Re: [線代] 正交矩陣多項式證明
k-2
ψ_(k+1) - xψ_k = -a_kψ_k + -b_kψ_(k-1) + Σc_iψ_i
i=0
0 <= i <= k-2：
-(xψ_k, ψ_i) = c_i(ψ_i, ψ_i) = -(ψ_k, xψ_i) = 0
=> c_i = 0 for 0 <= i <= k-2
-(xψ_k, ψ_k) = -a_k(ψ_k, ψ_k)
=> a_k = (xψ_k, ψ_k)/(ψ_k, ψ_k)
-(xψ_k, ψ_(k-1)) = -b_k(ψ_(k-1), ψ_(k-1)) = -(ψ_k, xψ_(k-1))
= -(ψ_k, ψ_k)
=> b_k = (ψ_k, ψ_k)/(ψ_(k-1), ψ_(k-1))
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