Sylvester不等式证明

Sylvester不等式

设A、B分别是 s × n 、 n × m s\times n、n\times m s×n、n×m,则 r a n k ( A B ) ≥ r a n k ( A ) + r a n k ( B ) − n rank(AB)\ge rank(A)+rank(B)-n rank(AB)≥rank(A)+rank(B)−n

证明

只需证 n + r a n k ( A B ) ≥ r a n k ( A ) + r a n k ( B ) n+rank(AB)\ge rank(A)+rank(B) n+rank(AB)≥rank(A)+rank(B)
n + r a n k ( A B ) = r a n k ( I n 0 0 A B ) n+rank(AB)=rank \begin{pmatrix}I_n & 0\\0 &AB \end{pmatrix} n+rank(AB)=rank(In​0​0AB​)

作分块矩阵的初等行变换

( I n 0 0 A B ) ⟶ ( I n 0 A A B ) ⟶ ( I n − B A 0 ) ⟶ ( I n B A 0 ) ⟶ ( B I n A 0 ) \begin{pmatrix} I_n&0\\0&AB\end{pmatrix}\longrightarrow \begin{pmatrix}I_n&0\\A&AB\end{pmatrix} \longrightarrow \begin{pmatrix}I_n&-B\\A&0\end{pmatrix}\\ \longrightarrow \begin{pmatrix}I_n&B\\A&0\end{pmatrix} \longrightarrow \begin{pmatrix}B&I_n\\A&0\end{pmatrix} (In​0​0AB​)⟶(In​A​0AB​)⟶(In​A​−B0​)⟶(In​A​B0​)⟶(BA​In​0​)

根据分块矩阵的初等行变换不改变矩阵的秩有：
r a n k ( I n 0 0 A B ) = ( B I n 0 A ) ≥ r a n k ( B ) + r a n k ( A ) rank \begin{pmatrix}I_n&0\\0&AB\end{pmatrix} =\begin{pmatrix}B&I_n\\0&A\end{pmatrix} \ge rank(B)+rank(A) rank(In​0​0AB​)=(B0​In​A​)≥rank(B)+rank(A)
因此

r a n k ( A B ) ≥ r a n k ( A ) + r a n k ( B ) − n rank(AB)\ge rank(A)+rank(B)-n rank(AB)≥rank(A)+rank(B)−n

posted @ 2022-06-06 20:35  lishangli  阅读(725)  评论(0)    收藏  举报