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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 $p$.

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    $\begingroup$ If $A=UP$ where $U$ is unitary and $P$ is positive definite, then we know that $\|A-U\| \le \|A-V\|$ for any unitary matrix $V$ (the norm above is any unitarily invariant norm). This may help extend the proof to your case, perhaps by using a unitary dilation of $B$... $\endgroup$ – Suvrit May 3 '15 at 14:25
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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]), 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.

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  • $\begingroup$ I've left the old answer because it includes ideas that were critical in helping me understand the problem, so they might be helpful to others too. $\endgroup$ – Suvrit May 9 '15 at 18:05
  • $\begingroup$ OK, this seems to be correct. $\endgroup$ – Omid Hatami May 9 '15 at 21:13
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EDIT The claim below is false (stupid error, as noted in the comments). I will leave the answer below as a potential approach; am checking if the idea can be fixed!


Here is a proof that holds under a slight additional hypothesis, namely, suppose that we also have $\|A^*B^*-I\| < \epsilon$. The conclusion is also stronger than the one requested, as under this additional hypothesis, the claim holds for all unitarily invariant norms, not just Schatten-$p$ norms.

Theorem ([Thm. IX.7.2, Bha97]). Let $A=UP$ be the polar decomposition of $A$. Then, for every unitarily invariant norm $\|\cdot\|$ and every unitary matrix $V$ \begin{equation*} \|A-U\| \le \|A-V\|. \end{equation*}

Applying this theorem to the direct sum $A\oplus (-A^*)$ we have \begin{equation*} \left\|\begin{pmatrix} A & 0\\ 0 & -A^* \end{pmatrix} - \begin{pmatrix} U & 0\\ 0 & -U^* \end{pmatrix}\right\| \le \left\|\begin{pmatrix} A & 0\\ 0 & -A^* \end{pmatrix} - V^*\right\| \end{equation*} for any unitary matrix $V$. Since $B$ is a contraction, we can dilate it to the unitary matrix \begin{equation*} V := \begin{pmatrix} B & (I-BB^*)^{1/2}\\ (I-B^*B)^{1/2} & -B^* \end{pmatrix}. \end{equation*}

Since the norms involved are unitarily invariant, we therefore obtain \begin{equation*} \|(A-U)\oplus(U-A)^*\| \le \left\|\begin{pmatrix} A & 0\\ 0 & -A^* \end{pmatrix}V - I\right\| = \|(AB-I)\oplus (A^*B^*-I)\|. \end{equation*} But notice that $s_j(A-U)=s_j(U^*-A^*)$, for each singular value $s_j$ for $1\le j \le n$. Therefore, we can conclude the following weak-majorization between the singular values: \begin{equation*} s(A-U) \prec_w \max\{s_j(AB-I),s_j(A^*B^*-I)\}_{j=1}^n. \end{equation*} From this under our hypotheses $\|AB-I\|<\epsilon$ and $\|A^*B^*-I\| < \epsilon$, it follows that $\|A-U\| < \epsilon$.

Note: It should be possible to get rid of the extra assumption, but am not sure as of now.

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    $\begingroup$ Could you please explicate in your answer above how the first inequality is obtain, especially why it is true for every unitary matrix $V$, and why the contraction B could be dilated to a unitary matrix? Thank you. $\endgroup$ – Hans May 6 '15 at 7:08
  • $\begingroup$ @Hans: I added the requested details. $\endgroup$ – Suvrit May 6 '15 at 20:59
  • $\begingroup$ I got the construction for my second question by seeking a symmetric solution for the *'s in your first version of $V$ which turns out to be exactly the answer you provided. I really appreciate you adding the details, and the references. $\endgroup$ – Hans May 7 '15 at 0:42
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    $\begingroup$ @Hans: the link that I put in the Lemma contains the entire proof; it is also based on block matrices, and ends up reducing to the case where we can use Lidskii's majorization: $$\lambda^{\downarrow}(X)-\lambda^{\downarrow}(Y) \prec \lambda(X-Y) \prec \lambda^{\downarrow}(X)-\lambda^{\uparrow}(Y)$$ for Hermitian matrices. If you are unable to locate the reference, I can try to supply details of that proof also as a part of my answer. $\endgroup$ – Suvrit May 7 '15 at 1:09
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    $\begingroup$ I have a counterexample. Let A=1 and $B=\cos \theta$. Then \begin{equation*} V = \begin{pmatrix} \cos \theta & \sin \theta\\ \sin\theta & -\cos \theta \end{pmatrix} \end{equation*} Then \begin{equation*} \left\|\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix} \begin{pmatrix} \cos \theta & \sin \theta\\ \sin\theta & -\cos \theta \end{pmatrix} - I\right\| = \left\| \begin{pmatrix} \cos \theta - 1 & \sin \theta\\ -\sin\theta & \cos \theta - 1 \end{pmatrix}\right\| \neq \left\| \begin{pmatrix} \cos \theta - 1 & 0 \\ 0 & \cos \theta - 1 \end{pmatrix}\right\| \end{equation*} $\endgroup$ – Omid Hatami May 8 '15 at 22:20

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