[Journal of mathematics, jiali] 247th-Question of 2013 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] $, and $ \ P (g (x) \ geq 1 $. proof: There is a unique polynomial sequence $ f_0 (x), F_1 (x), \ cdots, f_r (x) $, make $ \ P (f_ I (x) <\ P (g (x) $ or $ f_ I (x) = 0 $ for $0 \ Leq I \ Leq r$, and $ \ Bex f (x) = \ sum _ {I = 0} ^ r f_ I (x) G ^ I (X ). \ EEx $

Proof: by division with remainder, $ \ beex \ Bea f (x) & = f_0 (x) + Q_1 (x) g (x), \ quad f_0 (X) \ mbox {suitable for the conditions in the question, and} f_0, Q_1 \ mbox {unique}, \ Q_1 (x) & = f_1 (x) + Q_2 (X) g (x), \ quad F_1 (x) \ mbox {suitable for the conditions in the question, and} F_1, Q_2 \ mbox {unique }, \ cdots \ q _ {R-1} (x) & = F _ {R-1} (x) + f_r (x) g (x ), \ quad F _ {R-1} (x) \ mbox {suitable for the conditions in the question, and} f _ {R-1}, f_r \ mbox {unique }. \ EEA \ eeex $ so $ \ Bex f (x) = \ sum _ {I = 0} ^ r f_ I (x) G ^ I (X ). \ EEx $ recommended by $ \ P (g) \ geq 1 $ \ DPS {r = \ SEZ {\ P (f)/\ P (G )}} $. and by $ \ Bex \ P (f_ig ^ I) <(I + 1) \ P (g) <r \ P (g) \ Leq \ P (f ), \ Quad (0 \ Leq I \ Leq i-1) \ EEx $ \ Bex \ P (f) = \ P (f_rg ^ R) = \ P (f_r) + r \ P (G), \ EEx $ \ Bex \ P (f_r) = \ P (f)-r \ P (g) <\ P (g ). \ EEx $ in this question, we remember $ \ P (0) =-\ infty $.

 

2 ($ 15' $) calculates the rank of the following $ N $: $ \ Bex \ sev {\ BA {CCCC} 1 + A_1 & 1 & \ cdots & 1 \ 1 & 1 + A_2 & \ cdots & 1 \ vdots &\ vdots & \ ddots & \ vdots \ 1 & \ cdots & 1 + a_n \ EA }. \ EEx $

Answer: When all $ a_ I \ NEQ 0 $, $ \ beex \ Bea \ mbox {original format} & =\ sev {\ BA {CCCCC} 1 & 1 & 1 & \ cdots & 1 \ 0 & 1 + A_1 & 1 & \ cdots & 1 \ 0 & 1 & 1 + A_2 & \ cdots & 1 \ vdots & \ ddots & \ vdots \ 0 & 0 & 0 & \ cdots & 1 + a_n \ EA} \\\\\sev {\ba {CCCCC} 1 & 1 & 1 & \ cdots & 1 \-1 & A_1 & 0 & \ cdots & 0 \-1 & 0 & A_2 & \ cdots & 0 \ vdots & \ ddots & \ vdots \-1 & 0 & 0 & \ cdots & a_n \ EA} \\\\=\ sex {1 + \ sum _ {I = 1} ^ n \ frac {1} {a_ I }} \ prod _ {I = 1} ^ n a_ I. \ EEA \ eeex $ when a certain $ a_ I = 0 $, for example, $ A_1 = 0 $, $ \ Bex \ mbox {original format} = A_2 \ cdots a_n. \ EEx $ in short, original formula $ \ DPS {= \ prod _ {I = 1} ^ n a_ I + \ sum _ {I = 1} ^ n \ prod _ {j = 1, J \ NEQ I} ^ n a_j} $.

 

3 ($ 15' $) set $ N $ level real symmetric matrix $ A = (A _ {IJ}) $ to positive definite, $ B _1, \ cdots, B _n $ is any $ N $ non-zero real number. It proves that the matrix $ B = (A _ {IJ} B _ib_j) $ is also positive.

Proof: For $ \ forall \ x \ In \ BBR ^ N $, $ \ beex \ Bea \ sum _ {I, j} A _ {IJ} B _ib_jx_ix_j & =\ sum _ {I, j} A _ {IJ} (B _ix_ I) (B _jx_j) \\&=\ sum _ {I, j} A _ {IJ} y_iy_j \ Quad (y_ I = B _ix_ I) \ & \ geq 0. \ EEA \ eeex $ is equivalent to $ B $ semi-definite. it is determined by $ x \ NEQ 0 \ rA Y \ NEQ 0 $ $ B $ Dingding.

 

4 ($ 25') $ when discussing the value of the parameter $ A, B $, linear Equations $ \ Bex \ sedd {\ BA {rrrrrrrrl} X_1 & + & X_2 &-& 2x_3 & + & 3x_4 & = & 0, \ 2x & + & X_2 &-& 6x_3 & + & 4x_4 & = &-1, \ 3x_1 & + & 2x_2 & + & ax_3 & + & 7x_4 & = &-1, \ X_1 &-& X_2 &-& 6x_3 &-& x_4 & = & B \ EA} \ EEx $ solutions? No solution? When there is a solution, find the general solution.

Answer: by $ \ beex \ BEA (A, B) & =\ sex {\ BA {CCCCC} 1 & 1 &-2 & 3 & 0 \ 2 & 1 &-6 & 4 &-1 \ 3 & 2 & & 7 &-1 \ 1 &-1 &-6 &-1 & B \ EA} \ & \ RRA \ sex {\ BA {CCCCC} 1 & 1 &- 2 & 3 & 0 \ 0 &-1 &-2 &-2 &-1 \ 0 &-1 & A + 6 &-2 &-1 \ 0 & -2 &-4 &-4 & B \ EA} \ & \ RRA \ sex {\ BA {CCCCC} 1 & 1 &-2 & 3 & 0 \ 0 & 1 & 2 & 2 & 1 \ 0 & 0 & A + 8 & 0 & 0 \ 0 & 0 & 0 & B + 2 \ EA} \ EEA \ eeex $ Zhi

(1) When $ A \ NEQ-8, B \ NEQ-2 $, the original equation has no solution;

(2) When $ A =-8, B =-2 $) \ RRA \ sex {\ BA {CCCCC} 1 & 0 &-4 & 1 &-1 \ 0 & 1 & 2 & 1 \ 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ EA} \ EEx $ the solution to the original equation is $ \ Bex \ sex {\ BA {CCCCC}-1 \ 1 \ 0 \ 0 \ EA} + k \ sex {\ BA {CCCCC} 4 \-2 \ 1 \ 0 \ EA} + L \ sex {\ BA {CCCCC}-1 \-2 \ 0 \ 1 \ EA }, \ Quad (\ forall \ K, L ). \ EEx $

 

5 ($ 20' $) is a linear transformation of $ N $ Dimension Linear Space $ V $.

(1) proof: $ V $ any sub-space containing the value range $ \ SCRA v $ W $ is $ \ SCRA $-sub-space;

(2) In the kernel $ \ SCRA ^ {-1} (0) $, obtain a group of bases $ \ alpha_1, \ cdots, \ alpha_m $, and expand it to a group of $ \ Bex \ alpha_1, \ cdots, \ alpha_m, \ Alpha _ {m + 1}, \ cdots, \ alpha_n, \ EEx $ Q: What is the shape of the matrix $ \ SCRA $ under this group?

Proof:

(1) conclusion is drawn from $ \ Bex \ Alpha \ In w \ Ra \ SCRA \ Alpha \ In \ scra v \ subset w \ EEx $.

(2) $ \ beex \ Bea \ quad \ SCRA (\ alpha_1, \ cdots, \ alpha_n) = (\ alpha_1, \ cdots, \ alpha_n) \ sex {\ BA {cccccc} 0 & \ cdots & 0 & A _ {1, m + 1} & \ cdots & A _ {1N} \ vdots & \ vdots \ 0 & \ cdots & 0 & A _ {M, m + 1} & \ cdots & A _ {Mn} \ 0 & \ cdots & 0 & A _ {m + 1, m + 1} & \ cdots & A _ {m + 1, N }\\\ vdots & \ vdots \ 0 & \ cdots & 0 & A _ {n, m + 1} & \ cdots & A _ {NN} \ EA }. \ EEA \ eeex $

 

6 ($ 25' $) set $ V $ to $4 $ Dimension Euclidean space, $ \ ve_1, \ cdots, \ ve_4 $ to a set of standard orthogonal bases for $ V $. order $ \ Bex \ alpha_1 = \ ve_2 + \ ve_3 + \ ve_4, \ alpha_2 = \ ve_1 + \ ve_3 + \ ve_4, \\\ alpha_3 = \ ve_1 + \ ve_2 + \ ve_4, \\\ alpha_4 = \ ve_1 + \ ve_2 + \ ve_3. \ EEx $

(1) convert $ \ alpha_1, \ alpha_2, \ alpha_3 and \ alpha_4 into vector groups in Orthogonal units $ \ beta_1, \ beta_2, \ beta_3, \ beta_4 $;

(2) Calculate the transition matrix from base $ \ ve_1, \ ve_2, \ ve_3, \ ve_4 $ to base $ \ beta_1, \ beta_2, \ beta_3, \ beta_4 $;

(3) Orders $ W_1 = L (\ alpha_1, \ alpha_2), U_1 = W_1 ^ \ perp $; $ W_2 = L (\ alpha_2, \ alpha_4 ), u_2 = W_2 ^ \ perp $. use the base vector $ \ ve_1, \ ve_2, \ ve_3, \ ve_4 $ to represent the sub-space $ U_1 + u_2 $ and determine its dimension.

Answer:

(1) $ \ beex \ Bea \ beta_1 & =\ frac {\ alpha_1 }{|\ alpha_1 |}=\ frac {1 }{\ SQRT {3 }} (\ ve_2 + \ ve_3 + \ ve_4 ), \\\ beta_2 <=\ frac {\ alpha_2-(\ alpha_2, \ beta_1) \ beta_1 }{|\ alpha_2-(\ alpha_2, \ beta_1) \ beta_1 | }=\ frac {1} {\ SQRT {15} (3 \ ve_1-2 \ ve_2 + \ ve_3 + \ ve_4 ), \\\ beta_3 <=\ frac {\ alpha_3-(\ alpha_3, \ beta_1) \ beta_1-(\ alpha_3, \ beta_2) \ beta_2 }{| \ alpha_3-(\ alpha_3, \ beta_1) \ beta_1-(\ alpha_3, \ beta_2) \ beta_2 |}=\ frac {1 }{\ SQRT {35 }}( 3 \ ve_1 + 3 \ ve_2-4 \ ve_3 + \ ve_4 ), \\\ beta_4 <=\ frac {\ alpha_4-(\ alpha_4, \ beta_1) \ beta_1-(\ alpha_4, \ beta_2) \ beta_2-(\ alpha_4, \ beta_3) \ beta_3 }{| \ alpha_4-(\ alpha_4, \ beta_1) \ beta_1-(\ alpha_4, \ beta_2) \ beta_2-(\ alpha_4, \ beta_3) \ beta_3 |}=\ frac {1} {\ SQRT {7} (\ ve_1 + \ ve_2 + \ ve_3-2 \ ve_4 ). \ EEA \ eeex $

(2) $ \ BEX (\ beta_1, \ beta_2, \ beta_3, \ beta_4) = (\ ve_1, \ ve_2, \ ve_3, \ ve_4) \ sex {\ BA {CCCC} 0 & \ frac {3} {\ SQRT {15 }}& \ frac {3} {\ SQRT {35 }}& \ frac {1 }{\ SQRT {7 }}\\\ frac {1 }{\ SQRT {3 }}& \ frac {-2 }{\ SQRT {15 }}& \ frac {3 }{\ SQRT {35 }}& \ frac {1 }{\ SQRT {7 }}\\\ frac {1 }{\ SQRT {3 }}& \ frac {1 }{\ SQRT {15 }}& \ frac {-4 }{\ SQRT {35 }}& \ frac {1 }{\ SQRT {7 }}\\ frac {1 }{\ SQRT {3 }}& \ frac {1 }{\ SQRT {15 }}& \ frac {1 }{\ SQRT {35 }}& \ frac {-2 }{\ SQRT {7 }}\ EA }. \ EEx $

(2) by $ \ Bex W_1 = L (\ alpha_1, \ alpha_2) = L (\ beta_1, \ beta_2) \ EEx $ $ \ Bex U_1 = W_1 ^ \ perp = L (\ beta_3, \ beta_4 ). \ EEx $ again by $ \ beex \ Bea \ gamma_2 & =\ frac {\ alpha_2 }{|\ alpha_2 |}=\ frac {1 }{\ SQRT {3}} (\ ve_1 + \ ve_3 + \ ve_4 ), \ gamma_4 & =\ frac {\ alpha_4-(\ alpha_4, \ gamma_2) \ gamma_2 }{| \ alpha_4-(\ alpha_4, \ gamma_2) \ gamma_2 |}=\ frac {1 }{\ SQRT {15 }}( \ ve_1 + 3 \ ve_2 + \ ve_3-2 \ ve_4 ), \ gamma_1 & =\ frac {\ alpha_1-(\ alpha_1, \ gamma_2) \ gamma_2-(\ alpha_1, \ gamma_4) \ gamma_4 }{| \ alpha_1-(\ alpha_1, \ gamma_2) \ gamma_2-(\ alpha_1, \ gamma_4) \ gamma_4 |}=\ frac {1 }{\ SQRT {35 }}(-4 \ ve_1 + 3 \ ve_2 + \ ve_3 + 3 \ ve_4 ), \ gamma_3 & =\ frac {\ alpha_3-(\ alpha_3, \ gamma_2) \ gamma_2-(\ alpha_3, \ gamma_4) \ gamma_4-(\ alpha_3, \ gamma_1) \ gamma_1 }{| \ alpha_3-(\ alpha_3, \ gamma_2) \ gamma_2-(\ alpha_3, \ gamma_4) \ gamma_4-(\ alpha_3, \ gamma_1) \ gamma_1 |}=\ frac {1} {\ SQRT {7} (\ ve_1 + \ ve_2-2 \ ve_3 + \ ve_4 ). \ EEA \ eeex $ zhi$ $ \ Bex W_2 = L (\ alpha_2, \ alpha_4) = L (\ gamma_2, \ gamma_4 ), \ quad u_2 = W_2 ^ \ perp = L (\ gamma_1, \ gamma_3 ). \ EEx $ \ Bex U_1 + u_2 = L (\ beta_3, \ beta_4, \ gamma_1, \ gamma_3 ). \ EEx $ because $ \ BEX (\ beta_3, \ beta_4, \ gamma_1, \ gamma_3) = (\ ve_1, \ ve_2, \ ve_3, \ ve_4) \ sex {\ BA {CCCC} \ frac {3} {\ SQRT {35 }}& \ frac {1} {\ SQRT {7 }}& \ frac {-4} {\ SQRT {35 }}& \ frac {1 }{\ SQRT {7 }}\\ frac {3 }{\ SQRT {35 }}& \ frac {1} {\ SQRT {7 }}& \ frac {3 }{\ SQRT {35 }}& \ frac {1 }{\ SQRT {7 }}\\ frac {-4 }{\ SQRT {35 }}& \ frac {1 }{\ SQRT {7 }}& \ frac {1 }{\ SQRT {35 }}& \ frac {-2 }{\ SQRT {7 }}\\\ frac {1 }{\ SQRT {35 }}& \ frac {-2 }{\ SQRT {7 }}& \ frac {3} {\ SQRT {35 }}& \ frac {1} {\ SQRT {7 }}\ EA} \ EEx $ and (column transformation) $ \ Bex \ sex {\ BA {CCCC} 3 & 1 &-4 & 1 \ 3 & 3 & 1 \-4 & 1 & 1 &-2 \ 1 &-2 & 3 & 1 \ EA} \ RRA \ sex {\ BA {CCCC} 1 & \ & 1 & \ & 1 &\ \-1 &-1 & \ EA} \ EEx $ Zhi $ \ Bex U_1 + u_2 = L (\ ve_1, \ ve_2, \ ve_3), \ quad \ dim (U_1 + u_2) = 3. \ EEx $

 

7 ($ 20' $) set $ N $ square matrix $ A $ meet $ A ^ 2 = 2a $ and $ \ rank (A) = r$.

(1) proof: $ \ rank (A-2E) = N-r$;

(2) proof: $ A $ is similar to diagonal arrays;

(3) determine the determinant $ | A-E | $ value.

Answer:

(1) by $ A (A-2E) = 0 $ Zhi $ \ rank (A-2E) \ Leq N-r$; and by $-A + (A-2E) =-2E $ zhi$ \ rank (A-2E) \ geq N-r$.

(2) set the basic solution for $ AX = 0 $ to $ \ alpha_1, \ cdots, \ Alpha _ {n-r }$, $ (A-2E) the basic solution for x = 0 $ is $ \ Alpha _ {n-R + 1}, \ cdots, \ alpha_n $, it is easy to know that $ \ sed {\ alpha_ I }_{ I = 1} ^ N $ is a group of bases and $ \ Bex a (\ alpha_1, \ cdots, \ alpha_n) = (\ alpha_1, \ cdots, \ alpha_n) \ sex {\ BA {cccccc} 0 & \ ddots & \ & 0 & \ & 2 &&\ \ & \ ddots & \\& & 2 \ EA }. \ EEx $

(3) by $ \ BEX (A-E) (\ alpha_1, \ cdots, \ alpha_n) = (\ alpha_1, \ cdots, \ alpha_n) \ sex {\ BA {cccccc}-1 & \\& \ ddots & \\& &-1 & \\& & 1 & & \& & \ ddots & \\& & 1 \ EA} \ EEx $ $ | A-E | = (-1) ^ {n-r} $.

 

8 ($ 15' $) set $ A $ to the $ N $ square matrix on the number field $ \ BBP $, $ f (x), g (x) $ is the polynomial on the number field $ \ BBP $, and $ (f (x), g (x) = 1 $. for $ h (x) = f (x) g (x) $, use $ v_1, V_2, and V $ to represent the $ N $ homogeneous linear equations $ F () X = 0, g (a) x = 0, H (a) x = 0 $ solution space, here $ x = (x_1, \ cdots, X_n) ^ t $. proof: $ v = v_1 \ oplus V_2 $.

Proof: set by question, $ \ Bex UF + VG = 1. \ EEx $ for $ \ forall \ Alpha \ In V $, $ \ Bex \ alpha = U (a) f (a) \ Alpha + V () G (a) \ Alpha \ in V_2 + v_1. \ EEx $ so $ v = v_1 + V_2 $. and by $ \ Bex \ beta \ In v_1 \ cap V_2 \ Ra \ Beta = u (a) f (a) \ Beta + V (a) g () \ Beta = 0 \ EEx $ Zhi $ v = v_1 \ oplus V_2 $.