線性代數›Ch5 特徵值與對角化
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21. Rayleigh商、特徵值、Hermitian矩陣
#LA-05-021中Rayleigh商特徵值Hermitian矩陣
  1. Let AA be a Hermitian matrix with eigenvalues λ1≥λ2≥⋯≥λn\lambda_1 \ge \lambda_2 \ge \cdots \ge \lambda_n and orthonormal eigenvectors u1,…,un\mathbf{u}_1, \ldots, \mathbf{u}_n. For any nonzero vector x\mathbf{x} in Rn\mathbb{R}^n, the Rayleigh quotient ρ(x)\rho(\mathbf{x}) is defined by
ρ(x)=⟨Ax,x⟩⟨x,x⟩=xHAxxHx\rho(\mathbf{x}) = \frac{\langle A\mathbf{x}, \mathbf{x}\rangle}{\langle \mathbf{x}, \mathbf{x}\rangle} = \frac{\mathbf{x}^H A \mathbf{x}}{\mathbf{x}^H \mathbf{x}}

(a) If x=c1u1+⋯+cnun\mathbf{x} = c_1\mathbf{u}_1 + \cdots + c_n\mathbf{u}_n, show that

ρ(x)=∣c1∣2λ1+∣c2∣2λ2+⋯+∣cn∣2λn∥c∥2\rho(\mathbf{x}) = \frac{|c_1|^2\lambda_1 + |c_2|^2\lambda_2 + \cdots + |c_n|^2\lambda_n}{\|\mathbf{c}\|^2}

(10%)

(b) Show that λn≤ρ(x)≤λ1\lambda_n \le \rho(\mathbf{x}) \le \lambda_1. (5%)

(c) Show that max⁡x≠0ρ(x)=λ1\max_{\mathbf{x}\neq 0} \rho(\mathbf{x}) = \lambda_1 and min⁡x≠0ρ(x)=λn\min_{\mathbf{x}\neq 0} \rho(\mathbf{x}) = \lambda_n. (5%)

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