let $V$ be a finite dimensional inner product space. Let $T$ be a linear operator on $V$.

Prove that there exists an invertible linear operator $U$ such that

$U^{-1}TT^*U = T^*T $ where $T^*$ is the adjoint of $T$.

Of course we know that $TT^*$ itself is self adjoint, but what else should I do?

  • $\begingroup$ @PaulSinclair thanks, i corrected the mistake now $\endgroup$ – Phantom Nov 18 '15 at 2:59

Let $A$ be the matrix of $T$. Then $A=U\Sigma V^*$, where $\Sigma$ is diagonal and $U$, $V$ unitary (SVD decomposition), so $A^*A=V\Sigma^*\Sigma V^*$ and $AA^*=U\Sigma\Sigma^*U^*$. But $\Sigma$ is diagonal, so $\Sigma\Sigma^*=\Sigma^*\Sigma$ and thus $V^*A^*AV=U^*AA^*U$, which gives $AA^*=(VU^*)^{-1}A^*A(VU^*)$.


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