矩阵A的列空间和A的转置的零空间之间的关系

假设\(A\in\mathbb{R}^{m\times n}, m\ge n.\) 下面将证明以下结论。

1. \(\mathcal{R}(A) = \mathcal{N}(A^T)^{\perp}\)
2. \(\mathcal{N}(A^T) \oplus \mathcal{R}(A) = \mathbb{R}^m\)
3. \(\mathcal{N}(A)\oplus \mathcal{R}(A^T) = \mathbb{R}^n\)

证明：

1. \(“\mathcal{R}(A) \subset \mathcal{N}(A^T)^{\perp}”\):

   对任意的 \(\pmb{x}\in \mathcal{R}(A),\) 存在 \(\pmb{y}\), 使得 \(A \pmb{y} = \pmb{x}.\)
   对任意的\(\pmb{z} \in \mathcal{N}(A^{T})\), 有\(\pmb{x}^T \pmb{z} = \pmb{y}^T A^T \pmb{z} = 0.\)
   故 \(\mathcal{R}(A) \subset \mathcal{N}(A^T)^{\perp}\)

   \(“\mathcal{R}(A)^{\perp} \subset \mathcal{N}(A^T)”\):

   对任意的\(\pmb{x} \in \mathcal{R}(A)^{\perp}, \pmb{x} \perp \mathcal{R}(A),\) 即 \(\pmb{y}^T A^T \pmb{x}=0, \forall \pmb{y}.\) 从而有\(A^T \pmb{x} = 0,\) 因此\(\pmb{x} \in \mathcal{N}(A^T).\) 故 \(\mathcal{R}(A)^{\perp} \subset \mathcal{N}(A^T).\)

   因为 \(\mathcal{R}(A) \oplus \mathcal{R}(A)^{\perp} = \mathbb{R}^m, \mathcal{N}(A^T)\oplus \mathcal{N}(A^T)^{\perp} = \mathbb{R}^m,\) 所以有

   \[ \mathcal{R}(A) = \mathcal{N}(A^T)^{\perp}. \]

   \(\text{同理也有} \mathcal{R}(A)^{\perp} = \mathcal{N}(A^T).\)

2. 由1，

\[\mathcal{N}(A^T) \oplus \mathcal{R}(A) = \mathcal{N}(A^T) \oplus \mathcal{N}(A^T)^{\perp} = \mathbb{R}^{m}. \]

3. 由1，\(\mathcal{R}(A^T) = \mathcal{N}(A)^{\perp}\), 从而
   \[ \mathcal{N}(A)\oplus \mathcal{R}(A^T) = \mathcal{N}(A)\oplus \mathcal{N}(A)^{\perp} = \mathbb{R}^{n}. \]