→ Gabino: 是正交補空間 02/01 12:48
→ yupog2003: ∀x ∈N(A) => Ax=0 => A^tAx=A^t0=0 => x∈N(A^tA) 02/01 12:49
→ yupog2003: ∀x ∈N(A^tA) => A^tAx=0 => x^tA^tAx=x^t0=0 => 02/01 12:50
→ yupog2003: (Ax)^t(Ax)=0 => <Ax,Ax>=0 => ||(Ax)||^2=0 => Ax=0 02/01 12:51
→ YuxiWen: 因為ker(AtA)=ker(A) 02/01 12:51
→ yupog2003: =>x ∈N(A) 02/01 12:51
→ yupog2003: 由以上兩式得到ker(A^tA)⊆ker(A)且ker(A)⊆ker(A^tA) 02/01 12:52
→ yupog2003: =>ker(A)=ker(A^tA) 02/01 12:52
→ sandy900113: 懂了 感謝!! 02/01 12:53