06th sets of simulated exam for Mathematics Competition Training of Gannan Normal University

1. set $ F (\ Al, \ beta) $ to a linear space $ V $ to a non-degraded bilinear function. Example: $ \ Bex \ forall \ G \ In V ^ *, \ exists \ | \ Al \ In V, \ st F (\ Al, \ beta) = g (\ beta), \ quad \ forall \ beta \ In v. \ EEx $

Proof: (1) uniqueness: If $ \ Tilde \ Al $ is also suitable for the question, then $ \ beex \ Bea & \ quad F (\ Al, \ beta) = f (\ Tilde \ Al, \ beta), \ quad \ forall \ beta \ In V \ & \ rA F (\ al-\ Tilde \ Al, \ beta) = 0, \ quad \ forall \ beta \ In V \ & \ Ra \ al-\ Tilde \ Al = 0 \ quad \ sex {f \ mbox {non-return }}. \ EEA \ eeex $

(2) existence: Take a group of bases $ \ ve_1, \ cdots, \ ve_n $ for $ V $, and remember $ A _ {IJ} = f (\ ve_ I, \ ve_j) $, then for $ \ Bex \ Al = \ sum _ {I = 1} ^ n a_ I \ ve_ I, \ quad \ Beta = \ sum _ {I = 1} ^ n B _ I \ ve_ I, \ EEx $ \ Bex F (\ Al, \ beta) = \ sum _ {I, j = 1} ^ N A _ {IJ} a_ib_j = \ Al ^ ta \ beta. \ EEx $ by $ F $ non-degraded and $ \ beex \ Bea & \ quad F (\ Al, \ beta) = 0, \ quad \ forall \ beta \ In V \ & \ LRA \ Al ^ ta \ Beta = 0, \ quad \ forall \ beta \ In V \ & \ LRA \ Al ^ TA = 0 \ EEA \ eeex $ $ A $ reversible. for $ g \ In V ^ * $, take $ \ Bex \ Al = (\ ve_1, \ cdots, \ ve_n) (a ^ {-1 }) ^ t \ sex {\ BA {c} g (\ ve_1) \ vdots \ G (\ ve_n) \ EA} \ In V, \ EEx $ \ beex \ Bea F (\ Al, \ beta) & = (G (\ ve_1), \ cdots, g (\ ve_n )) A ^ {-1} \ cdot A \ cdot \ sex {\ BA {L} B _1 \ vdots \ B _n \ EA} \ & =\ sum _ {I = 1} ^ n B _ig (\ xi_ I) \\& = g (\ beta), \ quad \ forall \ beta \ In v. \ EEA \ eeex $

 

 

2. set $ \ SCRA $ to the orthogonal transformation on the Euclidean Space $ V $, and $ \ SCRA ^ m = \ scre \ (M> 1) $. note that $ W _ \ SCRA =\sed {\ Al \ In V; \ SCRA \ Al =\al }$, $ W _ \ SCRA ^ \ perp $ is its orthogonal complement, and $ \ Bex \ forall \ Al \ In V, \ exists \ | \ beta \ in W _ \ SCRA, \ gamma \ in W _ \ SCRA ^ \ perp, \ ST \ Al = \ Beta + \ Gamma. \ EEx $ test certificate: $ \ Bex \ Beta = \ frac {1} {m} \ sum _ {I = 1} ^ m \ SCRA ^ {I-1} \ Al. \ EEx $

Proof: Set $ \ Bex U =\sed {\ Al \ In V; \ Al + \ SCRA \ Al + \ cdots + \ SCRA ^ M-1} \ Al = 0 }. \ EEx $ for $ \ forall \ Al \ In V $, $ \ Bex \ Al = \ frac {1} {m} \ sum _ {I = 1} ^ m \ SCRA ^ {I-1} \ Al + \ sex {\ al- \ frac {1} {m} \ sum _ {I = 1} ^ m \ SCRA ^ {I-1} \ Al} \ equiv \ Beta + \ gamma, \ EEx $ \ beex \ Bea \ SCRA (\ beta) & =\ frac {1} {m} \ sum _ {I = 1} ^ m \ SCRA ^ I \ Al \ & =\ frac {1} {m }(\ SCRA \ Al + \ cdots + \ SCRA ^ M-1} \ Al + \ Al) \\&=\ beta, \ gamma + \ SCRA \ gamma + \ cdots + \ SCRA ^ M-1} \ gamma & = \ Al + \ SCRA \ Al + \ cdots + \ SCRA ^ {M-1} \ al-\ beta-\ SCRA \ beta-\ cdots-\ SCRA ^ M-1} \ beta \ & =\ Al + \ SCRA \ Al + \ cdots + \ SCRA ^ {M-1} \ al-m \ beta \ & = 0. \ EEA \ eeex $ so $ v = W _ \ SCRA + U $. it is also composed of $ \ Bex \ Al \ in W _ \ SCRA \ cap U \ rA 0 = \ Al + \ SCRA \ Al + \ cdots + \ SCRA ^ {M-1} \ al = m \ Al \ Ra \ Al = 0 \ EEx $ Zhi $ v = W _ \ SCRA \ oplus U $. finally, it consists of $ \ beex \ Bea \ Al \ in W _ \ SCRA, \ beta \ In U & \ Ra \ SCRA \ Al = \ Al \ quad \ sex {\ Al = \ SCRA ^ {-1} \ Al = \ SCRA ^ t \ Al} \\& \ Ra \ SEF {\ Al, \ beta }=\ SEF {\ SCRA ^ t \ Al, \ beta }=\ cdots =\ SEF {(\ SCRA ^ t) ^ {M-1} \ Al, \ beta }\\& \ quad \=\ frac {1} {m} \ SEF {\ Al + \ SCRA ^ t \ Al + \ cdots + \ sex {\ SCRA ^ {M-1 }}^ t \ Al, \ beta }\\& \ quad \=\ frac {1} {m} \ SEF {\ Al, (\ scre + \ SCRA + \ cdots + \ SCRA ^ M-1 }) \ beta} \ & \ quad \ = 0 \ EEA \ eeex $ Zhi $ W _ \ SCRA ^ \ perp = U $.

 

 

3. set $ W $ to the sub-space of the Euclidean Space $ V $, and define the distance from $ \ Al \ In V $ to $ W $ \ RD (\ Al, W) = | \ al-\ al' | $, where $ \ al' $ is an orthogonal projection of $ \ Al $ on $ W $. set $ \ al_1, \ cdots, \ al_m $ to a group of bases for $ W $. trial certificate: $ \ Bex \ RD (\ Al, W) =\ SQRT {\ cfrac {G (\ al_1, \ cdots, \ al_m, \ Al)} {G (\ al_1, \ cdots, \ al_m )}}, \ EEx $ where $ G (\ al_1, \ cdots, \ al_m) $ is the Gram matrix of $ \ al_1, \ cdots, \ al_m $.

Proof: $ \ beex \ Bea g (\ al_1, \ cdots, \ al_m, \ Al) & = g (\ al_1, \ cdots, \ al_m, \ al ') + g (G (\ al_1, \ cdots, \ al_m, \ al-\ al') \ & = g (\ al_1, \ cdots, \ al_m, \ al-\ al') \\\&=| \ al-\ al' | ^ 2G (\ al_1, \ cdots, \ al_m ). \ EEA \ eeex $

 

 

4. set $ A =\sex {\ BA {CCC} 1 & 0 & 0 \-1 & 0 & 1 \ 0 & 1 & 0 \ EA} $. try $ a ^ {100} $.

Proof: the polynomial of Yi Zhi $ A $ is $ F (\ lm) = | \ lm E-A | = (\ lm + 1) (\ LM-1) ^ 2 $. by the Hamilton-caylay theorem, $ F (a) = 0 $. for $ G (\ lm) =\lm ^ {100} $, based on the theory of moving and Division, $ \ Bex g (\ lm) = Q (\ lm) F (\ lm) + A \ lm ^ 2 + B \ lm + C. \ EEx $ Replace $ \ LM =-1 $, $ \ LM = 1 $ with the formula above, and $ \ LM = 1 $ with the formula above for the equations after the derivation, $ \ Bex a-B + c = 1, \ quad A + B + C = 1, \ quad2a + B = 100. \ EEx $ then $ \ beex \ Bea & \ quad A = 50, \ quad B = 0, \ quad C =-49 \\&\ Ra a ^ {100} = g () = 50a ^ 2-49e = \ sex {\ BA {CCC} 1 & 0 & 0 \-50 & 1 & 0 \-50 & 0 & 1 \ EA }. \ EEA \ eeex $

 

 

5. set the feature value of $ N $ level matrix $ A $ to $ \ lm_1, \ cdots, \ lm_n $. Test Certificate: $ A ^ * $ ($ A $ adjoint matrix) the feature value is $ \ Bex \ prod _ {j = 1, J \ NEQ I} ^ n \ lm_j, \ quad I = 1, 2, \ dots, N. \ EEx $

A: By the disturbance method, set $ A $ reversible, and $ \ beex \ Bea 0 & = | \ lm_ I E-A | \ cdot | a ^ * | \ & = | \ lm_ia ^ *-| A | E | \\& = | \ lm_ia ^ *-\ lm_1 \ cdots \ lm_n e |\\& = (-\ lm_ I) ^ n | \ lm_1 \ cdots \ lm _ {I-1} \ lm _ {I + 1} \ cdots \ lm_n E-A ^ * |. \ EEA \ eeex $

 

 

6. solve the following nonlinear equations $ \ beex \ Bea x ^ 3 + 3x ^ 2-13x-15 & = 0, \ x ^ 4-5x ^ 3-x ^ 2 + 17x + 12 & = 0, \ x ^ 5-2x ^ 4-3x ^ 3 + x ^ 2-2x-3 & = 0. \ EEA \ eeex $

Q: For polynomials $ f (x) and g (x) $, $ \ Bex f (x) = 0, \ quad g (x) = 0 \ LRA (f (x), g (x) = 0. \ EEx $ solve the problem. we only need to find the maximum Internet formula of the three polynomials before obtaining the root. the maximum formula of the three polynomials in the instant question is $ x ^ 2-2x + 3 $. so $ x =-1 $ or $ x = 3 $.

 

 

7. Find all non-zero third-order phalanx that satisfies $ A ^ 2 = 0 $.

A: The feature value of $ A ^ 2 = 0 $ Zhi $ A $ is $0 $, for $ A $, the Jordan standard type is $ \ Bex \ sex {\ BA {CCC} 0 & 1 & 0 \ 0 & 0 \ 0 & 0 & 0 \ EA} \ mbox {or} \ sex {\ BA {CCC} 0 & 1 & 0 \ 0 & 0 & 1 \ 0 & 0 \ EA }. \ EEx $ notice $ \ Bex \ sex {\ BA {CCC} 0 & 1 & 0 \ 0 & 0 \ 0 & 0 & 0 \ EA }\ sex {\ BA {CCC} 0 & 1 & 0 \ 0 & 0 \ 0 & 0 & 0 \ EA }=\ sex {\ BA {CCC} 0 & 0 & 0 \ 0 & 0 \ 0 & 0 & 0 \ EA }, \ EEx $ \ Bex \ sex {\ BA {CCC} 0 & 1 & 0 \ 0 & 0 \ 0 \ 0 & 0 \ EA} \ sex {\ BA {CCC} 0 & 1 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0 \ EA} \ NEQ \ sex {\ BA {CCC} 0 & 0 & 0 \ 0 & 0 \ 0 & 0 & 0 \ EA} \ EEx $ and $ A ^ 2 = 0 $ $ A $ only yes $ \ Bex \ sex {\ BA {CCC} 0 & 1 & 0 \ 0 & 0 \ 0 & 0 & 0 \ EA }. \ EEx $ then $ \ Bex a = t ^ {-1} \ sex {\ BA {CCC} 0 & 1 & 0 \ 0 & 0 \ 0 & 0 & 0 \ EA} t, \ Quad (T \ mbox {reversible }). \ EEx $ $ t ^ {-1} = t ^ */| T | $, We have (this is equivalent) $ \ Bex a = TT ^ * \ sex {\ BA {CCC} 0 & 1 & 0 \ 0 & 0 \ 0 & 0 & 0 \ EA} t, \ Quad (T \ NEQ 0, \ quad t \ mbox {reversible }). \ EEx $ note $ t = (T _ {IJ}) $, then $ \ Bex a = T \ sex {\ BA {CCC} t _ {21} (-T _ {23} t _ {32} + T _ {22} t _ {33 }) & T _ {22} (-T _ {23} t _ {32} + T _ {22} t _ {33 }) & T _ {23} (-T _ {23} t _ {32} + T _ {22} t _ {33 }) \ t _ {21} (T _ {23} t _ {31}-T _ {21} t _ {33 }) & T _ {22} (T _ {23} t _ {31}-T _ {21} t _ {33 }) & T _ {23} (T _ {23} t _ {31}-T _ {21} t _ {33 }) \ t _ {21} (-T _ {22} t _ {31} + T _ {21} t _ {32 }) & T _ {22} (-T _ {22} t _ {31} + T _ {21} t _ {32 }) & T _ {23} (-T _ {22} t _ {31} + T _ {21} t _ {32}) \ EA }. \ EEx $ then $ \ Bex a = T \ sex {\ BA {CCC} A (-c e + B f) & B (-c e + B F) & C (-c e + B f) \ A (c d-a f) & B (c d-a f) & C (c d-A F) \ A (-B d + a e) & B (-B d + a e) & C (-B D + A E) \ EA }, \ quad \ sex {T \ NEQ 0 }. \ EEx $

 

 

8. set $ f (x) $ to $ \ BBR $ first polynomial with no solid root. proof: $ g (x) $, $ h (x) $ exists, making $ \ Bex f (x) = G ^ 2 (x) + H ^ 2 (x), \ EEx $ and $ \ deg g (x)> \ deg h (x) $.

Proof: $ f (x) $ has a form $ \ Bex f (x) = \ prod _ {I = 1} ^ n (x ^ 2 + a_ix + B _ I ), \ quad a_ I ^ 2-4b_ I <0. \ EEx $ mathematical induction for $ N $. when $ n = 1 $, $ \ beex \ Bea f (x) & = x ^ 2 + a_1x + B _1 \\\\\sex {x + \ cfrac {A_1} {2 }}^ 2 + B _1-\ cfrac {A_1 ^ 2 }{ 4 }\\&=\ sex {x + \ cfrac {A_1} {2 }}^ 2 + \ sex {\ SQRT {B _1-\ cfrac {A_1 ^ 2} {4 }}^ 2 \ equiv \ PSI ^ 2 (x) + \ Phi ^ 2 (x ). \ EEA \ eeex $ if the conclusion is true for $ N $, $ \ beex \ Bea f (x) When $ n + 1 $) & = (x ^ 2 + a_1x + B _1) \ prod _ {I = 2} ^ {n + 1} (x ^ 2 + a_ix + B _ I) \\& = (\ PSI ^ 2 (x) + \ Phi ^ 2 (x) (G ^ 2 (x) + H ^ 2 (x )) \ quad \ sex {\ mbox {use inductive hypothesis }\\\&=\ SEZ {\ PSI (x) g (x) + \ PHI (x) h (x )} ^ 2 + \ SEZ {\ PSI (x) h (x)-\ PHI (x) g (x)} ^ 2. \ EEA \ eeex $

 

 

9. Set $ N $ to set the sum of all the first-level primary and child schemas of the level-2 Symmetric Matrix $ A $ to zero. It is proved that $ A $ is a zero matrix.

Proof: The feature value $ \ lm_ I $ of $ A $ is a real number. and $ \ Bex F (\ lm) \ equiv | \ lm E-A | = \ prod _ {I = 1} ^ N (\ lm-\ lm_ I ). \ EEx $ according to the question, $ F (\ lm) $ \ lm ^ {n-1} $, the coefficients of $ \ lm ^ {N-2} $ are the sum of all the first and second-order masters of $ A $, which is equal to zero. therefore, $ \ beex \ Bea \ lm_1 + \ cdots + \ lm_n & = 0, \\\ lm_1 \ lm_2 + \ cdots + \ lm_1 \ lm_n + \ cdots + \ lm _ {n-1} \ lm_n & = 0, \ (\ lm_1 + \ cdots + \ lm_n) ^ 2 & = (\ lm_1 + \ cdots + \ lm_n) ^ 2 \ & \ quad-2 (\ lm_1 \ lm_2 + \ cdots + \ lm_1 \ lm_n + \ cdots + \ lm _ {n-1} \ lm_n) \ & = 0, \ lm_1 = \ cdots = \ lm_n & = 0, \ A & = 0. \ EEA \ eeex $

 

 

10. Set $ A and B $ to two N $ positive definite matrices. proof:

(1) If $ AB = BA $, $ AB $ is also a positive matrix;

(2) If $ A-B $ is positive, $ B ^ {-1}-a ^ {-1} $ is also positive.

Proof: (1) Obviously, $ AB $ is symmetric. A reversible Matrix exists from $ A, B $ Zhengding. $ p, q $ makes $ A = P ^ TP, \ quad B = Q ^ TQ. $ and Q ^ {-1} = QP ^ tpq ^ t = (PQ ^ t) ^ t (PQ ^ t) \ EEx $ Zhi $ AB $ is similar to a positive matrix and has a positive feature value, which is positive. (2) $ \ beex \ Bea & \ quad A-B \ mbox {Positive Definite} \ & \ rA P ^ TP-Q ^ TQ \ mbox {Positive Definite} \ & \ rA Q ^ {-T} P ^ tpq ^ {-1}-e \ mbox {Positive Definite} \ & \ rA PQ ^ {-1} Q ^ {-t} P ^ T-E \ mbox {Positive Definite} \\& \ quad \ sex {Y (xy) y ^ {-1} = Yx \ rA XY, YX \ mbox {similar, with the same feature value }}\\& \ rA Q ^ {-1} Q ^ {-t}-P ^ {-1} P ^ {-t} \ mbox {positive Definite} \ & \ rA B ^ {-1}-a ^ {-1} \ mbox {Positive Definite }. \ EEA \ eeex $