07th sets of simulated exam for Mathematics Competition Training of Gannan Normal University

1. set the integer $ n \ geq 2 $, and $ A_1, A_2, \ cdots, a_n $ is different integers. prove that polynomials $ \ Bex f (x) = (x-a_1) \ cdots (x-a_2) + 1 \ EEx $ are not allowed in rational number fields.

Proof: Use the reverse verification method. if (any non-zero rational coefficient polynomial can be written as the product of a rational number and a primitive polynomial) $ \ Bex f (x) = g (x) h (x ), \ quad g (x) \ In \ bbz [X], \ h (x) \ In \ bbz [X], \ Quad 1 \ Leq \ deg g (x ), \ deg h (x) <n. \ EEx $ then $ \ Bex g (a_ I) H (a_ I) = f (a_ I) =-1. \ EEx $ notice $ G (a_ I), H (a_ I) \ In \ bbz $, and $ \ Bex g (a_ I) = \ pm 1, \ quad H (a_ I) = \ MP 1 \ rA g (a_ I) + H (a_ I) = 0. \ EEx $ this indicates the number of times $ <N $ polynomial $ g (x) + h (x) $ there are $ N $ different root types. this is in conflict with the basic theorem of modern mathematics. therefore, there is a conclusion.

 

 

2. (1) set $ A $ to $ N $ square matrix, and $ A $ idempotence (that is, $ A ^ 2 = A $). Test Certificate: $ \ rank () = \ tr (a) $.

(2) set $ A $ square matrix $ A_1, \ cdots, A_k $ to meet $ a_1 + \ cdots + A_k = e $. test Certificate: the required and sufficient conditions for square matrix $ A_1, \ cdots, A_k $ idempotence are $ \ Bex \ rank (A_1) + \ cdots + \ rank (A_k) = n. \ EEx $

Proof: (1) by $ A ^ 2 = A \ rA A (E-A) = 0 $ Zhi $ \ Bex \ rank (A) + \ rank (E-A) \ Leq n. \ EEx $ again by $ A + (E-A) = E $ Zhi $ \ Bex \ rank (A) + \ rank (E-A) \ geq n. \ EEx $ so $ \ Bex \ rank (A) + \ rank (E-A) = n. \ EEx $ note $ \ rank (A) = r$, $ (E-A) x = 0 $ basic solution $ \ Bex \ al_1, \ cdots, \ al_r; the basic solution for \ EEx $ AX = 0 $ is $ \ Bex \ Al _ {R + 1}, \ cdots, \ al_n. \ EEx $ it is easy to know $ \ al_1, \ cdots, \ al_n $ Linear Independence, $ \ Bex a (\ al_1, \ cdots, \ al_n) = (\ al_1, \ cdots, \ al_n) \ sex {\ BA {CC} e_r & 0 \ 0 & 0 \ EA }. \ EEx $ therefore $ A $ is similar to $ \ DPS {\ sex {\ BA {CC} e_r & 0 \ 0 & 0 \ EA} $. so, $ \ rank (A) = \ tr (a) $. (2) $ \ rA $: If $ a_ I ^ 2 = a_ I, \ I = 1, \ cdots, K $, then $ \ rank (a_ I) = \ tr (a_ I) $, and $ \ Bex \ sum _ {I = 1} ^ k \ rank (a_ I) = \ sum _ {I = 1} ^ k \ tr (a_ I) = \ tr \ sex {\ sum _ {I = 1} ^ K a_ I} = \ tr E = n. \ EEx $ \ la $: Conclusion when the first certificate $ k = 2 $ \ Bex a ^ 2 = A \ LRA \ rank () + \ rank (E-A) = n. \ EEx $ necessity: by (1), $ \ Bex \ rank (A) + \ rank (E-A) = \ tr (A) + \ tr (E-A) = \ tr E = n. \ EEx $ adequacy: Same as (1), $ \ Bex a (\ al_1, \ cdots, \ al_n) = (\ al_1, \ cdots, \ al_n) \ sex {\ BA {CC} e_r & 0 \ 0 & 0 \ EA }. \ EEx $ this indicates $ A ^ 2 = A $. general evidence. the necessity is obvious. to prove adequacy, you only need to prove $ \ rank (a_ I) + \ rank (E-A_ I) \ Leq N $ ($ \ geq $ apparently) as follows: $ \ beex \ Bea \ rank (E-A_ I) & =\ rank (a_1 + \ cdots + A _ {I-1} + A _ {I + 1} + \ cdots + A_k) \ & \ Leq \ rank (A_1) + \ cdots + \ rank (A _ {I-1}) + \ rank (A _ {I + 1 }) + \ cdots + \ rank (A_k) \\\& = N-\ rank (a_ I ). \ EEA \ eeex $

 

 

3. Set $ A, B $ to $ N $ square matrix, $ AB = BA = 0 $. test certificate:

(1) A positive integer $ K $ makes $ \ Bex \ rank (a ^ K + B ^ K) = \ rank (a ^ K) + \ rank (B ^ K); \ EEx $

(2) If $ \ rank (a ^ 2) = \ rank (a) $, then $ \ Bex \ rank (a + B) = \ rank () + \ rank (B ). \ EEx $

Proof: we verify first (2 ). take a group of base $ e_1, \ cdots, e_n $ and the linear transformation on $ \ SCRA $, $ \ scrb $ from $ \ BBF ^ N $. They are in the base $ e_1, \ cdots, the matrices under e_n $ are $ A $, $ B $, respectively. by $ AB = BA = 0 $ Zhi $ \ Bex \ SCRA \ scrb = \ scrb \ SCRA = \ scro. \ EEx $ by $ R (a ^ 2) = R () $ Zhi $ A ^ 2X = 0 \ lra ax = 0 $ (apparently $ \ Bex v_1 \ equiv \ sed {\ al; A \ Al = 0} \ subset \ sed {\ al; a ^ 2 \ Al = 0} \ equiv V_2, \ EEx $ but $ \ dim v_1 = N-\ rank (A) = N-\ rank (a ^ 2) = \ dim V_2 $ ), $ \ Bex \ Ker \ SCRA ^ 2 = \ Ker \ SCRA. \ EEx $ use formula above and dimension formula to know $ \ bee \ label {direct} \ im \ SCRA \ oplus \ Ker \ SCRA = \ BBF ^ n. \ EEE $ we have $ \ beex \ BEA (\ SCRA + \ scrb) \ im \ SCRA & =\ SCRA \ im \ SCRA \ quad \ sex {\ scrb \ SCRA = \ scro} \ & =\ im \ SCRA \ quad \ sex {\ subset: \ mbox {apparently };\\ supset: \ by \\ eqref {direct };\\ (\ SCRA + \ scrb) \ Ker \ SCRA & =\ scrb \ Ker \ SCRA \ & =\ im \ scrb \ quad \ sex {\ scrb \ SCRA = \ scro, \ eqref {direct }}. \ EEA \ eeex $ then $ \ beex \ Bea \ rank (a + B) & =\ dim \ im (\ SCRA + \ scrb) = \ dim (\ SCRA + \ scrb) \ BBF ^ n \ & = \ dim (\ SCRA + \ scrb) (\ im \ SCRA \ oplus \ Ker \ SCRA) \\&=\ dim [(\ SCRA + \ scrb) \ im \ SCRA + (\ SCRA + \ scrb) \ Ker \ SCRA] \\\&=\ dim (\ im \ SCRA + \ im \ scrb) \ & = \ dim \ im \ SCRA + \ dim \ im \ scrb \ & \ quad \ sex {\ im \ scrb \ subset \ Ker \ SCRA, \ quad \ im \ SCRA \ oplus \ Ker \ SCRA = \ BBF ^ n }\\&=\ rank (A) + \ rank (B ). \ EEA \ eeex $ Verify again (1 ). therefore, verify that $ K $ exists so that $ \ rank (a ^ K) = \ rank (a ^ {k + 1}) = \ cdots $. in fact, remember $ V_ I =\sed {\ al; a ^ I \ Al = 0} $, then $ v_1 \ subset V_2 \ subset \ cdots $. so $ \ Bex \ exists \ K, \ st V_K = V _ {k + 1 }. \ EEx $ so, $ \ beex \ Bea \ Al \ In V _ {k + 2} & \ rA a ^ {k + 2} \ Al = a ^ {k + 1} (\ al) = 0 \ & \ rA A \ Al \ In V _ {k + 1} = V_K \ & \ rA a ^ {k + 1} \ Al = a ^ K (A \ Al) = 0 \ & \ Ra \ Al \ In V _ {k + 1}, \ EEA \ eeex $, and so on. this indicates $ V_K = V _ {k + 1 }=\ cdots $, and $ \ Bex \ rank (a ^ K) = \ rank (a ^ {k + 1}) = \ cdots. \ EEx $ special, $ \ rank (a ^ K) = \ rank (a ^ {2 k}) $. permit (1 ). since $ AB ^ K = B ^ Ka = 0 $, $ \ rank (a ^ K) ^ 2) = \ rank (a ^ K) $, we set) that is, $ \ Bex \ rank (a ^ K + B ^ K) = \ rank (a ^ K) + \ rank (B ^ K ). \ EEx $

 

 

4. Set $ N $ square matrix $ A $ positive.

(1) $ B _1, \ cdots, B _n $ is any $ N $ non-zero real number. Test Certificate: matrix $ B = (A _ {IJ} B _ib_j) $ is also positive.

(2) $ B $ is $ n \ times M $ real matrix, and $ \ rank (B) = M $. Test Certificate: $ B ^ tab $ is also positive.

(3) $ B $ is $ N $ level positive definite matrix. Test Certificate: $ c = (A _ {IJ} B _ {IJ}) $ is also positive.

Proof: (1) for $ x = (x_1, \ cdots, X_n) ^ t \ NEQ 0 $, $ \ 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 \ sex {Y = (y_1, \ cdots, Y_n) ^ t, \ y_ I = B _ix_ I} \ &> 0. \ EEA \ eeex $ (2) to $ \ forall \ x = (x_1, \ cdots, X_n) ^ t \ NEQ 0 $, $ \ beex \ Bea x ^ t B ^ tabx & = y ^ Tay \ quad \ sex {Y = Bx \ NEQ 0 }\&> 0. \ EEA \ eeex $ (3) set $ B = C ^ TC $, where $ C $ is reversible, then $ \ Bex B _ {IJ} = \ sum_k C _ {Ki} C _ {kJ }, \ quad A _ {IJ} B _ {IJ} = \ sum_k C _ {Ki} A _ {IJ} C _ {kJ }. \ EEx $ pairs $ \ forall \ x = (x_1, \ cdots, X_n) ^ t \ NEQ 0 $, $ \ beex \ Bea \ sum _ {IJ} A _ {IJ} x_ix_j & =\ sum _ {ijk} C _ {Ki} x_ia _ {IJ} C _{ KJ} x_j \\&=\ sum_k \ sum _ {IJ} A _ {IJ} y ^ {(k )} _ Iy ^ {(k)} _ j \ quad \ sex {y ^ {(k)} _ I = C _ {Ki} X_ I, \ y ^ {(k )} \ NEQ 0 }\\&> 0. \ EEA \ eeex $

 

 

5. set $ v = \ BBC ^ {n \ times n} $ to represent $ N on the complex field $ \ BBC $. Linear Combination of the addition of the matrix and the multiplication of the number of Matrices space, $ A \ In V $, define the transform $ \ SCRA $ on $ V $ as follows: $ \ Bex \ SCRA (x) = AX-XA, \ quad \ forall \ x \ In v. \ EEx $ test certificate:

(1) $ \ SCRA $ is a linear transformation;

(2) $ \ SCRA (xy) = x \ SCRA (y) + \ SCRA (x) y $;

(3) $0 $ is an feature value of $ \ SCRA $;

(4) If $ A ^ K = 0 $, then $ \ SCRA ^ {2 k} = \ scro $.

Proof: (1) $ \ beex \ Bea \ SCRA (kx + ly) & = a (kx + ly)-(kx + ly) A \ & = K (AX-XA) + L (AY-YA) \ & = k \ SCRA (x) + L \ SCRA (y ). \ EEA \ eeex $ (2) $ \ beex \ Bea \ SCRA (xy) & = AXY-XYA \ & = (AX-XA) Y + x (AY-YA) \\&=\ SCRA (x) y + x \ SCRA (y ). \ EEA \ eeex $ (3) $ \ SCRA (e) = AE-EA = 0 $.

(4) use mathematical induction to learn easily $ \ Bex a ^ s (X) = \ sum _ {p + q = S + 1} C _ {pqs} a ^ pxq ^ Q. \ EEx $ then $ \ Bex \ SCRA ^ {2 k} (x) = \ sum _ {p + q = 2 k + 1} C _ {PQ, 2 k} a ^ PXA ^ q = 0. \ EEx $ this is because when $0 \ Leq P <K $, $ K + 2 \ Leq Q \ Leq 2 k + 1 $, $ A ^ q = 0 $; when $ k \ Leq p \ Leq 2 k + 1 $, $ A ^ p = 0 $.

 

 

6. set $ A, B $ to $ N $, and $ A $ is a non-zero semi-Definite Matrix. $ B $ is a positive definite matrix: $ | a + B |> | B | $.

Proof: There is a reversible matrix by $ B $ Zhengding. $ C $ makes $ \ Bex C ^ TBC = E. \ EEx $ you C ^ TAC $ is still non-zero semi-definite, and an orthogonal array $ p $ exists, making $ \ Bex P ^ TC ^ tacp = \ diag (\ lm_1, \ cdots, \ lm_n) \ NEQ 0. \ EEx $ then $ \ beex \ Bea | P ^ TC ^ t | \ cdot | a + B | \ cdot | CP | & = | E + \ diag (\ lm_1, \ cdots, \ lm_n) |\\&=\ prod _ {I = 1} ^ N (1 + \ lm_ I) \ quad \ sex {(\ lm_1, \ cdots, \ lm_n) \ NEQ 0 }\\&> 1, \\| a + B | &> | B |. \ EEA \ eeex $

 

 

7. set $ A, B $ to $ N $ square matrix, satisfying $ A ^ 2 = A $, $ B ^ 2 = B $, and $ e-(A + B) $ reversible, test certificate: $ R (A) = R (B) $.

Proof: By question 2nd $ \ Bex \ rank (A) + \ rank (E-A) = N, \ quad \ rank (B) + \ rank (E-B) = n. \ EEx $ again by $ (E-A)-B = (E-B)-A $ reversible knowledge $ \ Bex \ rank (E-A) + \ rank (B) \ geq N, \ quad \ rank (E-B) + \ rank (a) \ geq n. \ EEx $ so $ \ beex \ Bea \ rank (a) & = N-\ rank (E-A) \ Leq \ rank (B), \ rank (B) & = N-\ rank (E-B) \ Leq \ rank (). \ EEA \ eeex $ therefore $ \ rank (A) = \ rank (B) $.

 

 

8. set $ s =\sed {AB-BA; a, B \ In \ BBR ^ {n \ times N }}$, test certificate: $ S $ the number of dimensions of the zhangcheng sub-space $ W $ is $ n ^ 2-1 $, and a group of basis is obtained.

Answer: by $ \ beex \ Bea E _ {IJ} & = E _ {II} e _ {IJ}-E _ {IJ} e _ {II} \ Quad (I \ neq j ), \ E _ {11}-E _ {II} & = E _ {1i} e _ {I1}-E _ {I1} e _ {1i} \ Quad (2 \ Leq I \ Leq N) \ EEA \ eeex $ Zhi $ \ Bex \ dim w \ geq N (n-1) + (n-1) = n ^ 2-1. \ EEx $ again by $ \ tr (AB-BA) = 0 $ Zhi $ e \ not \ in W $, and $ \ Bex \ dim W = n ^ 2-1. \ EEx $