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Let $A, B$ be positive-definite matrices and $Q$ a unitary matrix, furthermore suppose $A=BQ$.

Prove or disprove: $A=B$.

I'm having a hard time figuring out where to begin. Thanks.

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I believe the problem is ill-posed. You certainly can't disprove $A=B$, since the identity matrix is a unitary matrix, and you also can't prove it, since you can choose $Q$ arbitrarily close to the identity, so that it won't destroy the positive-definiteness of $B$, and so you can't prove $A=B$, either. Are you sure you've listed all the premises? Are $A$ and $B$ real, complex, symmetric, Hermitian, ...? –  joriki Mar 22 '11 at 14:59

2 Answers 2

up vote 2 down vote accepted

(This is basically the same as user8268's proof, slightly rephrased/polished).

Let $\lambda, v$ be an eigenvalue-eigenvector of $Q$, then $Q v= \lambda v$. Then

$ \displaystyle A = B Q \Rightarrow v^* A v = v^* B Q v = \lambda \; v^* B v \Rightarrow \lambda = \frac{v^* A v}{v^* B v}$

But $A$ and $B$ are positive definite, hence both numerator and denominator are real and positive. Further, because eigenvalues of an orthogonal matrix have modulus one, we conclude that $\lambda =1$.

Then $Q$ is an orthogonal matrix with all its eigenvalues equal to one. Then, it must be the identity matrix (a quick way to see this is by its diagnolization; recall than an orthogonal matrix is normal, and hence diagonalizable).

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well, while polishing you should complex-conjugate your $v^T$'s :) –  user8268 Mar 22 '11 at 20:28

$Q$ must be $1$ (and hence $A=B$). If not, let $v$ be an eigenvector of $Q$, $Qv=\lambda v$, with $\lambda\neq1$. Then $\lambda$ is not a positive real. We have $(v,Av)=(v,BQv)=\lambda(v,Bv)$, hence $\lambda=(v,Av)/(v,Bv)$, which is a positive real.

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