[Journal of mathematics at home University] 308th questions about the Postgraduate Entrance Exam of Huazhong Normal University in 2006

 

1. ($ 14' $) calculate $ N $ level determining factor $ \ Bex D_n = \ sev {\ BA {CCCCC} x_1 + a_1 ^ 2 & a_1a_2 & a_1a_3 & \ cdots & a_1a_n \ a_2a_1 & X_2 + a_2 ^ 2 & a_2a_3 & \ cdots & a_2a_n \ a_3a_1 & a_3a_2 & X_3 + A_3 ^ 2 & \ cdots & a_3a_n \ vdots & \ ddots & \ vdots \ a_na_1 & a_na_2 & a_na_3 & \ cdots & x_n + a_n ^ 2 \ EA }, \ mbox {where} x_1x_2 \ cdots x_n \ NEQ 0. \ EEx $

 

Answer: $ \ beex \ Bea D_n & =\ sev {\ BA {CCCCC} 1 & A_1 & A_2 & \ cdots & a_n \ 0 & x_1 + a_1 ^ 2 & a_1a_2 &\ cdots & a_1a_n \ 0 & a_2a_1 & X_2 + A_2 ^ 2 & \ cdots & a_2a_n \ vdots & \ ddots & \ vdots \ 0 & a_na_1 & a_na_2 & \ cdots & x_n + a_n ^ 2 \ EA }\\\\\sev {\ba {CCCCC} 1 & A_1 & A_2 & \ cdots & a_n \-A_1 & X_1 & \ cdots & 0 \-A_2 & 0 & X_2 & \ cdots & 0 \ vdots & \ ddots & \ vdots \-a_n & 0 & 0 & \ cdots & x_n \ EA} \\\\=\ sev {\ BA {CCCCC} 1 + \ sum _ {I = 1} ^ n \ frac {a_ I ^ 2} {X_ I} & A_1 & A_2 & \ cdots & a_n \ 0 & X_1 & 0 & \ cdots & 0 \ 0 & 0 & X_2 & \ cdots & 0 \ \ vdots & \ ddots & \ vdots \ 0 & 0 & \ cdots & x_n \ EA} \ & = 1 + \ sum _{ I = 1} ^ n \ frac {a_ I ^ 2} {X_ I }. \ EEA \ eeex $

 

2. ($ 20' $) set $ a_ I = (A _ {I1}, A _ {I2}, \ cdots, A _ {In}) $, $ I = 1, 2, \ cdots, r$ and $ \ al_1, \ Al_2, \ cdots, \ al_r $ Linear Independence, $ \ Beta = (B _1, B _2, \ cdots, B _n) $. proof: the necessary conditions for linear correlation between $ \ al_1, \ Al_2, \ cdots, \ al_r, and \ beta $ are: linear Equations $ AX = 0 $ (here $ A = (A _ {IJ}) $) are solutions of equations $ \ Beta X = 0 $.

 

Proof: the solution space for the linear equations $ AX = 0 $ is $ v_1 $, the solution space for Linear Equations $ \ Bex \ sex {\ BA {CC} A \ beta \ EA} x = 0 \ EEx $ is $ V_2 $, then $ \ Bex V_2 \ subset v_1, \ quad \ dim v_1 = N-\ rank (), \ quad \ dim V_2 = N-\ rank \ sex {\ BA {CC} A \ beta \ EA }. \ EEx $ then $ \ beex \ Bea v_1 = V_2 & \ LRA \ rank () = \ rank \ sex {\ BA {CC} A \ beta \ EA} \ & \ LRA \ al_1, \ Al_2, \ cdots, \ al_r, \ beta \ mbox {linear correlation }. \ EEA \ eeex $

 

3. ($24 '$) set $ \ BBR $ to a real number field, $ V $ linear equations $ \ Bex \ BA {rrrrrrrrrrl} 2x_1 & + & 4x_2 &-& 2x_3 & + & 4x_4 &-& 7x_5 & = & 0 \ 2x_1 & & + & 2x_3 &-& 4x_4 &-& x_5 & = & 0 \ 3x_1 &-& X_2 & + & 4x_3 & + & 4x_4 &-& 4x_5 & = & 0 \ 4x_1 &-& 2x_2 & + & 6x_3 & + & 3x_4 &-4x_5 & = & 0 \ EA \ EEx $ is a set of all solutions.

(1) prove that $ V $ is the sub-space of $ \ BBR ^ 5 $ (space composed of column vectors;

(2) Calculate the basis and dimension of $ V $;

(3) Calculate the base and dimension of the orthogonal complement $ V ^ \ perp $ for $ V $ ($ \ BBR ^ 5 $ Inner Product $ (\ Al, \ beta) = \ Al ^ t \ beta $ ).

 

Answer:

(1) apparently.

(2) by $ \ Bex a =\sex {\ BA {CCCCC} 2 & 4 &-2 & 4 &-7 \ 2 & 0 & 2 &-4 &-1 \ \ 3 &-1 & 4 & 4 &-4 \ 4 &-2 & 6 & 3 &-4 \ EA} \ RRA \ sex {\ BA {CCCCC} 6 & 0 & 6 & 0 &-7 \ 0 & 6 &-6 & 0 &-5 \ 0 & 0 & 3 &-1 \ 0 & 0 & 0 & 0 & 0 & 0 \ EA} \ EEx $ a group of optional $ V $ bases is $ \ Bex \ sex {\ BA {CCCCC}-1 \ 1 \ 1 \ \ 0 \ 0 \ EA }, \ quad \ sex {\ BA {CCCCC} 7 \ 5 \ 0 \ 2 \ 6 \ EA }, \ EEx $ while $ V $ has a dimension of $2 $.

(3) $ \ dim V ^ \ perp = 3 $. set $ Y = (y_1, \ cdots, y_5) ^ t \ In V ^ \ perp $, then $ \ Bex \ BA {rrrrrrrrl}-Y_1 & + & Y_2 & + & y_3 & = & 0, \ 7y_1 & + & 5y_2 & + & 2y_4 & + & 6y_5 & = & 0. \ EA \ EEx $ a group of keys for $ V ^ \ perp $ \ Bex \ sex {\ BA {CCCCC} 6 \ 0 \ 6 \ 0 \ \-7 \ EA }, \ quad \ sex {\ BA {CCCCC} 0 \-6 \ 6 \ 0 \ 5 \ EA }, \ quad \ sex {\ BA {CCCCC} 0 \ 0 \ 0 \ 3 \-1 \ EA }. \ EEx $

 

4. ($ 32' $) set $ \ BBP $ to a number field, $ \ Bex v =\sed {f (x) \ In \ BBP [X]; \ f (x) = 0 \ mbox {or} \ P (f (x) <n }. \ EEx $ for any $ f9x) = A _ {n-1} x ^ {n-1} + \ cdots + a_1x + a_0 \ In V, $ rules $ \ Bex \ SCRA: \ quad f (x) \ mapsto A _ {n-1} x ^ {n-1 }. \ EEx $

(1) prove that $ \ SCRA $ is a linear transformation of $ V $;

(2) Evaluate the matrix $ \ SCRA $ under base $ X {n-1}, x ^ {N-2}, \ cdots, X, 1 $;

(3) calculate a group of cores $ \ SCRA ^ {-1} (0) $ for $ \ SCRA $;

(4) Evaluate all feature values and feature vectors of $ \ SCRA $.

 

Answer:

(1) apparently.

(2) $ \ Bex \ SCRA (x ^ {n-1}, x ^ {N-2}, \ cdots, 1) = (x ^ {n-1 }, x ^ {N-2}, \ cdots, 1) \ diag (, \ cdots, 0 ). \ EEx $

(3) $ \ SCRA ^ {-1} (0) = \ span \ sed {x ^ {N-2}, \ cdots, 1} $.

(4) The feature value of $ \ SCRA $ is $1 $ (single weight), $0 $ ($ n-1 $ weight ), the corresponding feature vectors are $ \ Bex x ^ {n-1}; \ quad x ^ {N-2}, \ cdots, 1. \ EEx $

 

5. ($ 20' $) set $ \ BBP $ to a number field, $ a, B \ In \ BBP ^ {n \ times N }$, $ c = AB-BA $, and $ BC = CB $. proof:

(1) For a natural number greater than $1 $, there are $ AB ^ K-B ^ Ka = kb ^ {k-1} C $;

(2) set $ F (\ lm) $ to the feature polynomial of $ B $, $ f' (\ lm) $ to the derivative of $ F (\ lm) $, then $ f' (B) C = 0 $.

 

Proof:

(1) Conclusion $ k \ geq 1 $ is true. it is proved by mathematical induction. when $ k = 1 $, the conclusion is obvious. if the conclusion is true when $ k = N $, then when $ k = n + 1 $, $ \ beex \ Bea AB ^ {n + 1}-B ^ {n + 1} A & = (AB ^ N) B-B (B ^ Na) \ & = (AB ^ N-B ^ Na) B-B (B ^ Na-AB ^ N) + B ^ NAB-Bab ^ n \ & = (NB ^ {n-1} c) B-B (-Nb ^ {n-1} c) + B ^ N (AB-BA) -(BA-AB) B ^ n \ quad + B ^ {n + 1} A-AB ^ {n + 1} \ & = Nb ^ nC + NB ^ nC + B ^ nC + CB ^ N-(AB ^ {n + 1}-B ^ {n + 1}) \ & = 2 (n + 1) B ^ nc-(AB ^ {n + 1}-B ^ {n + 1} ). \ EEA \ eeex $ then $ \ Bex AB ^ {n + 1}-B ^ {n + 1} A = (n + 1) B ^ NC. \ EEx $

(2)

(1) and Hamilton-caylay theorem, $ \ Bex 0 = AF (B)-f (B) A = f' (B) C. \ EEx $

 

6. ($ 20' $) set $ \ BBR $ to a real number field, $ A \ In \ BBR ^ {n \ times N }$, and $ A $ to a symmetric matrix.

(1) prove that $ A $'s adjoint matrix $ A ^ * $ is also a real symmetric matrix;

(2) What are the sufficient conditions for the contract between $ A $ and $ A ^ * $? And prove your conclusion.

 

Proof:

(1) The real matrix is changed from $ \ BBR $ to the number field $ A ^ * = (A _ {IJ}) $. by $ \ beex \ Bea a _ {IJ} & = (-1) ^ {I + J} \ sev {\ BA {cccccc} A _ {11} & \ cdots & A _ {1, J-1} & A _ {1, J + 1} & \ cdots & A _ {1N} \ vdots & \ vdots \ A _ {I-1, 1} & \ cdots & A _ {I-1, J-1} & A _ {I-1, J + 1} & \ cdots & A _ {I-1, n} \ A _ {I +} & \ cdots & A _ {I + 1, J-1} & A _ {I + 1, J + 1} & \ cdots & A _ {I + 1, N }\\\ vdots & \ vdots \ A _ {N1} & \ cdots & A _ {n, j-1} & A _ {n, J + 1} & \ cdots & A _ {NN} \ EA} \ & = (-1) ^ {I + J} \ sev {\ BA {cccccc} A _ {11} & \ cdots & A _ {J-1, 1} & A _ {J +} & \ cdots & A _ {N1} \ vdots & \ vdots \ A _ {1, i-1} & \ cdots & A _ {J-1, I-1} & A _ {J + 1, I-1} & \ cdots & A _ {n, i-1} \ A _ {1, I + 1} & \ cdots & A _ {J-1, I + 1} & A _ {J + 1, I + 1} & \ cdots & A _ {n, I + 1 }\\\ vdots & \ vdots \ A _ {1N} & \ cdots & A _ {J-1, n} & A _ {J + 1, n} & \ cdots & A _ {NN} \ EA} \ quad \ sex {A _ {IJ} = A _ {Ji }\\& = (-1) ^ {J + I} \ sev {\ BA {cccccc} A _ {11} & \ cdots & A _ {1, I-1} & A _ {1, I + 1} & \ cdots & A _ {1N} \ vdots & \ vdots \ A _ {J-1, 1} & \ cdots & A _ {J-1, I-1} & A _ {J-1, I + 1} & \ cdots & A _ {J-1, n} \ A _ {J +} & \ cdots & A _ {J + 1, I-1} & A _ {J + 1, I + 1} & \ cdots & A _ {J + 1, N }\\\ vdots & \ vdots \ A _ {N1} & \ cdots & A _ {n, i-1} & A _ {n, I + 1} & \ cdots & A _ {1N} \\\ EA} \ quad \ sex {| B ^ t |=| B |}\\& = _{ ji }. \ EEA \ eeex $ Zhi $ A ^ * $ symmetry.

(2) We know from $ A, a ^ * $ real symmetry that they are orthogonal and diagonal arrays, $ A $, $ A ^ * $ respectively contract with $ \ Bex \ diag \ sex {\ lm_1, \ cdots, \ lm_n }, \ quad \ diag \ sex {\ prod _ {I \ NEQ 1} \ lm_ I, \ cdots, \ prod _ {I \ NEQ n} \ lm_ I }. \ EEx $ A, a ^ * $ the contract is equivalent to the above two diagonal matrix contracts, the feature value of $ A $ is greater than $0 $, which guarantees the above two diagonal matrix contracts (both contracts are in the unit matrix ). therefore, $ A $ zhengding is a sufficient condition for the $ A, a ^ * $ contract.

 

7. ($ 20' $) set $ V $ to the $ N $ dimension linear space on the number field $ \ BBP $, $ \ ve_1, \ cdots, \ ve_r, \ ve _ {R + 1}, \ cdots, \ ve_n $ is a group of bases of $ V $, $ \ Bex v_1 = L (\ ve_1, \ cdots, \ ve_r), \ quad V_2 = L (\ ve _ {R + 1}, \ cdots, \ ve_n ). \ EEx $

(1) prove $ v = v_1 \ oplus V_2 $;

(2) set $ \ SCRA $ to a linear transformation of $ v_1 $. $ \ scrb $ is a linear transformation of $ V_2 $, evaluate the linear transformation of $ V $ \ SCRC $ so that $ v_1 and V_2 $ are all $ \ SCRC $-constant subspaces, and $ \ SCRC $ is in $ v_1, the display on V_2 $ is $ \ Bex \ SCRC | _ {v_1 }=\ SCRA, \ quad \ SCRC |_{ V_2 }=\ scrb. \ EEx $

 

Proof:

(1) apparently.

(2) $ \ Bex \ SCRC \ sex {\ sum _ {I = 1} ^ n K_ I \ ve_ I }=\ sum _ {I = 1} ^ r K_ I \ SCRA (\ ve_ I) + \ sum _ {I = R + 1} ^ n K_ I \ scrb (\ ve_ I) \ EEx $.

[Journal of mathematics at home University] 308th questions about the Postgraduate Entrance Exam of Huazhong Normal University in 2006