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

Source: Internet
Author: User

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 \$

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