Mixing
Transpose of $Y ={A}^{*} operatorname{diag} left( b right) {A} $
$begingroup$
What would the tranpose of $Y$ look like
$$Y ={A}^{*} operatorname{diag} left( b right) {A} $$
where $*$ is the conjugate tranpose, i.e., hermitian.
transpose
asked Jan 17 at 10:34
abina shrabina shr
698
$endgroup$
add a comment |
$begingroup$
What would the tranpose of $Y$ look like
$$Y ={A}^{*} operatorname{diag} left( b right) {A} $$
where $*$ is the conjugate tranpose, i.e., hermitian.
transpose
asked Jan 17 at 10:34
abina shrabina shr
698
$endgroup$
add a comment |
$begingroup$
What would the tranpose of $Y$ look like
$$Y ={A}^{*} operatorname{diag} left( b right) {A} $$
where $*$ is the conjugate tranpose, i.e., hermitian.
transpose
asked Jan 17 at 10:34
abina shrabina shr
698
$endgroup$
What would the tranpose of $Y$ look like
$$Y ={A}^{*} operatorname{diag} left( b right) {A} $$
where $*$ is the conjugate tranpose, i.e., hermitian.
transpose
transpose
asked Jan 17 at 10:34
abina shrabina shr
698
asked Jan 17 at 10:34
abina shrabina shr
698
asked Jan 17 at 10:34
abina shrabina shr
698
asked Jan 17 at 10:34
abina shrabina shr
698
asked Jan 17 at 10:34
abina shrabina shr
698
698
add a comment |
add a comment |
1 Answer
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$begingroup$
We have $operatorname{diag}(b)^*=operatorname{diag}(overline{b})$, hence
$Y^*=A^*operatorname{diag}(overline{b})A^{**}=A^*operatorname{diag}(overline{b})A$.
answered Jan 17 at 10:38
FredFred
46.9k1848
$endgroup$
$begingroup$
Is $Y^T=A^Toperatorname{diag}(overline{b})(A^{*})^T=A^Toperatorname{diag}(overline{b})conj(A)$?
$endgroup$
– abina shr
Jan 17 at 11:27
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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active
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active
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votes
$begingroup$
We have $operatorname{diag}(b)^*=operatorname{diag}(overline{b})$, hence
$Y^*=A^*operatorname{diag}(overline{b})A^{**}=A^*operatorname{diag}(overline{b})A$.
answered Jan 17 at 10:38
FredFred
46.9k1848
$endgroup$
$begingroup$
Is $Y^T=A^Toperatorname{diag}(overline{b})(A^{*})^T=A^Toperatorname{diag}(overline{b})conj(A)$?
$endgroup$
– abina shr
Jan 17 at 11:27
add a comment |
$begingroup$
We have $operatorname{diag}(b)^*=operatorname{diag}(overline{b})$, hence
$Y^*=A^*operatorname{diag}(overline{b})A^{**}=A^*operatorname{diag}(overline{b})A$.
answered Jan 17 at 10:38
FredFred
46.9k1848
$endgroup$
$begingroup$
Is $Y^T=A^Toperatorname{diag}(overline{b})(A^{*})^T=A^Toperatorname{diag}(overline{b})conj(A)$?
$endgroup$
– abina shr
Jan 17 at 11:27
add a comment |
$begingroup$
We have $operatorname{diag}(b)^*=operatorname{diag}(overline{b})$, hence
$Y^*=A^*operatorname{diag}(overline{b})A^{**}=A^*operatorname{diag}(overline{b})A$.
answered Jan 17 at 10:38
FredFred
46.9k1848
$endgroup$
We have $operatorname{diag}(b)^*=operatorname{diag}(overline{b})$, hence
$Y^*=A^*operatorname{diag}(overline{b})A^{**}=A^*operatorname{diag}(overline{b})A$.
answered Jan 17 at 10:38
FredFred
46.9k1848
answered Jan 17 at 10:38
FredFred
46.9k1848
answered Jan 17 at 10:38
FredFred
46.9k1848
answered Jan 17 at 10:38
FredFred
46.9k1848
46.9k1848
$begingroup$
Is $Y^T=A^Toperatorname{diag}(overline{b})(A^{*})^T=A^Toperatorname{diag}(overline{b})conj(A)$?
$endgroup$
– abina shr
Jan 17 at 11:27
add a comment |
$begingroup$
Is $Y^T=A^Toperatorname{diag}(overline{b})(A^{*})^T=A^Toperatorname{diag}(overline{b})conj(A)$?
$endgroup$
– abina shr
Jan 17 at 11:27
$begingroup$
Is $Y^T=A^Toperatorname{diag}(overline{b})(A^{*})^T=A^Toperatorname{diag}(overline{b})conj(A)$?
$endgroup$
– abina shr
Jan 17 at 11:27
$begingroup$
Is $Y^T=A^Toperatorname{diag}(overline{b})(A^{*})^T=A^Toperatorname{diag}(overline{b})conj(A)$?
$endgroup$
– abina shr
Jan 17 at 11:27
add a comment |
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