关于矩阵的秩的专题讨论

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

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$f命题：$设$Ain P^{n imes m},Bin P^{n imes s},Cin P^{m imes t},Din P^{s imes t}$，且$r(B)=s$，$AC+BD=0$，则$rleft( {egin{array}{*{20}{c}}C\Dend{array}} ight) = t Longleftrightarrow rleft( C ight) = t$

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$f命题：$设$n$阶矩阵$A$的特征多项式满足[fleft( lambda   ight) = gleft( lambda   ight)hleft( lambda   ight),left( {gleft( lambda   ight),hleft( lambda   ight)} ight) = 1]证明：$rleft( {gleft( A ight)} ight) = deg hleft( lambda   ight)，且rleft( {hleft( A ight)} ight) = deg gleft( lambda   ight)$

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$f命题：$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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$f命题：$设$A in {M_{m imes n}}left( F ight),B in {M_{n imes s}}left( F ight)$，且$W = left{ {BX in {F^s}left| {ABX = 0} ight.} ight}$，则$dim W = rleft( B ight) - rleft( {AB} ight)$

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$f命题：$设$A in {F^{m imes n}},B in {F^{p imes n}}$，令$S = left{ {Ax|Bx = 0} ight}$，则$S$作为${F^m}$的子空间的维数为$rleft( {egin{array}{*{20}{c}}A\Bend{array}} ight) - rleft( B ight)$

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$f命题：$设$sigma  in Lleft( V ight)$，$W$为$V$的一个有限维子空间，则[dim sigma left( W ight) = dim left( {Kerleft( sigma   ight) cap W} ight) = dim W]

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$f命题：$设实二次型$fleft( {{x_1}, cdots ,{x_n}} ight) = sumlimits_{i = 1}^n {{{left( {{a_{i1}}{x_1} + cdots + {a_{in}}{x_n}} ight)}^2}} $，证明二次型的秩等于$A = {left( {{a_{ij}}} ight)_{n imes n}}$的秩 

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

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

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$f命题：$

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$f命题1：$设$A in {M_{m imes n}}left( F ight),B in {M_{n imes p}}left( F ight)$，则$rleft( {AB} ight) le min left{ {rleft( A ight),rleft( B ight)} ight}$

 

$f命题2：$设$A in {M_{m imes n}}left( F ight)$，则$rleft( A ight) le m$且$rleft( A ight) le n$

 

$f命题3：$$A in {M_{m imes n}}left( F ight)$，且$P,Q$为可逆阵，则$rleft( A ight) = rleft( {PA} ight) = rleft( {AQ} ight) = rleft( {PAQ} ight)$

 

$f命题4：$$rleft( {egin{array}{*{20}{c}}A&0\0&Bend{array}} ight) = rleft( A ight) + rleft( B ight){ m{ }},{ m{ }}rleft( {egin{array}{*{20}{c}}A&C\0&Bend{array}} ight) ge rleft( A ight) + rleft( B ight)$

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$f命题5：$设$A,B in {M_{m imes n}}left( F ight)$，则$rleft( {A + B} ight) le rleft( A ight) + rleft( B ight)$

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$f命题6：$设$A in {M_{m imes n}}left( F ight),B in {M_{n imes p}}left( F ight)$，若$AB = 0$，则$rleft( A ight) + rleft( B ight) le n$ 

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$f命题7：$$rleft( {AB + A + B} ight) le rleft( A ight) + rleft( B ight)$ 

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$f命题8：$设$AB = BA$，则$rleft( {A,B} ight) le rleft( A ight) + rleft( B ight) - rleft( {AB} ight)$

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$f命题9：$$f（Frobenius公式）$$rleft( {ABC} ight) ge rleft( {AB} ight) + rleft( {BC} ight) - rleft( B ight)$

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$f命题10：$$f（Sylvester公式）$设$A in {M_{m imes n}}left( F ight),B in {M_{n imes p}}left( F ight)$，则$rleft( {AB} ight) geqslant rleft( A ight) + rleft( B ight) - n$

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$f命题11：$$f（秩降阶公式）$设$C in {M_{m imes n}}left( F ight),D in {M_{n imes m}}left( F ight),m = rleft( A ight) ge rleft( B ight) = n$，且$A,B$可逆，则[rleft( {A - D{B^{ - 1}}C} ight) = m - n + rleft( {B - C{A^{ - 1}}D} ight)] 

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$f命题12：$$f（幂等阵与秩）$${A^2} = A Leftrightarrow rleft( A ight) + rleft( {A - E} ight) = n$ 

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$f命题13：$$rleft( {{A^T}A} ight) = rleft( {A{A^T}} ight) = rleft( A ight) = rleft( {{A^T}} ight)$

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$f命题14：$设$rleft( {AB} ight) = rleft( B ight)$，则$rleft( {ABC} ight) = rleft( {BC} ight)$

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$f命题15：$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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$f命题16：$设$A$为$n$阶可逆阵，则${A^T}{ m{ = }} - A$当且仅当$rleft( {egin{array}{*{20}{c}}A&alpha \{{alpha ^T}}&0end{array}} ight) = rleft( A ight),forall alpha in {F^n}$

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$f命题17：$设$rleft( A ight) = rleft( {A,b} ight)$，则$rleft( {egin{array}{*{20}{c}}A&b\{ - {b^T}}&0end{array}} ight) = rleft( A ight)$，其中$A$为$n$阶反对称阵，$b$为$n$维列向量                     

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$f命题18：$设$A$是数域$F$上的$n$阶方阵，证明：对任意的正整数$k$，有$rankleft( {{A^{n + k}}} ight) = rankleft( {{A^n}} ight)$

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