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- Thread starter Adel Makram
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- #2

blue_leaf77

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- #3

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U and A are square matrices of the same rank, so UA is a square matrix too and it should be invertible. But even in this case, how to find A and its inverse from UA? In other words, can we decompose UA into U and A?

- #4

blue_leaf77

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I will give you a hint, what is the transpose conjugate of ##UA##?can we decompose UA into U and A?

- #5

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What if we find that UA is invertiabe, can we decompose it?Being square does not guarantee that it's invertible.

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blue_leaf77

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I appreciate your contribution to answer but frankly I have no clear idea about your words. My question is clear from the beginning and still I have no answer on it, will I be able to decompose UA into U and A where is U is a unique unitary and A a unique diagonal or no?

- #8

fresh_42

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@blue_leaf77's question in post #4 contains the answer. What can you say about ##(UA)^\dagger = \overline{(UA)}^\tau## and what happens, if you multiply this by ##UA##?I appreciate your contribution to answer but frankly I have no clear idea about your words. My question is clear from the beginning and still I have no answer on it, will I be able to decompose UA into U and A where is U is a unique unitary and A a unique diagonal or no?

- #9

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We will get ##A^2## because ##U^TU=1##, right@blue_leaf77's question in post #4 contains the answer. What can you say about ##(UA)^\dagger = \overline{(UA)}^\tau## and what happens, if you multiply this by ##UA##?

- #10

fresh_42

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If you interpret ##A^2=\overline{A}A##, then yes. You haven't said that ##A## is a real diagonal matrix, so ##\overline{A} \neq A## in general. This still doesn't give you a unique description of the diagonal elements, but some additional information. Maybe this can be used in ##1=(UA)(UA)^{-1}##.We will get ##A^2## because ##U^TU=1##, right

- #11

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But why is A, if it is real diagonal, not unique? If I get AIf you interpret ##A^2=\overline{A}A##, then yes. You haven't said that ##A## is a real diagonal matrix, so ##\overline{A} \neq A## in general. This still doesn't give you a unique description of the diagonal elements, but some additional information. Maybe this can be used in ##1=(UA)(UA)^{-1}##.

- #12

fresh_42

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Again, you have only said ##A=diag(a_1,\ldots,a_n)##, so ##a_i \in \mathbb{C}## and ##(UA)^\dagger \cdot (UA)= \overline{A}\cdot A## is all you can conclude. Esp. this gives you ##n## equations ##c_j=\overline{a}_j \cdot a_j## which cannot be solved uniquely without additional information. Even in the real case there are two solutions for each ##j\, : \,\pm \, a_j##But why is A, if it is real diagonal, not unique? If I get A^{2}then each diagonal element in A is ##\sqrt {A^2}##.

- #13

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So in special case where all elements of A are real and positive, then no additional information is required and A is solved.Again, you have only said ##A=diag(a_1,\ldots,a_n)##, so ##a_i \in \mathbb{C}## and ##(UA)^\dagger \cdot (UA)= \overline{A}\cdot A## is all you can conclude. Esp. this gives you ##n## equations ##c_j=\overline{a}_j \cdot a_j## which cannot be solved uniquely without additional information. Even in the real case there are two solutions for each ##j\, : \,\pm \, a_j##

- #14

fresh_42

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Yes.So in special case where all elements of A are real and positive, then no additional information is required and A is solved.

If all diagonal elements of ##A=diag(a_j)## are positive real numbers, then you can compute ##\overline{(UA)}^\tau## if you know ##UA##, then multiply both to ##\overline{(UA)}^\tau\cdot (UA)=\overline{A}^\tau \cdot A=A \cdot A = diag(a_j^2)## which has to be diagonal, if the given conditions for ##U## and ##A## are correct. Thus you have ##n## equations ##c_j :=\left(\overline{(UA)}^\tau(UA) \right)_{jj}=a_j^2## which determine ##A## and with it ##A^{-1}## and ##U=(UA)A^{-1}##.

- #15

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Yes.

If all diagonal elements of ##A=diag(a_j)## are positive real numbers, then you can compute ##\overline{(UA)}^\tau## if you know ##UA##, then multiply both to ##\overline{(UA)}^\tau\cdot (UA)=\overline{A}^\tau \cdot A=A \cdot A = diag(a_j^2)## which has to be diagonal, if the given conditions for ##U## and ##A## are correct. Thus you have ##n## equations ##c_j :=\left(\overline{(UA)}^\tau(UA) \right)_{jj}=a_j^2## which determine ##A## and with it ##A^{-1}## and ##U=(UA)A^{-1}##.

Or more generally: https://en.wikipedia.org/wiki/Polar_decomposition

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