Polar correction of a near-unitary matrix

For every vector $v$, $\mathrm{\parallel }Mv\mathrm{\parallel }\ge \mathrm{\parallel }Uv\mathrm{\parallel }-\mathrm{\parallel }(M-U)v\mathrm{\parallel }\ge (1-\epsilon )\mathrm{\parallel }v\mathrm{\parallel }$. Thus $M$ is injective and hence bijective in finite equal dimension. The eigenvalues of $M^{\ast }M$ are ${\sigma }_j^2$, where each singular value obeys $1-\epsilon \le {\sigma }_j\le 1+\epsilon$ by the triangle inequality and [F2].

By [F1], define $(M^{\ast }M{)}^{-1\mathrm{/}2}$ by applying the positive scalar inverse-square-root to its positive eigenvalues. Then $Q^{\ast }Q=I$, so $Q$ is unitary; moreover the singular-value decomposition, obtained by diagonalizing $M^{\ast }M$, gives $\mathrm{\parallel }Q-M\mathrm{\parallel }={\max }_j\mathrm{\mid }1-{\sigma }_j\mathrm{\mid }\le \epsilon$. Consequently $\mathrm{\parallel }Q-U\mathrm{\parallel }\le \mathrm{\parallel }Q-M\mathrm{\parallel }+\mathrm{\parallel }M-U\mathrm{\parallel }\le 2\epsilon$. Finally, $X=M^{\ast }M-I$ has norm at most $2\epsilon +{\epsilon }^2\le 17\mathrm{/}64<1\mathrm{/}2$; the binomial series for $(I+X{)}^{-1\mathrm{/}2}$ converges absolutely in operator norm, and its geometric tail bound gives effective truncations from effective entries of $M$.