P^(-1)AP怎么算?c1=(1/√3,1/√3,1/√3)',c2=(1/√2,-1/√2,0)',c3=(1/√6,1/√6,-2/√6)'令矩阵P=(c1,c2,c3),则P为正交矩阵,且P^-1AP=diag(5,-1,-1).正交变换X=PY,f=5y1^2-y2^2-y3^2.请问P^-1AP=diag(5,-1,-1).是怎么

Question

P^(-1)AP怎么算?
c1 = (1/√3,1/√3,1/√3)',
c2 = (1/√2,-1/√2,0)',
c3 = (1/√6,1/√6,-2/√6)'
令矩阵P = (c1,c2,c3),则P为正交矩阵,
且 P^-1AP = diag(5,-1,-1).
正交变换 X=PY,f = 5y1^2-y2^2-y3^2.
请问 P^-1AP = diag(5,-1,-1).是怎么算出来的啊

Answer 1

这里不是乘出来的
是根据矩阵可对角化定理直接得到的结论
P的列向量分别是特征值5,-1,-1的特征向量(线性无关或正交)
就一定有 P^-1AP = diag(5,-1,-1).