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

1. Dec 25, 2016

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?

2. Dec 25, 2016

### 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.

3. Dec 25, 2016

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. Dec 25, 2016

### blue_leaf77

Being square does not guarantee that it's invertible.
I will give you a hint, what is the transpose conjugate of $UA$?

5. Dec 25, 2016

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

6. Dec 25, 2016

### 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.

7. Dec 25, 2016

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. Dec 25, 2016

### Staff: Mentor

@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 $UA$?

9. Dec 25, 2016

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

10. Dec 25, 2016

### Staff: Mentor

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

11. Dec 25, 2016

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

12. Dec 25, 2016

### Staff: Mentor

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

13. Dec 25, 2016

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

14. Dec 25, 2016

### Staff: Mentor

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

15. Dec 25, 2016

### micromass

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