rank(A)=rank(A^TA)

 2020-06-30 

UTF8gbsn

本文简要证明命题

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

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

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

rank(AAT)=rank(ATA)

  • 首先,

    C=AAT\mathbf{C}=\mathbf{AA^T}

    C=AAT.
    那么我们可以看到.

    CT=rank(ATA)\mathbf{C^T}=rank(\mathbf{A^TA})

    CT=rank(ATA).

  • 根据矩阵转置不改变矩阵的秩可得.

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

    rank(AAT)=rank(ATA)

接下来证明

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

rank(ATA)=rank(A).

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

rank(ATA)=rank(A)

本轮的证明分为两步走, 先看第一步.

  • Nullspace of

    AA

    A included by nullspace of

    ATAA^{T}A

    ATA.

    ?x,Ax=0?ATAx=0\forall \mathbf{x}, A\mathbf{x}=0 \Rightarrow A^TA\mathbf{x}=0

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

    Nullspace(A)?Nullspace(ATA)Nullspace(A)\subseteq Nullspace(A^TA)

    Nullspace(A)?Nullspace(ATA)

  • Nullspace of

    ATAA^TA

    ATA include by nullspace of

    AA

    A

    ?x,ATAx=0?xTATAx=0?Ax=0\forall \mathbf{x}, A^TA\mathbf{x}=0 \Rightarrow \mathbf{x^T}A^{T}A\mathbf{x}=0 \Rightarrow A\mathbf{x}=0

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

    Nullspace(ATA)?Nullspace(A)Nullspace(A^TA)\subseteq Nullspace(A)

    Nullspace(ATA)?Nullspace(A).

  • Finally, we get

    Nullspace(A)=Nullspace(AAT)Nullspace(A)=Nullspace(AA^T)

    Nullspace(A)=Nullspace(AAT).

再来看第二步.

  • Assuming that

    AA

    A is a

    m×nm\times n

    m×n matrix and we can get that

    ATAA^TA

    ATA
    is a

    n×nn\times n

    n×n matrix.

  • Now, we have reached the final step.

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

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

    rank(ATA)+rank(Nullspace(ATA))=nrank(A^TA)+rank(Nullspace(A^TA))=n

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

  • At last, we get

    rank(A)=rank(ATA)rank(A)=rank(A^TA)

    rank(A)=rank(ATA)