Let $U$ and $V$ be unitary matrices and $A$ a positive definite matrix , for which $AU = VA$. Show that... a) Show that $UA  = AV$.
b) Show that $VA^2  = A^2V$.
c) Show that $VA  = AV$.
d) Show that $U = V$.
I did a, b and d, but I'm having trouble with c. I would appreciate some help.
a) $AU  = VA => (AU)^*  = (VA)^* => U^*A^* = A^*V^*$, since A is positive definite $A = A^*$, so $U^*A = AV^* => UU^*AV = UAV^*V$, since $U$ and $V$ are unitary $ UU^* = I$ and $V^*V = I$ so $AV = UA$.
b) $AU  = VA => AUA = VAA$, since $UA = AV => AAV = VAA => A^2V = VA^2$ .
d) $AU  = VA$ and $VA  = AV$ so $AU = AV$. Since A is positive definite it has an inverse, so $A^{-1}AU = A^{-1}AV => U = V$.
 A: For part c, start with part b.
$$
VA^2 = A^2V \implies VA^2V^* = A^2.
$$
Now, by the uniqueness of the positive (semi-) definite square root, we have
$$
VAV^* = A \implies VA = AV.
$$
A: I presume that this is home-work and that you are supposed to use elementary arguments, so here is a hint for proceeding:
Let  $x$ be any eigenvector of $A^2$, i.e. $A^2x=\lambda^2 x$ for some $\lambda>0$. Show that:
$$ A^2 x=\lambda^2 x \Rightarrow (A+\lambda I)(A-\lambda I)x = 0 \Rightarrow (A-\lambda I)x = 0 \Rightarrow A x =\lambda x.$$
Now use standard properties of symmetric matrices to show that (b) implies (c).
A: Here is another way that requires no square root trickery.
It is straightforward to check (using a)) that $A^k U = V A^k$ for $k=1,2,...$.
Let $p$ be the characteristic polynomial of $A$, then $p(0) \neq 0$ and if
$p(x) = p_0+p_1x+...+x^n$ then we have
$0 = p_0 I + p_1A+...+A^n $, so $I = -{1 \over p_0} (p_1A+...+A^n )$.
Then $V = -{1 \over p_0} (p_1VA+...+VA^n ) = -{1 \over p_0} (p_1AU+...+A^nU ) = U$.
b) & c) follow from this.
