rank(A)=rank(A^TA)

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本文簡要證明命題$rank(\mathbf{AA^T})=rank(\mathbf{A^TA})=rank(\mathbf{A})$,
此證明分爲兩步來完成.

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

• 首先,$\mathbf{C}=\mathbf{AA^T}$.
  那麼我們可以看到.$\mathbf{C^T}=rank(\mathbf{A^TA})$.

• 根據矩陣轉置不改變矩陣的秩可得.
  $rank(\mathbf{AA^T})=rank(\mathbf{A^TA})$

接下來證明 $rank(\mathbf{A^TA})=rank(\mathbf{A})$.

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

本輪的證明分爲兩步走, 先看第一步.

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

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

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

再來看第二步.

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

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

• At last, we get $rank(A)=rank(A^TA)$