Can we retrieve the inverse of matrix A in this example?

[Adel Makram]

Suppose we have a product formed by a multiplication of a unitary matrix U and a diagonal matrix A, can we retrieve the inverse of A without knowing either U or A?

 

[blue_leaf77]

If you have UA, check if it's invertible. If it's not, there is no point trying to find the inverse of A because it does not exist.

 

[Adel Makram]

If you have UA, check if it's invertible. If it's not, there is no point trying to find the inverse of A because it does not exist.

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]

Being square does not guarantee that it's invertible.

can we decompose UA into U and A?

I will give you a hint, what is the transpose conjugate of $UA$?

 

[Adel Makram]

Being square does not guarantee that it's invertible.

What if we find that UA is invertiabe, can we decompose it?

 

[blue_leaf77]

Probably I should also emphasized that $A$ is always unique but not necessarily diagonal. Which means, if you sought for a diagonal $A$ it may be to no avail.

 

[Adel Makram]

Probably I should also emphasized that $A$ is always unique but not necessarily diagonal. Which means, if you sought for a diagonal $A$ it may be to no avail.

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?

 

[fresh_42]

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'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]

@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 $UA$?

We will get $A^2$ because $U^TU=1$, right

 

[fresh_42]

We will get $A^2$ because $U^TU=1$, right

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]

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 $1=(UA)(UA)^{-1}$.

But why is A, if it is real diagonal, not unique? If I get A² then each diagonal element in A is $\sqrt {A^2}$.

 

[fresh_42]

But why is A, if it is real diagonal, not unique? If I get A² then each diagonal element in A is $\sqrt {A^2}$.

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]

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 $j\, : \,\pm \, a_j$

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]

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

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 $U=(UA)A^{-1}$.

 

[micromass]

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