Sylvester不等式证明

Sylvester不等式

设A、B分别是$s\times n、n\times m$,则$rank(AB)\ge rank(A)+rank(B)-n$

证明

只需证$n+rank(AB)\ge rank(A)+rank(B)$
n + r a n k ( A B ) = r a n k ( I n 0 0 A B ) n+rank(AB)=rank $$\left(\begin{array}{cc}{I}_n& 0\\ 0& AB\end{array}\right)$$ 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 ) $$\left(\begin{array}{cc}{I}_n& 0\\ 0& AB\end{array}\right)$$\longrightarrow $$\left(\begin{array}{cc}{I}_n& 0\\ A& AB\end{array}\right)$$ \longrightarrow $$\left(\begin{array}{cc}{I}_n& -B\\ A& 0\end{array}\right)$$\ \longrightarrow $$\left(\begin{array}{cc}{I}_n& B\\ A& 0\end{array}\right)$$ \longrightarrow $$\left(\begin{array}{cc}B& I_n\\ A& 0\end{array}\right)$$ (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 $$\left(\begin{array}{cc}{I}_n& 0\\ 0& AB\end{array}\right)$$ =$$\left(\begin{array}{cc}B& I_n\\ 0& A\end{array}\right)$$ \ge rank(B)+rank(A) rank(In​0​0AB​)=(B0​In​A​)≥rank(B)+rank(A)
因此

$$rank(AB)\ge rank(A)+rank(B)-n$$