1. 4 2 2 A = 2 4 2 , find orthonormal eigenvectors 2 2 4 过程上面有一点不懂得地方 λ1 = 8 [1] X = c1[1] [1] λ2 = λ3 = 2 [ 1] [ 0] X = c2[-1] + c3[ 1] [ 0] [-1] A 为实对称矩阵 相异特徵值对应之特徵向量正交 <X1 , X2> = 0 <X1 , X3> = 0 <X2 , X3> ≠0 利用 Gram-Schmidt 正交化法 X2 , X3 化为正交向量 ψ2 , ψ3 [ 1] <X3.ψ2> 取 ψ2 = X2 = [-1] , ψ3 = X3 - ---------- ψ2 [ 0] <ψ2,ψ2> [ 0] -1 [ 1] 1 [ 1] ψ3 = [ 1] - ---[-1] = ---[ 1] [-1] 2 [ 0] 2 [-2] ^^^^^ -1 1. ---如何算来的 2 当 λ2 = λ3 = 2 [ 1] 1 X = c2[-1] + c3*--- = c2*ψ2 + c3*ψ3 [ 0] 2 又 [1] X1 = [1] , {X1,ψ2,ψ3} 为正交特徵向量 [1] X1 1 [1] 取 U1 = ------ = ----[1] | X1 | √3[1] ψ2 1 [ 1] U2 = ------- = ----[-1] | ψ2 | √2 [ 1] ψ3 1 [ 1] U3 = ------- = ----[ 1] | ψ3 | √6 [-2] {U1,U2,U3} 为归一正交特徵向量 2. 我不懂"为什麽"上面要这样算出 U1,U2,U3 ... 然後另外一个解法是这样 ↓ ↓ [ 1] [ 1] X = c2[ 0] + c3[-2] 老师把箭头的地方圈起来 [-1] [ 1] ↑ ↑ 写这样正交 为什麽这样会正交@@ 3个问题 能请高手解答一下吗>"< --

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