
$O^tAOO^TBO = O^tABO, tag10$
in affirming (9). We close by simply setting
$S = O^tRO. tag11$
answered Mar 18 at 20:46
Robert LewisRobert Lewis
48.5k23167
$endgroup$
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
$O:V to V, tag 2$
$OO^t = O^tO = I, tag 3$
$OTO^t = D = textdiag (t_1, t_2, ldots, t_m), ; t_i in Bbb R, ; 1 le i le m, tag 4$
where
$m = dim_Bbb RV; tag 5$
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
$R = textdiag (rho_1, rho_2, ldots, rho_m); tag 7$
then
$R^n = textdiag (rho_1^n, rho_2^n, ldots, rho_m^n) =
textdiag (t_1, t_2, ldots, t_m) = D; tag 8$
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:
$O^tAOO^TBO = O^tABO, tag10$
in affirming (9). We close by simply setting
$S = O^tRO. tag11$
answered Mar 18 at 20:46
Robert LewisRobert Lewis
48.5k23167
$endgroup$
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
$O:V to V, tag 2$
$OO^t = O^tO = I, tag 3$
$OTO^t = D = textdiag (t_1, t_2, ldots, t_m), ; t_i in Bbb R, ; 1 le i le m, tag 4$
where
$m = dim_Bbb RV; tag 5$
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
$R = textdiag (rho_1, rho_2, ldots, rho_m); tag 7$
then
$R^n = textdiag (rho_1^n, rho_2^n, ldots, rho_m^n) =
textdiag (t_1, t_2, ldots, t_m) = D; tag 8$
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:
$O^tAOO^TBO = O^tABO, tag10$
in affirming (9). We close by simply setting
$S = O^tRO. tag11$
answered Mar 18 at 20:46
Robert LewisRobert Lewis
48.5k23167
answered Mar 18 at 20:46
Robert LewisRobert Lewis
48.5k23167
answered Mar 18 at 20:46
Robert LewisRobert Lewis
48.5k23167
answered Mar 18 at 20:46
Robert LewisRobert Lewis
48.5k23167
48.5k23167
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
$begingroup$
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
1
1
$begingroup$
Ah, and the diagonalisability property follows from the Spectral Theorem, right?
$endgroup$
– Kevin Allen
Mar 18 at 22:41
$begingroup$
Ah, and the diagonalisability property follows from the Spectral Theorem, right?
$endgroup$
– Kevin Allen
Mar 18 at 22:41
1
1
$begingroup$
@KevinAllen: yes, more or less; see en.wikipedia.org/wiki/Spectral_theorem.
$endgroup$
– Robert Lewis
Mar 19 at 16:40
$begingroup$
@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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$begingroup$
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