質問: この行列の固有値と、各固有値に属する固有ベクトルの求め方を教えて下さい。
回答: A= (2...0...-2) (1...0...-1) (1...0...-1) |2-λ...0...-2|=0 |1...-λ...-1| |1...0...-1-λ| λ=0,0,1 λ=0のとき 2x1+0x2-2x3=0 x2=0、x3=1 x1=1 ベクトル X= (1) (0) (1) λ=0 x1+0x2-x3=0 x2=1、x3=0 x1=0 X= (0) (1) (0) λ=1のとき x1-0x2-2x3=0 x1-x2-x3=0 x1+0x2+0x3=0 x2=1 x1-2x3=0 x1-x3=1 x3=1 x1=2 X= (2) (1) (1) P= (1...0...2) (0...1...1) (1...0...1) P−1= (-1...0...2) (-1...1...1) (1...0...-1) D=P−1AP= (0...0...0) (0...0...0) (0...0...1)