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本文简要证明命题

rank(AAT)=rank(ATA)=rank(A),
此证明分为两步来完成.

rank(AAT)=rank(ATA)

• 首先,

  $\mathbf{C}=\mathbf{A}{\mathbf{A}}^{\mathbf{T}}$

  C=AAT.
  那么我们可以看到.

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

  CT=rank(ATA).

• 根据矩阵转置不改变矩阵的秩可得.

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

  rank(AAT)=rank(ATA)

接下来证明

rank(ATA)=rank(A).

rank(ATA)=rank(A)

本轮的证明分为两步走, 先看第一步.

• Nullspace of

  $A$

  A included by nullspace of

  $A^{T}A$

  ATA.

  $\mathrm{?}\mathbf{x},A\mathbf{x}=0?A^{T}A\mathbf{x}=0$

  ?x,Ax=0?ATAx=0
  which means that

  $Nullspace(A)?Nullspace(A^{T}A)$

  Nullspace(A)?Nullspace(ATA)

• Nullspace of

  $A^{T}A$

  ATA include by nullspace of

  $A$

  A

  $\mathrm{?}\mathbf{x},A^{T}A\mathbf{x}=0?{\mathbf{x}}^{\mathbf{T}}{A}^{T}A\mathbf{x}=0?A\mathbf{x}=0$

  ?x,ATAx=0?xTATAx=0?Ax=0
  which means that

  $Nullspace(A^{T}A)?Nullspace(A)$

  Nullspace(ATA)?Nullspace(A).

• Finally, we get

  $Nullspace(A)=Nullspace(A{A}^T)$

  Nullspace(A)=Nullspace(AAT).

再来看第二步.

• Assuming that

  $A$

  A is a

  $m\times n$

  m×n matrix and we can get that

  $A^{T}A$

  ATA
  is a

  $n\times n$

  n×n matrix.

• Now, we have reached the final step.

  $rank(A)+rank(Nullspace(A))=n$

  rank(A)+rank(Nullspace(A))=n

  $rank(A^{T}A)+rank(Nullspace(A^{T}A))=n$

  rank(ATA)+rank(Nullspace(ATA))=n

• At last, we get

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

  rank(A)=rank(ATA)