关于矩阵的秩的专题讨论

$\bf命题:$设$f(x),g(x) \in F[x],A \in {F^{n \times n}}$,且$(f(x),g(x))=1$,则$$r(f(A))+r(g(A))=n+r(f(A)g(A))$$

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$\bf命题:$设$A\in P^{n\times m},B\in P^{n\times s},C\in P^{m\times t},D\in P^{s\times t}$,且$r(B)=s$,$AC+BD=0$,则$r\left( {\begin{array}{*{20}{c}}C\\D\end{array}} \right) = t \Longleftrightarrow r\left( C \right) = t$

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$\bf命题:$设$n$阶矩阵$A$的特征多项式满足\[f\left( \lambda  \right) = g\left( \lambda  \right)h\left( \lambda  \right),\left( {g\left( \lambda  \right),h\left( \lambda  \right)} \right) = 1\]证明:$r\left( {g\left( A \right)} \right) = \deg h\left( \lambda  \right),且r\left( {h\left( A \right)} \right) = \deg g\left( \lambda  \right)$

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$\bf命题:$设$A,B_{i}\in M_{n}(F),rank(A)<n$,且$A=B_{1}B_{2}\cdots B_{k},B_{i}^{2}=B_{i},i=1,2,\cdots,k$,则$$rank(E-A)\leq k(n-rank(A))$$

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$\bf命题:$设$A \in {M_{m \times n}}\left( F \right),B \in {M_{n \times s}}\left( F \right)$,且$W = \left\{ {BX \in {F^s}\left| {ABX = 0} \right.} \right\}$,则$\dim W = r\left( B \right) - r\left( {AB} \right)$

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$\bf命题:$设$A \in {F^{m \times n}},B \in {F^{p \times n}}$,令$S = \left\{ {Ax|Bx = 0} \right\}$,则$S$作为${F^m}$的子空间的维数为$r\left( {\begin{array}{*{20}{c}}A\\B\end{array}} \right) - r\left( B \right)$

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$\bf命题:$设$\sigma  \in L\left( V \right)$,$W$为$V$的一个有限维子空间,则\[\dim \sigma \left( W \right) = \dim \left( {Ker\left( \sigma  \right) \cap W} \right) = \dim W\]

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$\bf命题:$设实二次型$f\left( {{x_1}, \cdots ,{x_n}} \right) = \sum\limits_{i = 1}^n {{{\left( {{a_{i1}}{x_1} + \cdots + {a_{in}}{x_n}} \right)}^2}} $,证明二次型的秩等于$A = {\left( {{a_{ij}}} \right)_{n \times n}}$的秩 

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$\bf命题:$设$A$为秩为$s$的$n$阶实对称阵,且$x'Ax \ne 0,\forall x \in {R^n}$,记$B = A - {\left( {x'Ax} \right)^{ - 1}}Axx'A$,证明:秩$B=s-1$

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$\bf命题:$设$f$为数域$F$上$n$维线性空间$V$上的一个双线性函数,则${L_f}:\alpha  \to {\alpha _L}$是$V$到${V^*}$(对偶空间)一个线性映射,其中${\alpha _L}:\beta  \to f\left( {\alpha ,\beta } \right),\forall \beta  \in V$,则$\dim \left( {{\mathop{\rm Im}\nolimits} {L_f}} \right)=f$的秩($f$的度量矩阵的秩)

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$\bf命题:$

附录1

$\bf命题1:$设$A \in {M_{m \times n}}\left( F \right),B \in {M_{n \times p}}\left( F \right)$,则$r\left( {AB} \right) \le \min \left\{ {r\left( A \right),r\left( B \right)} \right\}$

 

$\bf命题2:$设$A \in {M_{m \times n}}\left( F \right)$,则$r\left( A \right) \le m$且$r\left( A \right) \le n$

 

$\bf命题3:$$A \in {M_{m \times n}}\left( F \right)$,且$P,Q$为可逆阵,则$r\left( A \right) = r\left( {PA} \right) = r\left( {AQ} \right) = r\left( {PAQ} \right)$

 

$\bf命题4:$$r\left( {\begin{array}{*{20}{c}}A&0\\0&B\end{array}} \right) = r\left( A \right) + r\left( B \right){\rm{ }},{\rm{ }}r\left( {\begin{array}{*{20}{c}}A&C\\0&B\end{array}} \right) \ge r\left( A \right) + r\left( B \right)$

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$\bf命题5:$设$A,B \in {M_{m \times n}}\left( F \right)$,则$r\left( {A + B} \right) \le r\left( A \right) + r\left( B \right)$

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$\bf命题6:$设$A \in {M_{m \times n}}\left( F \right),B \in {M_{n \times p}}\left( F \right)$,若$AB = 0$,则$r\left( A \right) + r\left( B \right) \le n$ 

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$\bf命题7:$$r\left( {AB + A + B} \right) \le r\left( A \right) + r\left( B \right)$ 

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$\bf命题8:$设$AB = BA$,则$r\left( {A,B} \right) \le r\left( A \right) + r\left( B \right) - r\left( {AB} \right)$

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$\bf命题9:$$\bf(Frobenius公式)$$r\left( {ABC} \right) \ge r\left( {AB} \right) + r\left( {BC} \right) - r\left( B \right)$

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$\bf命题10:$$\bf(Sylvester公式)$设$A \in {M_{m \times n}}\left( F \right),B \in {M_{n \times p}}\left( F \right)$,则$r\left( {AB} \right) \geqslant r\left( A \right) + r\left( B \right) - n$

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$\bf命题11:$$\bf(秩降阶公式)$设$C \in {M_{m \times n}}\left( F \right),D \in {M_{n \times m}}\left( F \right),m = r\left( A \right) \ge r\left( B \right) = n$,且$A,B$可逆,则\[r\left( {A - D{B^{ - 1}}C} \right) = m - n + r\left( {B - C{A^{ - 1}}D} \right)\] 

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$\bf命题12:$$\bf(幂等阵与秩)$${A^2} = A \Leftrightarrow r\left( A \right) + r\left( {A - E} \right) = n$ 

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$\bf命题13:$$r\left( {{A^T}A} \right) = r\left( {A{A^T}} \right) = r\left( A \right) = r\left( {{A^T}} \right)$

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$\bf命题14:$设$r\left( {AB} \right) = r\left( B \right)$,则$r\left( {ABC} \right) = r\left( {BC} \right)$

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$\bf命题15:$设$A$为$n$阶方阵,且$r(A)=r(A^2)$,则对任意的自然数$k$都有$r(A^k)=r(A)$ 

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$\bf命题16:$设$A$为$n$阶可逆阵,则${A^T}{\rm{ = }} - A$当且仅当$r\left( {\begin{array}{*{20}{c}}A&\alpha \\{{\alpha ^T}}&0\end{array}} \right) = r\left( A \right),\forall \alpha \in {F^n}$

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$\bf命题17:$设$r\left( A \right) = r\left( {A,b} \right)$,则$r\left( {\begin{array}{*{20}{c}}A&b\\{ - {b^T}}&0\end{array}} \right) = r\left( A \right)$,其中$A$为$n$阶反对称阵,$b$为$n$维列向量                     

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$\bf命题18:$设$A$是数域$F$上的$n$阶方阵,证明:对任意的正整数$k$,有$rank\left( {{A^{n + k}}} \right) = rank\left( {{A^n}} \right)$

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posted on 2014-04-18 12:00  142857  阅读(1050)  评论(0编辑  收藏  举报

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