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Adel Makram

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In summary: 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

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Adel Makram

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blue_leaf77

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Adel Makram

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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?blue_leaf77 said:

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blue_leaf77

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

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Adel Makram

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What if we find that UA is invertiabe, can we decompose it?blue_leaf77 said:Being square does not guarantee that it's invertible.

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blue_leaf77

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Adel Makram

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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?blue_leaf77 said:

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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##?Adel Makram said: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?

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Adel Makram

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We will get ##A^2## because ##U^TU=1##, rightfresh_42 said:@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##?

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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}##.Adel Makram said:We will get ##A^2## because ##U^TU=1##, right

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Adel Makram

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But why is A, if it is real diagonal, not unique? If I get Afresh_42 said: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}##.

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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##Adel Makram said: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}##.

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Adel Makram

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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.fresh_42 said: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##

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Yes.Adel Makram said: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}##.

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fresh_42 said: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

No, not all matrices have an inverse. For a matrix to have an inverse, it must be a square matrix and its determinant must not be zero.

To find the inverse of a matrix, we can use various methods such as Gaussian elimination, Cramer's rule, or the adjugate matrix method. The method used depends on the size and complexity of the matrix.

The inverse of a matrix is important in many mathematical and scientific applications. It allows us to solve systems of linear equations, find the solution to a matrix equation, and perform operations such as division on matrices.

Yes, we can retrieve the inverse of any square matrix as long as it is non-singular (its determinant is not zero). However, the process may become more complex and computationally intensive for larger matrices.

Yes, the inverse of a matrix is unique. This means that a matrix cannot have more than one inverse. If a matrix has an inverse, it can only have one unique solution.

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