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

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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 \$

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