Given polar decompositions $A=US$ and $A'=U' S'$, prove that $A' A^{-1}$ unitary implies $S=S'$

Question

This is exercise I was given at today's exam, and I couldn't do this. Since I'm curious how it's done I'm asking you for hints.

Given two polar decomposition of invertible matrices

$A = US$

$A'= U'S'$

Where $U, U'$ is a unitary matrix and $S,S'$ is a positive-semidefinite Hermitian matrix.

Prove that if $A'A^{-1}$ is unitary, then $S=S'$.

Answer 1

I use a slightly different notation, let:

$$A_1=U_1S_1$$ $$A_2=U_2S_2$$

and, we have: $U_i^TU_i=U_iU_i^T=I$ and $S_i\succeq 0$. Now let: $A_1A_2^{-1}$ be unitary, then:

$$A_2^{-T}A_1^TA_1A_2^{-1}=I$$ $$\Rightarrow (U_2S_2)^{-T}(U_1S_1)^T(U_1S_1)(U_2S_2)^{-1}=I$$

$$\Rightarrow U_2^{-T}S_2^{-T}S_1^TU_1^TU_1S_1S_2^{-1}U_2^{-1}=I$$

$U_1^TU_1=I$ therefore:

$$\Rightarrow U_2^{-T}S_2^{-T}S_1^TS_1S_2^{-1}U_2^{-1}=I$$

Multiply by $U_2^{T}$ from left and by $U_2$ from right:

$$\Rightarrow S_2^{-T}S_1^TS_1S_2^{-1}=U_2^{T}IU_2=I$$

Now you must have:

$S_2^{-T}S_1^TS_1S_2^{-1}=I$ (and as a result $S_2^{-T}S_1^T=S_2S_1^{-1}$)

Which holds iff $S_1=S_2$. In this case $S_2^{-T}S_1^T=S_2S_1^{-1}=I$

Answer 2

OK, I think this'll work:

First of all note that with

$A = US, \tag{1}$

$A' = U'S', \tag{2}$

and $A$ invertible we may write

$A^{-1} = S^{-1}U^{-1}, \tag{3}$

whence

$A'A^{-1} = U'S'S^{-1}U^{-1}. \tag{4}$

Now we are given that there is a unitary $V$ such that $A'A^{-1} = V$, so that by (4)

$U'S'S^{-1}U^{-1} = V, \tag{5}$

or

$S'S^{-1} = U'^{-1}VU = W \tag{6}$

for some unitary $W$ and then

$S' = WS. \tag{7}$

In light of (7) we have that

$S'^\dagger S' = S^\dagger W^\dagger W S = S^\dagger S, \tag{8}$

using the unitarity of $W$, $W^\dagger W = I$. From (8) we see that

$S'^2 = S'S' = S'^\dagger S' = S^\dagger S = SS = S^2, \tag{9}$

which in turn implies

$S' = \sqrt{S'^\dagger S'} = \sqrt{S^\dagger S} = S, \tag{10}$

since a positive semi-definite Hermitian matrix has a unique positive semi-definite Hermitian square root; see this Wikipedia page for more; note we are using the fact that $B^\dagger B$ is positive semidefinite Hermitian for any matrix $B$, which is in turn easy to see: $(B^\dagger B)^\dagger = B^\dagger (B^\dagger)^\dagger = B^\dagger B$ and $\langle x, B^\dagger Bx \rangle = \langle Bx, Bx, \rangle \ge 0$ for all vectors $x$.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!