Let $U$ and $V$ be unitary matrices and $A$ a positive definite matrix , for which $AU = VA$. Show that...

Question

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 $A^{-1}AU = A^{-1}AV => U = V$.

Answer 1

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).

Answer 2

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 $$ VAV^* = A \implies VA = AV. $$

Answer 3

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.