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.
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.
(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).
$Q$ 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.