Mixing
Conjugate symmetry of an inner product
$begingroup$
I want to prove the following:
$langle A,B rangle = overline{langle B,A rangle}$
where
$langle A,B rangle := tr(AB^{*}),, and ,, A,B in mathbb{C}^{n times n} $
Note:
The bar denotes the complex conjugate of all entries of a matrix and the "*" the adjoint matrix.
Edit:
$langle A,B rangle = tr(AB^{*}) = tr(overline{(BA^{*})^T}) = tr(overline{BA^{*}}) = overline{tr(BA^*)} = overline{langle B,A rangle}$
The third equality holds since A and B are square matrices.
linear-algebra complex-numbers proof-writing inner-product-space
asked Jan 15 at 19:10
Fo Young Areal LoFo Young Areal Lo
345
$endgroup$
add a comment |
$begingroup$
I want to prove the following:
$langle A,B rangle = overline{langle B,A rangle}$
where
$langle A,B rangle := tr(AB^{*}),, and ,, A,B in mathbb{C}^{n times n} $
Note:
The bar denotes the complex conjugate of all entries of a matrix and the "*" the adjoint matrix.
Edit:
$langle A,B rangle = tr(AB^{*}) = tr(overline{(BA^{*})^T}) = tr(overline{BA^{*}}) = overline{tr(BA^*)} = overline{langle B,A rangle}$
The third equality holds since A and B are square matrices.
linear-algebra complex-numbers proof-writing inner-product-space
asked Jan 15 at 19:10
Fo Young Areal LoFo Young Areal Lo
345
$endgroup$
1
$begingroup$
Write down an explicit formula for $langle A,Brangle $ in terms of $A_{ij}, B_{ij}$, the entries of $A$ and $B$.
$endgroup$
– Christopher A. Wong
Jan 15 at 19:14
$begingroup$
I did it like this: $langle A,B rangle = tr(AB^*) = sum_{j=1}^{n} (AB^*)_{jj}$
$endgroup$
– Fo Young Areal Lo
Jan 15 at 19:37
$begingroup$
There is not need to use the elements of the matrices. Did you see my answer?
$endgroup$
– GReyes
Jan 15 at 19:52
$begingroup$
Yes I did but for me personally it didn't really help me. I did prove it now, I will add an edit to my original answer.
$endgroup$
– Fo Young Areal Lo
Jan 15 at 20:26
add a comment |
$begingroup$
I want to prove the following:
$langle A,B rangle = overline{langle B,A rangle}$
where
$langle A,B rangle := tr(AB^{*}),, and ,, A,B in mathbb{C}^{n times n} $
Note:
The bar denotes the complex conjugate of all entries of a matrix and the "*" the adjoint matrix.
Edit:
$langle A,B rangle = tr(AB^{*}) = tr(overline{(BA^{*})^T}) = tr(overline{BA^{*}}) = overline{tr(BA^*)} = overline{langle B,A rangle}$
The third equality holds since A and B are square matrices.
linear-algebra complex-numbers proof-writing inner-product-space
asked Jan 15 at 19:10
Fo Young Areal LoFo Young Areal Lo
345
$endgroup$
I want to prove the following:
$langle A,B rangle = overline{langle B,A rangle}$
where
$langle A,B rangle := tr(AB^{*}),, and ,, A,B in mathbb{C}^{n times n} $
Note:
The bar denotes the complex conjugate of all entries of a matrix and the "*" the adjoint matrix.
Edit:
$langle A,B rangle = tr(AB^{*}) = tr(overline{(BA^{*})^T}) = tr(overline{BA^{*}}) = overline{tr(BA^*)} = overline{langle B,A rangle}$
The third equality holds since A and B are square matrices.
linear-algebra complex-numbers proof-writing inner-product-space
linear-algebra complex-numbers proof-writing inner-product-space
asked Jan 15 at 19:10
Fo Young Areal LoFo Young Areal Lo
345
asked Jan 15 at 19:10
Fo Young Areal LoFo Young Areal Lo
345
edited Jan 15 at 20:33
Fo Young Areal Lo
asked Jan 15 at 19:10
Fo Young Areal LoFo Young Areal Lo
345
asked Jan 15 at 19:10
Fo Young Areal LoFo Young Areal Lo
345
asked Jan 15 at 19:10
Fo Young Areal LoFo Young Areal Lo
345
345
1
$begingroup$
Write down an explicit formula for $langle A,Brangle $ in terms of $A_{ij}, B_{ij}$, the entries of $A$ and $B$.
$endgroup$
– Christopher A. Wong
Jan 15 at 19:14
$begingroup$
I did it like this: $langle A,B rangle = tr(AB^*) = sum_{j=1}^{n} (AB^*)_{jj}$
$endgroup$
– Fo Young Areal Lo
Jan 15 at 19:37
$begingroup$
There is not need to use the elements of the matrices. Did you see my answer?
$endgroup$
– GReyes
Jan 15 at 19:52
$begingroup$
Yes I did but for me personally it didn't really help me. I did prove it now, I will add an edit to my original answer.
$endgroup$
– Fo Young Areal Lo
Jan 15 at 20:26
add a comment |
1
$begingroup$
Write down an explicit formula for $langle A,Brangle $ in terms of $A_{ij}, B_{ij}$, the entries of $A$ and $B$.
$endgroup$
– Christopher A. Wong
Jan 15 at 19:14
$begingroup$
I did it like this: $langle A,B rangle = tr(AB^*) = sum_{j=1}^{n} (AB^*)_{jj}$
$endgroup$
– Fo Young Areal Lo
Jan 15 at 19:37
$begingroup$
There is not need to use the elements of the matrices. Did you see my answer?
$endgroup$
– GReyes
Jan 15 at 19:52
$begingroup$
Yes I did but for me personally it didn't really help me. I did prove it now, I will add an edit to my original answer.
$endgroup$
– Fo Young Areal Lo
Jan 15 at 20:26
1
1
$begingroup$
Write down an explicit formula for $langle A,Brangle $ in terms of $A_{ij}, B_{ij}$, the entries of $A$ and $B$.
$endgroup$
– Christopher A. Wong
Jan 15 at 19:14
$begingroup$
Write down an explicit formula for $langle A,Brangle $ in terms of $A_{ij}, B_{ij}$, the entries of $A$ and $B$.
$endgroup$
– Christopher A. Wong
Jan 15 at 19:14
$begingroup$
I did it like this: $langle A,B rangle = tr(AB^*) = sum_{j=1}^{n} (AB^*)_{jj}$
$endgroup$
– Fo Young Areal Lo
Jan 15 at 19:37
$begingroup$
I did it like this: $langle A,B rangle = tr(AB^*) = sum_{j=1}^{n} (AB^*)_{jj}$
$endgroup$
– Fo Young Areal Lo
Jan 15 at 19:37
$begingroup$
There is not need to use the elements of the matrices. Did you see my answer?
$endgroup$
– GReyes
Jan 15 at 19:52
$begingroup$
There is not need to use the elements of the matrices. Did you see my answer?
$endgroup$
– GReyes
Jan 15 at 19:52
$begingroup$
Yes I did but for me personally it didn't really help me. I did prove it now, I will add an edit to my original answer.
$endgroup$
– Fo Young Areal Lo
Jan 15 at 20:26
$begingroup$
Yes I did but for me personally it didn't really help me. I did prove it now, I will add an edit to my original answer.
$endgroup$
– Fo Young Areal Lo
Jan 15 at 20:26
add a comment |
1 Answer
1
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oldest
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$begingroup$
Just observe that $(AB^*)^*=B^{**}A^*=BA^*$ and therefore the matrices $AB^*$ and $BA^*$ are complex conjugate of each other. Their diagonal elements are thus conjugate and their sums are as well, that is, $tr(AB^*)=overline{tr(BA^*)}$
answered Jan 15 at 19:24
GReyesGReyes
1,43515
$endgroup$
add a comment |
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active
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votes
$begingroup$
Just observe that $(AB^*)^*=B^{**}A^*=BA^*$ and therefore the matrices $AB^*$ and $BA^*$ are complex conjugate of each other. Their diagonal elements are thus conjugate and their sums are as well, that is, $tr(AB^*)=overline{tr(BA^*)}$
answered Jan 15 at 19:24
GReyesGReyes
1,43515
$endgroup$
add a comment |
$begingroup$
Just observe that $(AB^*)^*=B^{**}A^*=BA^*$ and therefore the matrices $AB^*$ and $BA^*$ are complex conjugate of each other. Their diagonal elements are thus conjugate and their sums are as well, that is, $tr(AB^*)=overline{tr(BA^*)}$
answered Jan 15 at 19:24
GReyesGReyes
1,43515
$endgroup$
add a comment |
$begingroup$
Just observe that $(AB^*)^*=B^{**}A^*=BA^*$ and therefore the matrices $AB^*$ and $BA^*$ are complex conjugate of each other. Their diagonal elements are thus conjugate and their sums are as well, that is, $tr(AB^*)=overline{tr(BA^*)}$
answered Jan 15 at 19:24
GReyesGReyes
1,43515
$endgroup$
Just observe that $(AB^*)^*=B^{**}A^*=BA^*$ and therefore the matrices $AB^*$ and $BA^*$ are complex conjugate of each other. Their diagonal elements are thus conjugate and their sums are as well, that is, $tr(AB^*)=overline{tr(BA^*)}$
answered Jan 15 at 19:24
GReyesGReyes
1,43515
answered Jan 15 at 19:24
GReyesGReyes
1,43515
answered Jan 15 at 19:24
GReyesGReyes
1,43515
answered Jan 15 at 19:24
GReyesGReyes
1,43515
1,43515
add a comment |
add a comment |
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1
$begingroup$
Write down an explicit formula for $langle A,Brangle $ in terms of $A_{ij}, B_{ij}$, the entries of $A$ and $B$.
$endgroup$
– Christopher A. Wong
Jan 15 at 19:14
$begingroup$
I did it like this: $langle A,B rangle = tr(AB^*) = sum_{j=1}^{n} (AB^*)_{jj}$
$endgroup$
– Fo Young Areal Lo
Jan 15 at 19:37
$begingroup$
There is not need to use the elements of the matrices. Did you see my answer?
$endgroup$
– GReyes
Jan 15 at 19:52
$begingroup$
Yes I did but for me personally it didn't really help me. I did prove it now, I will add an edit to my original answer.
$endgroup$
– Fo Young Areal Lo
Jan 15 at 20:26