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討論串[線代] eigenvalue
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Let T be a linear operator on a finite-dimensional vector space over an. algebraically. closed field F. Let f be a polynomial over F. Prove that c is
(還有34個字)
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|u|^2=|v|^2=α=55,(u'.v)=(v'.u)=β=35, where ' means transpose. A=I+u.v'+v.u'. w1,w2,w3,u,v are a basis. such that wi⊥u, wi⊥v for i=1,2,3. A.wi=wi, i=1,
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[1] [5]. [2] [4] [v^T]. u = [3] v = [3] A = I + [u v][u^T]. [4] [2]. [5] [1]. find all the eigenvalue of A. 請問這題除了暴力把A矩陣算出來. 再慢慢求eigenvalue以外. 還有什麼比較好
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