rank(A)=rank(A^TA)

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本文简要证明命题$rank(\mathbf{A}{\mathbf{A}}^{\mathbf{T}})=rank({\mathbf{A}}^{\mathbf{T}}\mathbf{A})=rank(\mathbf{A})$,
此证明分为两步来完成.

$rank(\mathbf{A}{\mathbf{A}}^{\mathbf{T}})=rank({\mathbf{A}}^{\mathbf{T}}\mathbf{A})$

• 首先,$\mathbf{C}=\mathbf{A}{\mathbf{A}}^{\mathbf{T}}$.
  那么我们可以看到.${\mathbf{C}}^{\mathbf{T}}=rank({\mathbf{A}}^{\mathbf{T}}\mathbf{A})$.

• 根据矩阵转置不改变矩阵的秩可得.
  $rank(\mathbf{A}{\mathbf{A}}^{\mathbf{T}})=rank({\mathbf{A}}^{\mathbf{T}}\mathbf{A})$

接下来证明 $rank({\mathbf{A}}^{\mathbf{T}}\mathbf{A})=rank(\mathbf{A})$.

$rank({\mathbf{A}}^{\mathbf{T}}\mathbf{A})=rank(\mathbf{A})$

本轮的证明分为两步走, 先看第一步.

• Nullspace of $A$ included by nullspace of $A^{T}A$.
  $$\forall \mathbf{x},A\mathbf{x}=0\Rightarrow A^{T}A\mathbf{x}=0$$
  which means that $Nullspace(A)\subseteq Nullspace(A^{T}A)$

• Nullspace of $A^{T}A$ include by nullspace of $A$
  $$\forall \mathbf{x},A^{T}A\mathbf{x}=0\Rightarrow {\mathbf{x}}^{\mathbf{T}}{A}^{T}A\mathbf{x}=0\Rightarrow A\mathbf{x}=0$$
  which means that $Nullspace(A^{T}A)\subseteq Nullspace(A)$.

• Finally, we get $Nullspace(A)=Nullspace(A{A}^T)$.

再来看第二步.

• Assuming that $A$ is a $m\times n$ matrix and we can get that $A^{T}A$
  is a $n\times n$ matrix.

• Now, we have reached the final step.
  $$rank(A)+rank(Nullspace(A))=n$$
  $$rank(A^{T}A)+rank(Nullspace(A^{T}A))=n$$

• At last, we get $rank(A)=rank(A^{T}A)$