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Let $ A, B $ positively definite matrices, and $ Q $ a unitary matrix. Prove that if $ A = BQ $, then $ A = B $.

The idea I've had - show that all eigenvalues of $ Q $ equal to $ 1 $, and therefore $ Q = I_{n} $, but I'm not sure this direction is even correct.

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Let's see... we want that $\mathbf B\mathbf Q=(\mathbf B\mathbf Q)^\ast$, no? We then have $\mathbf B\mathbf Q=\mathbf Q^\ast\mathbf B$, or $\mathbf Q\mathbf B\mathbf Q=\mathbf B$. Then... – J. M. Nov 13 '11 at 22:19
up vote 2 down vote accepted

According to Wikipedia, "there is no agreement in the literature on the proper definition of positive-definite for non-Hermitian matrices.". If we merely require $\Re(x^*Ax)\gt0$, your statement doesn't hold, since $B=I$ and $A=Q$ a two-dimensional rotation matrix with small rotation angle would be a counterexample. So I'll assume that for $M$ to be positive-definite you're requiring $x^*Mx$ to be always real and positive, which implies that $M$ is Hermitian.

So $A$ and $B$ are Hermitian. As a unitary matrix, $Q$ is diagonalized by a unitary matrix $U$. Transforming $A$ and $B$ by $U$ still leaves them Hermitian, and the transformed matrices satisfy $A'=B'Q'$, with $Q'$ diagonal and unitary. Since $A'$ and $B'$ are positive definite, their diagonal elements are positive. But their ratios are the diagonal elements of $Q'$, so $Q'$, and hence $Q$, has to be the unit matrix.

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