08th sets of simulated exam for Mathematics Competition Training of Gannan Normal University

1. set $ A, B $ to $ N $ square matrix, $ \ rank (a) <N $, and $ A = B _1 \ cdots B _k $, $ B _ I ^ 2 = B _ I $, $ I = 1, \ cdots, K $. exam: $ \ Bex \ rank (E-A) \ Leq k \ SEZ {n-\ rank ()}. \ EEx $

Proof: $ \ beex \ Bea \ rank (E-A) & =\ rank (E-B_1 \ cdots B _k) \ & = \ rank (E-B_1 + B _1 (E-B_2 \ cdots B _k )) \ & \ Leq \ rank (E-B_1) + \ rank (E-B_2 \ cdots B _k) \ & = N-\ rank (B _1) + \ rank (E-B_2 \ cdots B _k) \ & \ Leq N-\ rank (A) + \ rank (E-B_2 \ cdots B _k) \\& \ Leq \ cdots \\\& \ Leq k \ SEZ {n-\ rank ()}. \ EEA \ eeex $

 

 

2. note $ \ lm_1 (a) \ geq \ lm_2 (a) \ geq \ cdots \ geq \ lm_n (a) $ is the feature value of $ N $ level real symmetric matrix $ A $. test certificate:

(1) If $ A and B $ are real symmetric matrices and the real number $0 \ Leq A \ Leq 1 $, then for $ I = 1, 2, \ cdots, N $, $ \ Bex a \ lm_ I (A) + (1-A) \ lm_n (B) \ Leq \ lm_ I (AA + (1-A) B) \ Leq A \ lm_ I () + (1-A) \ lm_1 (B); \ EEx $

(2) If $ B $ is semi-definite, then $ \ Bex \ lm_1 (a + B) \ geq \ lm_1 (A), \ quad \ lm_n (A + B) \ geq \ lm_n (). \ EEx $

Proof: (1) notes $ A \ lm_ I (A) = \ lm_ I (AA), \ A \ geq 0 $, we only need, B $ proof of real symmetry: $ \ Bex \ lm_ I (A) + \ lm_n (B) \ Leq \ lm_ I (a + B) \ Leq \ lm_ I () + \ lm_1 (B ). \ EEx $ there is a reversible matrix by $ A $ Real Symmetric knowledge $ P = (\ al_1, \ cdots, \ al_n) $ make $ \ Bex ap = p \ diag (\ lm_1 (A), \ cdots, \ lm_n ()). \ EEx $ it is also known by $ A + B $ that there is a reversible array of real symmetry $ q = (\ beta_1, \ cdots, \ beta_n) $ make $ \ BEX (a + B) q = Q \ diag (\ lm_1 (A + B), \ cdots, \ lm_n (A + B )). \ EEx $ note $ \ Bex v_1 = span \ sed {\ al_ I, \ cdots, \ al_n}, \ quad V_2 = span \ sed {\ beta_1, \ cdots, \ beta_ I}, \ EEx $ by $ \ beex \ Bea \ dim (v_1 \ cap V_2) & =\ dim v_1 + \ dim V_2-\ dim (v_1 + V_2) \ & \ geq (n-I + 1) + I-n \\&> 0 \ EEA \ eeex $ \ Bex \ exists \ 0 \ NEQ \ gamma \ In v_1 \ cap V_2, \ st | \ gamma | = 1. \ EEx $ then $ \ beex \ Bea \ lm_ I (a + B) & \ Leq \ SEF {(a + B) \ gamma, \ gamma }\\\\=\ SEF {A \ gamma, \ gamma }+ \ SEF {B \ gamma, \ gamma }\\\& \ Leq \ lm_ I () + \ lm_1 (B ). \ EEA \ eeex $ Similarly, note $ \ Bex W_1 = \ span \ sed {\ al_1, \ cdots, \ al_ I }, \ quad W_2 = \ span \ sed {\ beta_ I, \ cdots, \ beta_n }, \ EEx $ then $ \ Bex \ exists \ 0 \ NEQ \ Delta \ In W_1 \ cap W_2, \ st | \ Delta | = 1. \ EEx $ then $ \ beex \ Bea \ lm_ I (a + B) & \ geq \ SEF {(a + B) \ delta, \ Delta }\\\\=\ SEF {A \ delta, \ Delta }+ \ SEF {B \ delta, \ Delta }\\\\\=\ geq \ lm_ I () + \ lm_n (B ). \ EEA \ eeex $ (2) if $ B $ is semi-definite, then $ \ lm_ I (B) \ geq 0, \ I = 1, 2, \ cdots, N $. therefore, from (1), $ \ beex \ Bea \ lm_1 (a + B) & \ geq \ lm_1 (A) + \ lm_n (B) \ geq \ lm_1 (A), \\\ lm_n (a + B) & \ geq \ lm_n (A) + \ lm_n (B) \ geq \ lm_n (B ). \ EEA \ eeex $

 

 

3. set $ A $ to $ N $ order reversible matrix. Test Certificate: orthogonal matrix exists $ p $, positive matrix $ u, v $ to make $ \ Bex a = Ru = VR. \ EEx $

Proof: from $ A $ reversible knowledge $ A ^ ta $ positive definite, there is an orthogonal array $ q $ to make $ \ Bex a ^ TA = Q ^ t \ diag (\ lm_1, \ cdots, \ lm_n) Q. \ EEx $ get $ \ Bex d = \ diag (\ SQRT {\ lm_1}, \ cdots, \ SQRT {\ lm_n }), \ quad P = AQ ^ TD ^ {-1}, \ EEx $ then $ \ Bex A = (aq ^ TD ^ {-1}) DQ \ equiv PDQ, \ EEx $ and $ p $ and $ q $ Are Orthogonal Arrays: $ \ Bex P ^ TP = d ^ {-1} QA ^ Taq ^ TD ^ {-1} = E. \ EEx $ then $ \ beex \ Bea Q & = (PQ) (q ^ tdq) \ equiv Ru \ & = (PDP ^ t) (PQ) \ equiv VR. \ EEA \ eeex $

 

 

4. set $ A and B $ to level $ M $ and level $ N $ matrices respectively. test evidence: $ A, B $ a matrix equation $ AX = XB $ only has zero solutions.

Proof: $ \ rA $: by $ AX = XB $ Zhi $ \ Bex a ^ 2X = a (ax) = a (XB) = (ax) B = (XB) B = XB ^ 2, \ EEx $ and so on, while $ \ Bex F (a) x = XF (B), \ EEx $ where $ F (\ lm) $ is the feature polynomial of $ A $. by the Hamilton-caylay theorem, $ \ Bex 0 = XF (B ). \ EEx $ because $ A and B $ have no common feature values, while $ F (B) $ reversible, $ x = 0 $. $ \ rA $: Use reverse identification. if $ A, B $ has a common feature value $ \ lm $, then by $ \ Bex | \ lm E-B | = (\ lm E-B) ^ t | = | \ lm E-B ^ t | \ EEx $ know $ B ^ t $ also has feature value $ \ lm $. set $ \ Al, \ beta $ to $ A, B ^ t $, which belongs to feature vectors with feature values $ \ lm $, then $ \ beex \ Bea a \ Al & =\ lm \ Al, \ B ^ t \ beta & =\ lm \ beta, \ A \ Al \ beta ^ t & = \ lm \ Al \ beta ^ t = \ Al (\ lm \ beta) ^ t = \ Al (B ^ t \ beta) ^ t = \ Al \ beta ^ t B. \ EEA \ eeex $ hence $ AX = XB $ has a non-zero solution $ \ Al \ beta ^ t $. this is a conflict. therefore, there is a conclusion.

 

 

5. set $ \ lm_1, \ cdots, \ lm_n $ to all feature values of $ N $ level matrix $ A $, but $-\ lm_ I \ (I = 1, 2, \ cdots, n) $ is not the feature value of $ A $. It defines a linear transformation of $ \ BBR ^ {n \ times n} $ \ Bex \ SCRA (x) = a ^ Tx + XA, \ quad \ forall \ x \ In \ BBR ^ {n \ times n }. \ EEx $

(1) test: $ \ SCRA $ is a reversible linear transformation;

(2) For any real symmetric matrix $ C $, a unique real symmetric matrix $ B $ must exist, making $ A ^ TB + BA = C $.

Proof: (1) formula by dimension $ \ Bex \ dim \ SCRA ^ {-1} (0) + \ dim \ im \ SCRA = n \ EEx $ only proof $ \ Bex \ SCRA (X) = 0 \ LRA a ^ Tx =-XA = x (-a) \ rA x = 0. \ EEx $, which can be obtained immediately by question 4th ($ A ^ t $ has the same feature value as $ A $ ).

(2) by (1) knowing that a unique Matrix exists $ B $ makes $ A ^ TB + BA = C $. $ B ^ Ta + A ^ TB ^ t = C. $ also known by uniqueness $ B ^ t = B $.

 

 

6. set $ A $ to $ N $ square matrix, and $0 $ to $ M $ of $ A $, with a credential: $ \ Bex \ rank (a ^ m) = N-M. \ EEx $

Proof: by Jordan standard theory, there is a reversible array $ p $ to make $ \ Bex P ^ {-1} AP =\sex {\ BA {cccccc} J_1 & \\\& \ ddots && & \& & j_s & \\& & J _ {S + 1} & \\& & \ ddots &\\&&&&& j_t \ EA }, \ EEx $ \ Bex J_ I = \ sex {\ BA {CCCC} \ lm_ I & 1 & \\\& \ ddots & \ ddots \\\& \ ddots & 1 \\& & \ lm_ I \ EA }, \ EEx $ and $ J_1, \ cdots, j_s $ corresponding $ \ lm_ I = 0 $. therefore, $ \ Bex P ^ {-1} a ^ MP = \ sex {\ BA {cccccc} 0 & \ ddots &&&&\\ & amp; 0 & amp; \\& & amp; J _ {S + 1} ^ M & amp; \\& amp; & amp; \ ddots & amp; \& & amp; j_t ^ m \ EA} \ Ra \ rank (a ^ m) = N-M. \ EEx $

 

 

7. Test Certificate: $ N $ level matrix $ X, Y $ makes $ XY-YX = e $.

Proof: Use the reverse identification method. If there is, then $ \ Bex 0 = \ tr (XY-YX) = \ tr E = n. \ EEx $ is a contradiction. Therefore, there is a conclusion.

 

 

8. set $ q $ to $ N $ positive sequence matrix, and $ x $ to $ N $ dimension column vector. Example: $ \ Bex 0 \ Leq x ^ t (q + XX ^ t) ^ {-1} x <1. \ EEx $

Proof: obtain $ \ beex \ BEA (q + XX ^ t) in the form of the Newman power series) ^ {-1} & =\ SEZ {q (E + q ^ {-1} XX ^ t )} ^ {-1 }\\& = (E + q ^ {-1} XX ^ t) ^ {-1} Q ^ {-1} \\\& =\ sum _ {k = 0} ^ \ infty (-1) ^ K (Q ^ {-1} XX ^ t) ^ KQ ^ {-1 }\\&=\ SEZ {e + \ sum _ {k = 1} ^ \ infty (x ^ TQ ^ {-1} X) ^ {k-1} Q ^ {-1} XX ^ t} Q ^ {-1} \ & = Q ^ {-1}-\ frac {q ^ {-1} XX ^ TQ ^ {-1 }}{ 1 + x ^ TQ ^ {-1} x }. \ EEA \ eeex $ and then verify it directly. original Certificate question: $ \ beex \ Bea x ^ t (q + XX ^ t) ^ {-1} X & = x ^ t \ sex {q ^ {-1}-\ frac {q ^ {-1} XX ^ TQ ^ {-1 }}{ 1 + x ^ TQ ^ {-1} X }}x \ & = x ^ TQ ^ {-1} X-\ frac {(x ^ TQ ^ {-1} x) ^ 2} {1 + x ^ TQ ^ {-1} X }\\\\=\ frac {x ^ TQ ^ {-1} x} {1 + x ^ TQ ^ {-1} X }\\& \ in [0, 1 ). \ EEA \ eeex $

 

 

9. set $ N $ level antisymmetric matrix $ A = (A _ {IJ}) $ to the determinant of $1 $. for any $ x $, calculate the determinant of $ B = (A _ {IJ} + x) $.

Answer: $ \ beex \ Bea | B | & =\ sev {\ BA {CCCC} 1 & X & \ cdots & X \ 0 & A _ {11} + X &\ cdots & A _ {1N} + x \ vdots & \ ddots & \ vdots \ 0 & A _ {N1} + X & \ cdots & _{ nn} + x \ EA }\\\\=\ sev {\ BA {CCCC} 1 & X & \ cdots & X \-X & A _ {11} & \ cdots & A _ {1N }\\\ vdots & \ ddots & \ vdots \-X & A _ {N1} & \ cdots & A _ {NN} \ EA }\\&=\ sev {\ BA {CC} 1 & X \ Al ^ t \-x \ Al & A \ EA} \ quad \ sex {\ Al = (1, 1, \ cdots, 1) ^ t }. \ EEA \ eeex $ by $ \ Bex \ sex {\ BA {CC} 1 &-x \ Al ^ ta ^ {-1} \ 0 & E \ EA} \ sex {\ BA {CC} 1 & X \ Al ^ t \-x \ Al & A \ EA} \ sex {\ BA {CC} 1 & 0 \\- XA ^ {-1} \ Al & E \ EA }=\ sex {\ BA {CC} ^ 2 \ Al ^ ta ^ {-1} \ Al & 0 \\ 0 & A \ EA} \ EEx $ $ \ beex \ Bea | B | & = (^ 2 \ Al ^ ta ^ {-1} \ Al) | A | \\& = _x ^ 2 \ Al ^ ta ^ {-1} \ Al \\\& = 1 + x ^ 2 \ beta ^ ta \ beta \ quad \ sex {\ Beta = a ^ {-1} \ Al }\\& = 1. \ EEA \ eeex $

 

10. Set $ A and B $ to two N $ positive definite matrices. proof:

(1) If $ AB = BA $, $ AB $ is also a positive matrix;

(2) If $ A-B $ is positive, $ B ^ {-1}-a ^ {-1} $ is also positive.

Proof: (1) Obviously, $ AB $ is symmetric. A reversible Matrix exists from $ A, B $ Zhengding. $ p, q $ makes $ A = P ^ TP, \ quad B = Q ^ TQ. $ and Q ^ {-1} = QP ^ tpq ^ t = (PQ ^ t) ^ t (PQ ^ t) \ EEx $ Zhi $ AB $ is similar to a positive matrix and has a positive feature value, which is positive.

(2) $ \ beex \ Bea & \ quad A-B \ mbox {Positive Definite} \ & \ rA P ^ TP-Q ^ TQ \ mbox {Positive Definite} \ & \ rA Q ^ {-T} P ^ tpq ^ {-1}-e \ mbox {Positive Definite} \ & \ rA PQ ^ {-1} Q ^ {-t} P ^ T-E \ mbox {Positive Definite} \\& \ quad \ sex {Y (xy) y ^ {-1} = Yx \ rA XY, YX \ mbox {similar, with the same feature value }}\\& \ rA Q ^ {-1} Q ^ {-t}-P ^ {-1} P ^ {-t} \ mbox {positive Definite} \ & \ rA B ^ {-1}-a ^ {-1} \ mbox {Positive Definite }. \ EEA \ eeex $