In this question $\|A\|\_p$ is the normalized $p$-th Schatten norm which is defined to be $\left(\mathbb E\_{i} \lambda\_i^p\right)^{1/p}$, where $\lambda\_i$ are singular values of matrix $A$.


Suppose that $A, B\in M\_n(\mathbb C)$ are matrices with operator norm at most 1. Suppose that $\|AB-I\|\_p < \varepsilon$. Can you prove that there is a unitary matrix $U$ such that $\|A-U\|\_p < \varepsilon$?


I already know proof for this when $p=1$ or $2$ or $\infty$, but not for any other $p$.



On closest unitary matrix
It turns out the the previous answer had the right ingredients, but in the wrong combination. Here is a cleaner proof.


**Notation:** Let $s\_j(X)$ denote the $j$-th singular value of a matrix $X$ (we assume that singular values are arranged in decreasing order). Similarly, let $\lambda\_j(X)$ denote the $j$-th eigenvalue of a Hermitian matrix $X$. Let $S(X)$ denote the diagonal matrix of singular values of $X$.



> 
> **Lemma 1:** If $B \in M\_n(\mathbb{C})$ is a contraction, then for any $A \in M\_n(\mathbb{C})$ we have
>  \begin{equation\*}
>  s\_j(A) \ge s\_j(AB),\qquad 1\le j \le n.
> \end{equation\*}
> 
> 
> 


*Proof:* Since $B$ is a contraction, we have
\begin{eqnarray\*}
 I &\ge& BB^\*\\
 AA^\* &\ge& ABB^\*A^\*\\
 \lambda\_j(AA^\*) &\ge& \lambda\_j(ABB^\*A^\*)\\
 \lambda\_j^{1/2}(AA^\*) &\ge& \lambda\_j^{1/2}(ABB^\*A^\*)\\
 s\_j(A) &\ge& s\_j(AB).
\end{eqnarray\*}



> 
> **Theorem 2:** Let $A, B \in M\_n(\mathbb{C})$ be contractions. Then,
>  \begin{equation\*}
>  \|I - S(A)\| \le \|I-S(AB)\| \le \|I-AB\|,
> \end{equation\*}
>  for any unitarily invariant norm $\|\cdot\|$.
> 
> 
> 


*Proof:* Using Lemma 1 and that $A$ is a contraction, we have $0 \le 1 - s\_j(A) \le 1 - s\_j(AB)$ for all $j$. Consequently, it follows that
\begin{equation\*}
 \|I-S(A)\| \le \|I-S(AB)\|,
\end{equation\*}
for any unitarily invariant norm. Now using a corollary of Lidkskii's majorization (see e.g., [Theorem IV.3.4 in [Bha97]](https://books.google.com/books?id=lh4BCAAAQBAJ)), it follows that the following inequality
\begin{equation\*}
 \|I-S(AB)\| = \|S(I)-S(AB)\| \le \|I-AB\|,
\end{equation\*}
holds for all unitarily invariant norms.



> 
> **Corollary 3:** If for contractions $A$ and $B$, we have $\|AB-I\| < \epsilon$, then there exists a unitary matrix $U$ such that $\|A-U\| < \epsilon$.
> 
> 
> 


*Proof:* Let $A=UP$ be the polar decomposition of $A$. Then for any unitarily invariant norm $\|\cdot\|$, 
\begin{equation\*}
 \|A-U\|=\|P-I\|=\|S(A)-I\|.
\end{equation\*}
Combining this equality with Theorem 2, the result is immediate.


