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请你仔细解决这道题这对我很很重要：若 $n > 1$ 正整数, $x, y \in \mathbb{...

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سؤال

请你仔细解决这道题这对我很很重要：若 n>1n > 1 正整数, x,y∈Cnx, y \in \mathbb{C}^n, y∗x≠0y^* x \ne 0. 证明 ∥In−xy∗y∗x∥2=∥x∥2∥y∥2∣y∗x∣\left\| I_n - \frac{xy^*}{y^*x} \right\|_2 = \frac{\|x\|_2 \|y\|_2}{|y^*x|}

سؤال

请你仔细解决这道题这对我很很重要：若 n>1n > 1 正整数, x,y∈Cnx, y \in \mathbb{C}^n, y∗x≠0y^* x \ne 0. 证明 ∥In−xy∗y∗x∥2=∥x∥2∥y∥2∣y∗x∣\left\| I_n - \frac{xy^*}{y^*x} \right\|_2 = \frac{\|x\|_2 \|y\|_2}{|y^*x|}

Python

python
def compute(x,y):
    A = np.eye(len(x)) - np.outer(x, y.conj())/(y.conj().T @ x)
    return la.norm(A,2)
for _ in range(5):
    x = np.random.randn(5)+1j*np.random.randn(5)
    y = np.random.randn(5)+1j*np.random.randn(5)
    if abs(y.conj().T @ x)>1e-6:
        break
val = compute(x,y)
ratio = la.norm(x)*la.norm(y)/abs(y.conj().T @ x)
val, ratio

(2.609518459835824, 2.609518459835824)

Python

python
def eigenvalues_AstA(x,y):
    A = np.eye(len(x)) - np.outer(x, y.conj())/(y.conj().T @ x)
    return np.linalg.eigvals(A.conj().T @ A)
eigs = eigenvalues_AstA(x,y)
eigs

array([ 6.80958659e+00+2.28072117e-16j, -4.58559401e-16-1.09160789e-16j,
        1.00000000e+00-5.55111512e-17j,  1.00000000e+00-2.22044605e-16j,
        1.00000000e+00+8.32667268e-17j])

Python

python
def find_vector(x,y):
    n = len(x)
    u = x/la.norm(x)
    v = y/la.norm(y)
    beta = np.vdot(v,u)
    A = np.eye(n) - np.outer(u, v.conj())/beta
    # compute eigenvectors of A^*A
    vals, vecs = la.eig(A.conj().T @ A)
    # find vector with eigenvalue approx s^2
    large_index = np.argmax(np.abs(vals))
    return vals[large_index], vecs[:,large_index]
s2, vec = find_vector(x,y)
s2

(6.809586592223928-3.1115253030788085e-17j)

Python

python
vec_proj_u = np.vdot(x/la.norm(x), vec)
vec_proj_v = np.vdot(y/la.norm(y), vec)
vec_proj_u, vec_proj_v

((2.7755575615628914e-17+2.7755575615628914e-17j),
 (-0.3811171190859335+0.8413667147184378j))

الفكرة

Computing eigenvalues and singular valuesComputing eigenvalues and singular values

For matrix AA, we have the relation A∗A=I−xy∗y∗x−yx∗x∗y+∥x∥2∣y∗x∣2yy∗A^* A = I - \frac{xy^*}{y^*x} - \frac{yx^*}{x^*y} + \frac{\|x\|^2}{|y^*x|^2} yy^*. This results in a rank-2 perturbation of identity. Nontrivial eigenvalues are found on the span of xx and yy, while the orthogonal complement is invariant with eigenvalue 1. To compute the largest singular value, I need to compute the eigenvalues of the 2x2 matrix AA restricted to the span.For matrix AA, we have the relation A∗A=I−xy∗y∗x−yx∗x∗y+∥x∥2∣y∗x∣2yy∗A^* A = I - \frac{xy^*}{y^*x} - \frac{yx^*}{x^*y} + \frac{\|x\|^2}{|y^*x|^2} yy^*. This results in a rank-2 perturbation of identity. Nontrivial eigenvalues are found on the span of xx and yy, while the orthogonal complement is invariant with eigenvalue 1. To compute the largest singular value, I need to compute the eigenvalues of the 2x2 matrix AA restricted to the span.

Clarifying matrix computation

I’m trying to resolve an issue with the matrix UU by re-evaluating its actions on the basis vectors e1e1, e2e2, and considering their properties. I think I initially misunderstood certain terms, like δ\delta and δ∗\delta^\ast, especially since their modulus properties matter. The calculations point towards a complication with how I expect things to behave. It seems I need to adjust my approach to correctly account for how this matrix acts on the basis vectors, particularly with why Ue1U e1 should match existing assumptions.

Resolving matrix UU behavior

I’m working on resolving the behavior of matrix UU acting on the eigenvectors e1e1 and e2e2. I've been verifying my assumptions, especially checking how units like δ\delta and α\alpha interact with each other. I confirmed that Ue1=e1U e1 = e1 and calculated the matrix representation for UU, which seems correct now. After some adjustments, I’m ready to check the eigenvalues of A∗AA^*A, so I’m moving forward with further computations for the singular values and eigenvalue determination.

Thought for 3m 4s

الإجابة

证明思路提纲

  1. 把问题写成求最大奇异值
    记

    A  =  In−xy∗y∗x,∥A∥2=σmax⁡(A)=λmax⁡(A∗A).A \;=\;I_n-\frac{xy^{*}}{y^{*}x},\qquad \|A\|_2=\sigma_{\max}(A)=\sqrt{\lambda_{\max}(A^{*}A)} .

    （这里 σmax⁡\sigma_{\max} 表示最大奇异值，λmax⁡\lambda_{\max} 表示最大特征值。2‑范数等于最大奇异值是教材定义。）

  2. 把 A∗AA^{*}A 写成秩‑2 的显式形式
    设

    U=xy∗y∗x(秩 1),A=I−U.U=\frac{xy^{*}}{y^{*}x}\quad(\text{秩 }1),\qquad A=I-U .

    则

    A∗A=I−U−U∗+U∗U.A^{*}A=I-U-U^{*}+U^{*}U .

    其中

    U∗U=∥x∥22∣y∗x∣2  yy∗(仍为秩 1). U^{*}U =\frac{\|x\|_2^{2}}{|y^{*}x|^{2}}\;yy^{*}\quad(\text{仍为秩 }1).

    因此 A∗AA^{*}A 最多是 秩‑2 的扰动，只需在 span⁡{x,y}\operatorname{span}\{x,y\} 上求特征值。

  3. 选一组正交基并求 AA 的矩阵表达式
    令

    u=x∥x∥2,v=y∥y∥2,α=v∗u=y∗x∥x∥2∥y∥2,ρ=∥x∥2∥y∥2∣y∗x∣=1∣α∣  (≥1).u=\frac{x}{\|x\|_2},\qquad v=\frac{y}{\|y\|_2},\qquad \alpha =v^{*}u=\frac{y^{*}x}{\|x\|_2\|y\|_2},\qquad \rho=\frac{\|x\|_2\|y\|_2}{|y^{*}x|}=\frac1{|\alpha|}\;(\ge1).

    取

    w=v−αu1−∣α∣2w=\frac{v-\alpha u}{\sqrt{1-|\alpha|^{2}}}

    得到正交基 {u,w}\{u,w\}（当 yy 与 xx 共线时，下面的极限形式仍然成立）。

    在此基下

    U=[1β00],A=I−U=[0−β01],β=1−∣α∣2α.U= \begin{bmatrix} 1 & \beta\\[2pt] 0 & 0 \end{bmatrix}, \qquad A=I-U= \begin{bmatrix} 0 & -\beta\\[2pt] 0 & 1 \end{bmatrix}, \qquad \beta=\frac{\sqrt{1-|\alpha|^{2}}}{\alpha}.

  4. 计算 A∗AA^{*}A 的非平凡特征值

    A∗A=[0001+∣β∣2],⟹λ非平凡=1+∣β∣2.A^{*}A= \begin{bmatrix} 0 & 0\\[2pt] 0 & 1+|\beta|^{2} \end{bmatrix}, \quad\Longrightarrow\quad \lambda_{\text{非平凡}}=1+|\beta|^{2}.

    而

    1+∣β∣2=1+1−∣α∣2∣α∣2=1∣α∣2=ρ2.1+|\beta|^{2}=1+\frac{1-|\alpha|^{2}}{|\alpha|^{2}} =\frac{1}{|\alpha|^{2}} =\rho^{2}.

    其余方向：

    • 在 span⁡{x}\operatorname{span}\{x\} 上 A∗AA^{*}A 的特征值为 00；
    • 在与 yy 正交的 (n−2)(n-2)-维子空间上 A∗A=IA^{*}A=I，特征值为 11。

  5. 取平方根得到最大奇异值

    ∥A∥2=λmax⁡(A∗A)=ρ2=ρ=∥x∥2∥y∥2∣y∗x∣. \|A\|_2=\sqrt{\lambda_{\max}(A^{*}A)}=\sqrt{\rho^{2}}=\rho =\frac{\|x\|_2\|y\|_2}{|y^{*}x|}.

  6. 极端情形检查

    • 若 yy 与 xx 共线，则 ∣α∣=1,  ρ=1|\alpha|=1,\;\rho=1，此时
      A=I−xx∗∥x∥22A=I-\frac{xx^{*}}{\|x\|_2^{2}}
      为正交投影的补，显然 ∥A∥2=1\|A\|_2=1。
    • 由柯西‑施瓦兹不等式 ∣y∗x∣≤∥x∥2∥y∥2|y^{*}x|\le\|x\|_2\|y\|_2，故 ρ≥1\rho\ge1，与奇异值分布一致。

因此

∥ In−xy∗y∗x∥2=∥x∥2 ∥y∥2∣y∗x∣  \boxed{\displaystyle \left\|\,I_n-\frac{xy^{*}}{y^{*}x}\right\|_2 =\frac{\|x\|_2\,\|y\|_2}{|y^{*}x|}\;}

证毕。

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