⑵ |λE-A|=﹙λ-1﹚﹙λ-3﹚﹙λ-4﹚
λ=1 -x+y+z=0 x-2y=0 α1=﹙2/√6,1/√6,1/√6﹚′
λ=3 x+y+z=0 x=0 α2=﹙0,1/√2,-1/√2﹚′
λ=4 2x+y+z=0 x+y =0 α3=﹙1/√3,-1/√3,-1/√3﹚′
T=﹙α1 α2 α3﹚ T′AT=diag﹙1,3,4﹚
这是体力活,⑴ 留给楼主自己辛苦吧!
⑵ |λE-A|=﹙λ-1﹚﹙λ-3﹚﹙λ-4﹚
λ=1 -x+y+z=0 x-2y=0 α1=﹙2/√6,1/√6,1/√6﹚′
λ=3 x+y+z=0 x=0 α2=﹙0,1/√2,-1/√2﹚′
λ=4 2x+y+z=0 x+y =0 α3=﹙1/√3,-1/√3,-1/√3﹚′
T=﹙α1 α2 α3﹚ T′AT=diag﹙1,3,4﹚
这是体力活,⑴ 留给楼主自己辛苦吧!