关于给定算子的最接近酉算子的证明问题（Linear Algebra Done Right 7F习题27）

Hey there, let's walk through this proof step by step—this is a neat result about finding the unitary operator closest to a given linear operator in the operator norm, straight from Axler's Linear Algebra Done Right.

First, let's recap the given setup to make sure we're on the same page:

• Let $V$ be a nonzero finite-dimensional inner product space over $\mathbb{F}$, where $\mathbb{F}$ is either $\mathbb{R}$ or $\mathbb{C}$.
• Let $T \in \mathcal{L}(V)$ (an operator on $V$) with singular values $s_1, \dots, s_n$.
• We have orthonormal bases ${e_1, \dots, e_n}$ and ${f_1, \dots, f_n}$ of $V$ such that for every $v \in V$:
  $$Tv = s_1\langle v, e_1\rangle f_1 + \dots + s_n\langle v, e_n\rangle f_n$$
• Define the operator $S \in \mathcal{L}(V)$ by:
  $$Sv = \langle v, e_1\rangle f_1 + \dots + \langle v, e_n\rangle f_n$$
• We already know $S$ is unitary, and that $|T-S| = \max{|s_1 - 1|, \dots, |s_n - 1|}$.

Our goal is to show that for any unitary operator $E \in \mathcal{L}(V)$, $|T-E| \geq |T-S|$.

Proof Walkthrough #

Let's break this down using properties of unitary operators, operator norms, and singular value decompositions:

1. Simplify the problem using unitary invariance
   Since $S$ is unitary, the operator norm is unitarily invariant—meaning $|UAU^| = |A|$ for any operator $A$ and unitary $U$. Let's apply this to $T-E$:
   $$|T-E| = |S^(T-E)| = |S^*T - S^*E|$$
   Notice that $S^*E$ is also unitary (the product of two unitary operators is unitary). Let's define:

• $A = S^T$: This is a diagonal operator with respect to the basis ${e_1, \dots, e_n}$, since $S^T e_i = s_i e_i$ (you can verify this by applying $S^$ to $Te_i = s_i f_i$, and recalling $S^ f_i = e_i$ because $S$ is unitary).
• $U = S^*E$: A unitary operator, as mentioned.

Now our problem reduces to showing: for any unitary operator $U$, $|A - U| \geq \max{|s_i - 1|}$, where $A$ is the diagonal operator with entries $s_1, \dots, s_n$.

2. Estimate the norm using basis vectors
   The operator norm is defined as $|B| = \sup_{|v|=1} |Bv|$. For each index $i$, take the basis vector $e_i$ (which has norm 1). Let's compute $|(A - U)e_i|^2$:
   $$
   \begin{align*}
   |(A - U)e_i|^2 &= |Ae_i - Ue_i|^2 \
   &= |s_i e_i - Ue_i|^2 \
   &= |s_i e_i|^2 + |Ue_i|^2 - 2\operatorname{Re}\langle s_i e_i, Ue_i\rangle
   \end{align*}
   $$
   Since $U$ is unitary, $|Ue_i| = |e_i| = 1$, and $|s_i e_i| = s_i$. Substituting these in:
   $$
   |(A - U)e_i|^2 = s_i^2 + 1 - 2s_i \cdot \operatorname{Re}\langle e_i, Ue_i\rangle
   $$

3. Bound the inner product term
   For any unitary operator $U$, the entry $u_{ii} = \langle e_i, Ue_i\rangle$ satisfies $|\operatorname{Re}(u_{ii})| \leq |u_{ii}| \leq 1$ (by Cauchy-Schwarz inequality, since $|e_i| = |Ue_i| = 1$). Therefore:
   $$
   -2s_i \cdot \operatorname{Re}\langle e_i, Ue_i\rangle \geq -2s_i \cdot 1
   $$
   Substituting back into our norm squared expression:
   $$
   |(A - U)e_i|^2 \geq s_i^2 + 1 - 2s_i = (s_i - 1)^2
   $$
   Taking square roots (since norms are non-negative), we get:
   $$
   |(A - U)e_i| \geq |s_i - 1|
   $$

4. Conclude the result
   Since $|A - U|$ is the supremum of $|(A - U)v|$ over all unit vectors $v$, it must be at least as large as $|(A - U)e_i|$ for every $i$. This means:
   $$
   |A - U| \geq \max{|s_1 - 1|, \dots, |s_n - 1|} = |T-S|
   $$
   Translating back to our original operators, this gives us:
   $$
   |T-E| \geq |T-S|
   $$
   Which is exactly what we needed to prove!

备注：内容来源于stack exchange，提问作者Tim