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Egalitate de ranguri

 
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alex2008
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Joined: 19 Oct 2008
Posts: 489
Location: Tulcea

PostPosted: Sat Dec 11, 2010 2:36 pm    Post subject: Egalitate de ranguri Reply with quote

Fie A\in \mathcal{M}_n(\mathbb{C}). Sa se demonstreze ca:

\text{rang}(A)=\text{rang}(A\cdot \overline{^tA})
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Radu Titiu
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Joined: 28 Sep 2007
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Location: Mures \Bucuresti

PostPosted: Tue Dec 14, 2010 1:12 am    Post subject: Reply with quote

Admitem cunoscut faptul ca  rang(A) este numarul de linii ( sau coloane ) liniar independente ale matricei  A .

Notez  \<x,y\>=\sum_{i=1}^n x_i \overline{y_i} ,produsul scalar din  \mathbb{C}^n .

Daca in plus mai notez  L_1 ,L_2 ,...,L_n liniile matricei  A , rezulta imediat ca  A \cdot ^t \overline{A}=L , unde matricea  L are componentele  (\< L_i,L_j\>) .
Pentru a demonstra ceea ce vrem este suficient sa demonstram ca matricea  A are tot atatea linii liniar indepentede cate are matricea  L .

Daca notez cu  M_1,...,M_n liniile matrice  A \cdot ^t \overline{A} , atunci e usor de verificat ca  \sum_{i=1}^k a_i M_i =\left( \<\sum_{i=1}^k a_i L_i ,L_1\> ,...,\<\sum_{i=1}^k a_i L_i ,L_k\> \right ) ,pentru  k \leq n(*)

Mai mult se verifica usor ca are loc relatia
 \sum_{i=1}^k a_i L_i =0 daca si numai daca  \< \sum_{i=1}^k a_i L_i, L_j\>=0 pentru orice  j=\overline{1,k} .
Dar din aceasta ultima relatie si din observatia (*) rezulta ca  A si  L au acelasi numar de linii liniar indepentente.


O eventuala alta solutie se poate da folosind putina teorie de spatii vectoriale ;problema se reduce la a arata ca  \dim Ker (A) =\dim Ker (A \cdot ^t \overline{A}

Dar  \dim Ker (A \cdot ^t \overline{A}=\dim Ker ^t \overline{A} .Deci e suficient sa dem ca  \dim Ker (A)=\dim Ker (^t \overline{A}) , care e echivalent cu faptul ca  rang(A)=rang(^t \overline{A}) , si care rezulta din faptul ca rangul e nr de linii / coloane liniar independente.
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Beniamin Bogosel
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Joined: 07 Mar 2008
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Location: Chambery, Franta

PostPosted: Fri Dec 17, 2010 7:59 pm    Post subject: Reply with quote

O solutie mai simpla se poate da folosind o teorema arhicunoscuta folosita deseori la olimpiada nationala de matematica, si anume:

O matrice A \in \mathcal{M}(\mathbb{R}), de rang r se poate scrie sub forma A=P\begin{pmatrix}I_r & 0 \\ 0 & 0\end{pmatrix}Q, unde P,Q sunt matrici inversabile.

O demonstratie rapida a acestei teoreme este aceea ca fie care transformare elementara a unei linii sau a unei coloane poate fi descrisa ca fiind inmultirea matricii la dreapta sau la stanga cu o matrice inversabila. Astfel se obtin P,Q.

Revenind la teorema noastra, notam r=rang(A). Atunci X=A\cdot \overline{^t A}=P\begin{pmatrix}I_r & 0 \\ 0 & 0\end{pmatrix}Q\overline{^t Q}\begin{pmatrix}I_r & 0 \\ 0 & 0\end{pmatrix}\overline{^t P}.

Deoarece inmultirea cu matrici inversabile pastreaza rangul, avem rang(X)=rang(\begin{pmatrix}I_r & 0 \\ 0 & 0\end{pmatrix}Q\overline{^t Q}\begin{pmatrix}I_r & 0 \\ 0 & 0\end{pmatrix})=r, deoarece matricea din mijloc este inversabila, iar inmultirea cu \begin{pmatrix}I_r & 0 \\ 0 & 0\end{pmatrix} la stanga si la dreapta anuleaza coloanele si liniile de indice mai mare decat r.
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