Fredholm二择一定理(矩阵语言)
写在前面
Fredholm定理是泛函分析中的一个定理,本文给出的是用矩阵语言表述的阉割版。
定理内容
设 A∈Fm×n,A \in \mathbb{F}^{m \times n},A∈Fm×n,l:Rn→Rm\mathscr{l}: \R^{n} \to \R^{m}l:Rn→Rm。对于给定的 b∈Rmb \in \R^{m}b∈Rm,方程 Ax=bAx = bAx=b 是否有解?
(1)b∈Im(A)⇔b∈(Ker(AT))⊥;b\in\operatorname{Im}(A)\Leftrightarrow b\in\operatorname({Ker}(A^{T}))^{\perp};b∈Im(A)⇔b∈(Ker(AT))⊥;
(2)x∈Ker(A)⇔x∈(Im(AT))⊥.(2)x\in\operatorname{Ker}(A)\Leftrightarrow x\in\operatorname({Im}(A^{T}))^{\perp}.(2)x∈Ker(A)⇔x∈(Im(AT))⊥.
证明
只证(1).
充分性:
b∈Im(A) ⟺ ∃x∈Rn,Ax=b.b\in\o ...
关于一道高等代数习题的解答
写在前面
本题解法来自帅气的高代助教,感兴趣的同学可以移步他的b站账号:@野生月兔
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