[Journal of mathematics, jiali] 313rd-Question of 2014 advanced algebra postgraduate exams of South China University of Technology

 

 

1. ($ 15' $) set $ \ BBP $ to a number field, $ f (x), g (x) \ In \ BBP [x] $. proof: $ \ BEX (f (x), g (x) = 1 \ LRA (f (x ^ N), g (x ^ n) = 1, \ EEx $ here, $ N $ is a given natural number.

Proof: $ \ rA $: by $ (f (x), g (x) = 1 $ Zhi $ \ Bex \ exists \ U (x ), V (x), \ st u (x) f (x) + V (x) g (x) = 1. \ EEx $ then $ \ Bex u (x ^ N) f (x ^ n) + V (x ^ N) g (x ^ n) = 1. \ EEx $ and any polynomial except $ f (x ^ N) $, $ g (x ^ N) $ is all except $1 $. therefore, $ (f (x ^ N), g (x ^ n) = 1 $. $ \ la $: Set $ h (x) $ to $ f (x), g (x) $ as the first public factor, then $ h (x ^ N) $ is the formula of $ f (x ^ N) $, $ g (x ^ N) $. (f (x ^ N), g (x ^ n) = 1 $, we have $ h (x ^ N) \ mid 1 $, $ h (x ^ n) = 1 $, $ h (x) = 1 $. this indicates $ (f (x), g (x) = 1 $.

 

2. ($ 20' $) Calculate the determinant: $ \ Bex D_n =\sev {\ BA {cccccc} x_1y_1 & x_1y_2 & x_1y_3 & \ cdots & x_1y _ {n-1} & {\ x_1y_2 & x_2y_2 & x_2y_3 & \ cdots & x_2y _ {n-1} & x_2y_n \ x_1y_3 & x_2y_3 & x_3y_3 & \ cdots & x_3y _ {n-1} & \\\ vdots & \ ddots & \ vdots & \ vdots \ x_1y _ {n-1} & x_2y _ {n-1} & x_3y _ {n-1} & \ cdots & X _ {n-1} y _ {n-1} & X _ {n-1} Y_n \ x_1y_n & x_2y_n & x_3y_n & \ cdots & X _ {n-1} Y_n & x_ny_n \ EA }. \ EEx $

Answer: The level of $ N $ is determined by $ D_n $, then $ \ beex \ Bea D_n & = Y_n \ sev {\ BA {cccccc} x_1y_1 & x_1y_2 & x_1y_3 & \ cdots & x_1y _ {n-1} & X_1 \ x_1y_2 & x_2y_2 & x_2y_3 & \ cdots & x_2y _ {n-1} & X_2 \ x_1y_3 & x_2y_3 & x_3y_3 & \ cdots & x_3y _ {n-1} & X_3 \ vdots &\ vdots & \ ddots & \ vdots \ x_1y _ {n-1} & x_2y _ {n-1} & x_3y _ {n-1} & \ cdots & X _ {n-1} y _ {n-1} & X _ {n-1} \ x_1y_n & x_2y_n & x_3y_n & \ cdots & X _ {n-1} Y_n & x_n \ EA }, \ y _ {n-1} D_n & = Y_n \ sev {\ BA {cccccc} x_1y_1 & x_1y_2 & x_1y_3 & \ cdots & x_1y _ {n-1} & x_1y _ {n-1 }\ \ x_1y_2 & x_2y_2 & x_2y_3 & \ cdots & x_2y _ {n-1} & x_2y _ {n-1} \ x_1y_3 & x_2y_3 & x_3y_3 & \ cdots & x_3y _ {n-1} & x_3y _ {n-1 }\\\ vdots & \ ddots & \ vdots \ x_1y _ {n-1} & x_2y _ {n-1} & x_3y _ {n-1} & \ cdots & X _ {n-1} y _ {n-1} & X _ {n-1} y _ {n-1} \ x_1y_n & x_2y_n & x_3y_n & \ cdots & X _{ n-1} Y_n & x_ny _ {n-1} \ EA }\\& = Y_n \ sev {\ BA {cccccc} x_1y_1 & x_1y_2 & x_1y_3 & \ cdots & x_1y _ {n-1 }& 0 \ x_1y_2 & x_2y_2 & x_2y_3 & \ cdots & x_2y _ {n-1} & 0 \ x_1y_3 & x_2y_3 & x_3y_3 & \ cdots & x_3y _ {n-1} & 0 \\\ vdots & \ ddots & \ vdots \ x_1y _ {n-1} & x_2y _ {n-1} & x_3y _ {n-1} & \ cdots & X _ {n-1} y _ {n-1} & 0 \ x_1y_n & x_2y_n & x_3y_n & \ cdots & X _ {n-1} Y_n & x_ny _ {n-1}-X _ {n-1} -Y_n \ EA }\\\& \ quad \ sex {\ mbox {Column n''-column n-1 ''}}\\& = Y_n (x_ny _ {n-1}-x _ {n-1} Y_n) D _ {n-1 }. \ EEA \ eeex $ with a recursive formula, evaluate $ D_n $: $ \ beex \ Bea Y _ {N-2} y _ {n-1} D_n & = Y_n (x_ny _ {n-1}-X _ {n-1} Y_n) \ cdot Y _ {N-2} D _ {n-1} \ & = Y_n (x_ny _ {n-1}-X _ {n-1} Y_n) \ cdot Y _ {n-1} (X _ {n-1} y _ {N-2}-X _ {N-2} y _ {n-1}) D _ {N-2 }, \\\ cdots <=\ cdots \\\ prod _ {I = 1} ^ {n-1} y_ I \ cdot D_n & =\ prod _ {I = 2} ^ n y_ I (x_iy _ {I-1}-X _ {I-1} y_ I ), \ D_n & = x_1y_n \ prod _ {I = 1} ^ n Y_n \ prod _ {I = 1} ^ N (x_iy _ {I-1}-X _ {I-1} y_ I ). \ EEA \ eeex $ the last step is true for $ Y_1 \ cdots Y _ {n-1} \ NEQ 0 $. However, if some $ y_j = 0 $, we use $ \ ve $ to replace $ y_j $. After applying the above conclusion, we can make $ \ ve \ to 0 $, that is, the final result is also correct.

 

3. ($ 20' $) proof:

(1) If $ \ lm \ NEQ 0 $ is an feature value of matrix $ A $, then $ \ DPS {\ frac {1} {\ lm} | A |}$ is an feature value of $ A ^ * $;

(2) If $ \ Al $ is a feature vector of $ A $, $ \ Al $ is also a feature vector of $ A ^ * $.

Proof:

(1) set $ \ Al \ NEQ 0 $ to the feature vector of $ A $ that belongs to the feature value $ \ lm \ NEQ 0 $, then $ \ Bex a \ Al = \ lm \ Al \ rA | A | \ Al = a ^ * A \ Al = \ lm a ^ * \ Al \ rA a ^ * \ Al = \ frac {1} {\ lm} | A | \ Al. \ EEx $ so $ \ DPS {\ frac {1} {\ lm} | A |}$ is an feature value of $ A ^ * $.

(2) If $ \ rank (A) = N $, then $ \ Al $ corresponds to the feature value $ \ lm \ neq0 $, instead of (1 ), $ \ Al $ is a feature vector of $ A ^ * $ that belongs to the feature value $ \ DPS {\ frac {1 }{\ lm} | A |}$. if $ \ rank (A) = n-1 $, the same as above, only $ \ LM = 0 $ must be considered. $ A \ Al = 0 $, $ \ Al $ is a basic solution of $ A $. by $ AA ^ * = 0 $ Zhi $ \ Bex a ^ * = (k_1 \ Al, \ cdots, K_n \ Al) = \ Al (k_1, \ cdots, k_n), \ EEx $ \ Bex a ^ * \ Al = \ Al (k_1, \ cdots, k_n) \ Al = \ sex {\ sum _ {I = 1} ^ n K_ I \ al_ I} \ Al \ quad \ sex {\ Al = (\ al_1, \ cdots, \ al_n) ^ t }, \ EEx $ \ Al $ is a feature vector of $ A ^ * $ that belongs to the feature value $ \ DPS {\ sum _ {I = 1} ^ n K_ I \ al_ I} $. if $ \ rank (A) = n-2 $, $ A ^ * = 0 $, $ \ Al $ is $ A ^ * $, which belongs to the feature vector with the feature value $0 $.

 

4. ($ 20' $) set $ \ Bex W = \ sed {f (x); F (1) = 0, f (x) \ In \ BBR [x] _ n }, \ EEx $ here $ \ BBR [x] _ n $ represents the linear space of the number of times on the real number field $ \ BBR $ is less than $ N $.

(1) proof: $ W $ is the linear subspaces of $ \ BBR [x] _ n $;

(2) Calculate the dimension of $ W $ and a group of bases.

Answer:

(1) apparently true.

(2) set $ \ Bex f (x) = A _ {n-1} x ^ {n-1} + \ cdots + a_0 \ in W, \ EEx $ then $ \ Bex a _ {n-1} + \ cdots + a_0 = 0, \ EEx $ basic solution $ \ Bex \ sex {\ BA {c}-1 \ 1 \ 0 \ vdots \ 0 \ EA }, \ quad \ sex {\ BA {c}-1 \ 0 \ 1 \ vdots \ 0 \ EA}, \ quad \ cdots, \ quad \ sex {\ BA {c}-1 \ 0 \ 0 \ vdots \ 1 \ EA }. \ EEx $ therefore, $ W $ has a dimension of $ n-1 $ with a base $ \ Bex-x ^ {n-1} + x ^ {N-2 }, -x ^ {n-1} + x ^ {n-3}, \ quad \ cdots, \ quad-x ^ {n-1} + 1. \ EEx $

 

5. ($ 20' $) known homogeneous linear equation $ \ Bex \ sedd {\ BA {rrrrrrc} (a_1 + B) x_1 & + & a_2x_2 & + & \ cdots & + & a_nx_n & = & 0, \ a_1x_1 & + & (A_2 + B) x_2 & + & \ cdots & + & a_nx_n & = & 0, \\\ cdots & \ a_1x_1 & + & a_2x_2 & + & \ cdots & + & (a_n + B) x_n & = & 0, \ EA} \ EEx $ where $ \ DPS {\ sum _ {I = 1} ^ n a_ I \ NEQ 0} $. when we discuss the relationship between $ A_1, \ cdots, a_n, and B $,

(1) The equations have only zero solutions?

(2) Are there non-zero solutions to the equations? When there is a non-zero solution, a basic solution of this equations is obtained.

Answer: coefficient Matrix $ \ Bex a =\sex {\ BA {CCC} a_1 + B & \ cdots & a_n \\ vdots & \ ddots & \ vdots \ A_1 & \ cdots & a_n + B \ EA} \ RRA \ sex {\ BA {CCCC} a_1 + B & A_2 & \ cdots & a_n \-B & \ cdots & 0 \\ \ vdots & \ ddots & \ vdots \-B & 0 & \ cdots & B \ EA }. \ EEx $ when $ B = 0 $, $ \ Bex a \ RRA \ sex {\ BA {CCC} A_1 & \ cdots & a_n \ 0 & \ cdots & 0 \ 0 & \ cdots & 0 \ EA}. \ EEx $ set by $ \ DPS {\ sum _ {I = 1} ^ n a_ I \ NEQ 0} $ a_$ a_n \ NEQ 0 $, $ \ rank (A) = 1 $, $ AX = 0 $ basic solution $ \ Bex \ sex {\ BA {c}-a_n \ 0 \ 0 \ vdots \ A_1 \ EA}, \ quad \ sex {\ BA {c} 0 \-a_n \ 0 \ vdots \ A_2 \ EA}, \ quad \ cdots, \ quad \ sex {\ BA {c} 0 \ vdots \ 0 \-a_n \ A _ {n-1} \ EA }. \ EEx $ when $ B \ NEQ 0 $, $ \ Bex a \ RRA \ sex {\ BA {CCCC} B + \ sum _ {I = 1} ^ n a_ I & 0 & \ cdots & 0 \-1 & 1 & \ cdots & 0 \ vdots & \ ddots & \ vdots \-1 & 0 & \ cdots & 1 \ EA }. \ EEx $ at this time, when $ \ DPS {B + \ sum _ {I = 1} ^ n a_ I \ NEQ 0} $, $ AX = 0 $ only has zero solution; when $ \ DPS {B + \ sum _ {I = 1} ^ n a_ I = 0} $, $ AX = 0 $ has a non-zero solution, the basic solution is $ \ Bex \ sex {\ BA {c} 1 \ vdots \ 1 \ EA }. \ EEx $

 

6. ($ 20' $) set a quadratic form $ \ Bex f (x) = X_1 ^ 2 + x_2 ^ 2 + X_3 ^ 2 + 2ax_1x_2 + 2bx_2x_3 + 2x_1x_3 \ EEx $ regular transformer $ x = py $ to standard type $ F = Y_2 ^ 2 + y_3 ^ 2 $, $ x = (x_1, x_2, X_3) ^ t $, $ Y = (y_1, Y_2, y_3) ^ t $ is a three-dimensional column vector, and $ p $ is a third-order orthogonal matrix.

(1) Evaluate the values of constants $ A and B $;

(2) orthogonal matrix $ p $.

Answer:

(1) From question, $ f (x) = x ^ tax $, $ \ Bex a =\sex {\ BA {CCC} 1 & A & 1 \ A & 1 & B \ 1 & B & 1 \ EA }, \ quad P ^ tap = \ diag \ sed {0, 1, 2 }. \ EEx $ \ Bex | E-A | = 0 \ ra-2AB = 0 \ rA AB = 0; \ quad | 2e-a | = 0 \ ra-(A + B) = 0. \ EEx $ and $ a = B = 0 $.

(2) by $ \ Bex 0e-a \ RRA \ sex {\ BA {CCC} 1 & 0 & 1 \ 0 & 0 \ 0 & 0 & 0 \ EA} \ EEx $ $ A $ feature vectors with feature values $0 $ \ Bex \ al_1 = \ sex {\ BA {c}-1 \ 0 \ 1 \ EA }. \ EEx $ by $ \ Bex E-A \ RRA \ sex {\ BA {CCC} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 0 \ EA} \ EEx $ $ A $ the feature vector that belongs to the feature value $1 $ is $ \ Bex \ Al_2 = \ sex {\ BA {c} 0 \ 1 \ \ 0 \ EA }. \ EEx $ by $ \ Bex 2e-a \ RRA \ sex {\ BA {CCC} 1 & 0 &-1 \ 0 & 1 & 0 \ 0 & 0 & 0 & 0 \ EA} \ EEx $ $ A $ the feature vector that belongs to the feature value $2 $ is $ \ Bex \ Al_3 = \ sex {\ BA {c} 1 \ 2 \ 1 \ EA }. \ EEx $ \ al_1, \ Al_2, \ Al_3 $ standard orthogonal form $ \ beex \ Bea \ beta_1 & =\ frac {\ al_1 }{|\ al_1 |}=\ sex {\ BA {CCC}-\ frac {1 }{\ SQRT {2 }}\\ 0 \\ frac {1 }{\ SQRT {2 }}\ EA }, \\\ beta_2 <=\ frac {\ Al_2-\ SEF {\ Al_2, \ beta_1 }\ beta_1 }{| \ Al_2-\ SEF {\ Al_2, \ beta_1 }\ beta_1 |}=\ sex {\ BA {CCC} 0 \ 1 \ 0 \ EA }, \\\ beta_3 & =\ frac {\ Al_3-\ SEF {\ Al_3, \ beta_1} \ beta_1-\ SEF {\ Al_3, \ beta_2 }\ beta_2 }{| \ Al_3-\ SEF {\ Al_3, \ beta_1} \ beta_1-\ SEF {\ Al_3, \ beta_2 }\ beta_2 |}=\ sex {\ BA {CCC} \ frac {1 }{\ SQRT {2 }}\ \ 0 \\ frac {1 }{\ SQRT {2 }}\ EA }. \ EEA \ eeex $ \ Bex P = (\ beta_1, \ beta_2, \ beta_3) = \ sex {\ BA {CCC}-\ frac {1} {\ SQRT {2 }}& 0 & \ frac {1} {\ SQRT {2 }}\ \ 0 & 1 & 0 \ frac {1 }{\ SQRT {2 }}& 0 & \ frac {1 }{\ SQRT {2 }}\ EA }, \ EEx $ P ^ tap = \ diag \ sed {0, 1, 2} $.

 

7. ($ 20' $) set $ W $ to the number field $ \ BBP $ upper $ N $ Dimension Linear Space $ V $'s sub-space, $ \ SCRA $ is a linear transformation of $ V $. $ \ scra w $ represents a subset of the images of the vectors in $ W $, make $ W_0 = W \ cap \ SCRA ^ {-1} (0) $, proof: $ \ Bex \ dim W = \ dim \ scra w + \ dim W_0. \ EEx $

Proof: Set $ \ al_1, \ cdots, \ al_r $ to a group of bases of $ W_0 $, and expand it to a group of bases of $ W $ \ al_1, \ cdots, \ al_r, \ beta_1, \ cdots, \ beta_s $. then $ \ bee \ label {313_7_eq} \ SCRA \ beta_1, \ cdots, \ SCRA \ beta_r \ mbox {linear independence, and it is a group of bases of} \ SCRA w \ mbox }. \ EEE $ \ Bex \ dim W = S + T = \ dim W_0 + \ dim \ SCRA v. \ EEx $ forward certificate \ eqref {313_7_eq }. on the one hand, $ \ beex \ Bea & \ quad \ sum _ {I = 1} ^ t K_ I \ SCRA \ beta_ I = 0 \ & \ Ra \ SCRA \ sex {\ sum _{ I = 1} ^ t K_ I \ beta_ I} = 0 \ & \ Ra \ sum _ {I = 1} ^ t K_ I \ beta_ I \ In w \ cap \ SCRA ^ {- 1} (0) = W_0 \ & \ Ra \ sum _ {I = 1} ^ t K_ I \ beta_ I = \ sum _ {j = 1} ^ r l_j \ al_j \ & \ rA K_ I = 0 \ quad \ sex {1 \ Leq I \ Leq t }; \ EEA \ eeex $ on the other hand, for $ \ forall \ Al \ in W $, $ \ Bex \ SCRA \ Al = \ SCRA \ sex {\ sum _ {j = 1} ^ rl_j \ al_j + \ sum _ {I = 1} ^ t K_ I \ beta_ I }=\ sum _ {j = 1} ^ tk_j \ SCRA \ beta_j. \ EEx $

 

8. ($15 '$) set $ A and C $ to $ N $ order positive definite matrix. If the matrix equation $ AX + XA = C $ has a unique solution $ B $. test Certificate: $ B $ Zhengding.

Proof: the conversion of $ AB + BA = C $ is $ \ Bex B ^ Ta + AB ^ t = C. \ EEx $ uniqueness of the solution, $ B ^ t = B $. therefore, $ B $ is a symmetric array. to verify that $ B $ is a positive definite array, you only need to verify any feature value of $ B $ \ lm> 0 $. set $ \ Al \ NEQ 0 $ to the corresponding feature vector, then $ \ beex \ Bea 0 & <\ Al ^ TC \ Al \ & = \ Al ^ t (AB + BA) \ Al \ & = \ Al ^ TA (B \ Al) + (B \ Al) ^ ta \ Al \ & = \ lm \ Al ^ ta \ Al + \ lm \ Al ^ ta \ Al \ & = 2 \ lm \ Al ^ ta \ Al. \ EEA \ eeex $

[Journal of mathematics, jiali] 313rd-Question of 2014 advanced algebra postgraduate exams of South China University of Technology