sledgehammer exercise

 Practice 2.51Through the previous contents of the book, we know that the below is divided into two parts, and beside divides the frame into the left and right parts. Therefore,below is defined as follows:(Define (below Painter1 painter2) (Let (split-point (make-vect 0.0 0.5)) (Let (Paint-top (Transform-painter painter2 Split-point(Make-vect 1.0 0.5) (make-vect 0.0 1.0))) (Paint-down (transform-painterpainter1 (make-vect 0.0 0.0)(Make-vect 1.0 0.0) split-point))) (lambda (frame) (Paint-top f

provision an instance of that custom list definition in the particleSite where the feature is activated. Which approach shocould you recommend?A. Set the scope of feature a to web and the scope of Feature B to site. Add feature a to the set of activation dependencies in Feature B.B. Set the scope of feature a to web and the scope of Feature B to site. Add Feature B to the set of activation dependencies in feature.C. Set the scope of feature a to site and the scope of Feature B to Web. Add featu

DecimalFormat exercise..., decimalformat exercise Package test; import java. text. decimalFormat; class FormatDemo {public void format1 (String pattern, double value) {// This method is used to format and display a number. DecimalFormat df = null; // declare the DecimalFormat Class Object df = new DecimalFormat (pattern); // instantiate the object and input the template String str = df. format (value); //

17. (ando-Zhan) is set to $ A, B \ In M_n $ semi-definite, $ \ Sen {\ cdot} $ is an undo norm, then $ \ Bex \ Sen {(a + B) ^ r} \ Leq \ Sen {A ^ r + B ^ r }, \ Quad (0 Proof: (1 ). verify when $0 (2 ). when $ r \ geq 1 $, make $ g (t) = t ^ r = f ^ {-1} (t) $, $ F (t) = t ^ \ frac {1} {r} $, then by (1), $ \ Bex S (F [G (A) + g (B)]) \ prec_w S (F [G (A)] + F [G (A)]) = S (A + B ). \ EEx $ by $ G $ non-negative incremental convex function, theorem 3.21, and ing theorem similar to Theorem 1.3

Label: Style Color SP on BS ad amp size nbsp 15. (fan-Hoffman) set $ A, H \ In M_n $, where $ h $ is the Hermite matrix, then $ \ Bex \ Sen {A-\ re a} \ Leq \ Sen {A-H} \ EEx $ is set to any Unio constant norm. Proof: (1 ). first, it is proved that $ \ Bex \ Sen {\ cdot} \ mbox {is an undo norm }, X \ In M_n \ Ra \ Sen {x }=\ Sen {x ^ *}. \ EEx $ in fact, $ x $ has the same singular value as $ x ^ * $, while $ \ Bex S (x) \ prec S (x ^ *) \ prec S (X ). \ EEx $ the fan-dominated principl

Label: Style Color ar SP on BS ad amp ef 1. (fan-Hoffman ). set $ A \ In M_n $, note $ \ re a = (a + A ^ *)/2 $. then $ \ Bex \ lm_j (\ re a) \ Leq s_j (A), \ quad j = 1, \ cdots, N. \ EEx $ Proof: For $ x \ In \ BBC ^ N $, $ \ beex \ Bea x ^ * (\ re a) suitable for $ \ Sen {x} = 1 $) X = x ^ * \ frac {A + A ^ *} {2} X \ =\ frac {1} {2} (x ^ * AX + x ^ * A ^ * X) \\=\ Re (x ^ * Ax) \ quad \ sex {z \ In \ BBC \ rA Z ^ * = \ bar z} \ \ Leq | x ^ * ax |\\ = | \ SEF {ax, x} | \\ \ Leq \ S

{ll} A _ {II }, I = J \\\ cfrac {1} {n} \ DPS {\ sum _ {k = 0} ^ {n-1} \ Omega ^ {(I-j) k} A _ {IJ }}=\ cfrac {1} {n} \ cdot \ cfrac {1-\ Omega ^ {(I-j) N }}{ 1-\ Omega ^ {I-j} A _ {IJ} = 0, I \ NEQ J \ EA }\\ = A _ {IJ} \ Delta _ {IJ }. \ EEA \ eeex $ by \ eqref {4_4_diag}, you can call this operation to know any Unio constant norm $ \ Sen {\ cdot} $, $ \ Bex \ Sen {\ diag (A _ {11}, \ cdots, A _ {NN })} \ Leq \ frac {1} {n} \ sum _ {k = 0} ^ {n-1} \ Sen {u ^ Kau ^ {* k }}=\ Sen {}. \ EEx $

1. $ A \ in M_n $ is called an orthogonal projection matrix. If $ A $ is A Hermite matrix and its idempotence is: $ \ bex A ^ * = A ^ 2. \ eex $ Proof: If $ A, B \ in M_n $ is an orthogonal projection matrix, then $ \ sen {A-B} _ \ infty \ leq 1 $.Proof: By $ A ^ * = A $ Zhi $ A $, you can perform the right-to-right corner. the feature value of $ A ^ 2 = A $ Zhi $ A $ is $0 $ or $1 $. therefore, $ Makes $ \ bex A = U ^ * \ diag (I _r, 0) U, \ quad r = \ rank (). \ eex $, $ \ beex \ bea x ^ * Ax

-\ beta _ {\ sigma (I )} |=\ sqrt {\ frac {28} {13 }}, \\\ sigma =\sed {2, 3, 1} \ ra \ max_ I | \ al_ I-\ beta _ {\ sigma (I) }|=\ sqrt {\ frac {28} {13 }}, \\\ sigma =\sed {3, 1, 2} \ ra \ max_ I | \ al_ I-\ beta _ {\ sigma (I )} |=\ sqrt {\ frac {28} {13 }}, \\\ sigma =\sed {3, 2, 1} \ ra \ max_ I | \ al_ I-\ beta _ {\ sigma (I) }|=\ sqrt {\ frac {28} {13 }}. \ eea \ eeex $ hence $ \ bex \ rd (\ sigma (A), \ sigma (B) = \ sqrt {\ frac {28} {13 }}. \ eex $ the feature value of $ (A-B) ^ * (

Label: style io color ar sp for on bs line3. (Bhatia-Davis) if $ A, B \ in M_n $ is A matrix, then $ \ bex \ rd (\ sigma (A), \ sigma (B )) \ leq \ sen {A-B} _ \ infty. \ eex $Proof: [see R. Bhatia, C. Davis, A bound for the spectral variation of a unitary operator, Linear and Multilinear Algebra, 15 (1984),-76.][Zhan Xiang matrix theory exercise reference] exercise 5.3

8. set $ k \ Leq m \ Leq N $. what kind of matrix $ A \ in M _ {m, n} $ each diagonal line exactly contains $ K $ zero elements? Answer: Theorem 2.5 (k \ "onig ), each diagonal line of $ A $ contains $ K $ zero elements $ \ LRA $ A $ has a zero submatrix of $ r \ times S $, $ R + S = N + k $; $ A $ a diagonal line contains $ k + 1 $ zero elements $ \ LRA $ any $ r \ times S $ level submatrix of $ A $ non-zero, $ R + S = N + k + 1 $. therefore, each diagonal line of $ A $ contains $ K $ zero

Label: Style Color SP on BS size nbsp C element3. Set $ a, B \ In M_n $, $ A $ positive definite, $ B $ semi-positive, and the logarithm element is positive, then $ A \ circ B $ positive.Proof: By Schur theorem, $ A \ circ B $ is semi-definite, and its feature value $ \ geq 0 $. to prove that $ A \ circ B $ is positive, only $ \ det (A \ circ B)> 0 $ ($ \ rA $ any feature value $> 0 $) must be proved ). this can be directly caused by the Oppenheim inequality $ \ Bex \ det (A \ circ B) \ geq \ de

Label: Style Color SP on BS amp AD Size DP 9. Note $ \ DPS {M = \ sex {n \ atop K }}$. Composite Matrix ing $ c_k (\ cdot): M_n \ To m_m $ is it a single shot? Is it a full shot? A: When $ k = 1 $, $ c_k (a) $ is each element of $ A $. therefore, $ c_k $ is a single shot or full shot. when $ k \ geq 2 $, generally, $ c_k $ is not a single shot, for example, $ \ Bex \ sex {\ BA {CCCC} 1 0 \ cdots 0 \ 0 \ cdots 0 \ vdots \ ddots \ vdots \ 0 0 \ cdots 0 \ EA }, \ quad \ sex {\ BA {CC

14. if ing $ F: M_n \ To M_n $ sorts the elements of each matrix in $ M_n $ in a fixed mode, $ F $ is called a replacement operator. what kind of replacement operator does not change the feature value of the Matrix? Keep the rank unchanged? Answer: The replacement operator $ F $ keeps the feature value of the matrix unchanged. If only the replacement matrix $ p $ exists, $ \ Bex F (a) = pap ^ t, \ quad \ forall \ A \ In M_n; \ EEx $ or $ \ Bex F (a) = PA ^ TP ^ t, \ quad \ forall \ A \ In M

2. (oldenburgere) set $ A \ In M_n $, $ \ rock (a) $ to indicate the spectral radius of $ A $, that is, the creator of the modulus of the feature value of $ A $. proof: $ \ Bex \ vlm {k} a ^ K = 0 \ LRA \ rock (a) Proof: $ \ rA $: Based on Jordan standard theory, there is a reversible array $ p $, make $ \ Bex P ^ {-1} AP =\sex {\ BA {CCC} \ lm_1 * \\\\ ddots \\\ \ lm_n \ EA }, \ EEx $ and $ \ Bex P ^ {-1} a ^ Kp = \ sex {\ BA {CCC} \ lm_1 ^ K * \ ddots \\ \ lm_n ^ k \ EA }. \ EEx $ thus,

10. set $ A and B $ to the same-order semi-Definite Matrix, $0 \ Leq s \ Leq 1 $. proof: $ \ Bex \ Sen {A ^ Sb ^ s} _ \ infty \ Leq \ Sen {AB} _ \ infty ^ s. \ EEx $ Proof: (1 ). first, it is proved that the spectral norm of $ A $ is the maximum singular value of $ A $. in fact, $ \ beex \ Bea \ Sen {A} _ \ infty ^ 2 =\ Max _ {\ Sen {x} _ 2 = 1} \ Sen {ax} _ 2 ^ 2 \ =\ Max _ {\ Sen {x} _ 2 = 1} x ^ * a ^ * ax \ =\ Max _ {\ Sen {x }_ 2 = 1} x ^ * VV ^ * a ^ * u ^ * uavv ^ * x \ =\ Max

Label: Style Color SP on BS amp ad EF size 8. it is proved that each semi-positive definite matrix has a unique square root of the semi-positive definite number. That is, if $ A \ geq 0 $, there is a unique $ B \ geq 0 $ that satisfies $ B ^ 2 = A $. Proof: $ A \ geq 0 $ indicates the existence of U $, so that $ \ Bex U ^ * Au = \ diag (\ lm_1, \ cdots, \ lm_n ), \ quad \ lm_ I \ geq 0. \ EEx $ \ Bex B = U \ diag (\ SQRT {\ lm_1}, \ cdots, \ SQRT {\ lm_n}) U ^ *, \ EEx $ B \ geq 0 $, an

Label: Style Color SP on BS ad size as nbsp 5. Without the Weierstrass theorem, we can directly prove that the function operation (3.6) of the Hermite matrix has nothing to do with the specific spectral decomposition. Proof: $ h $ also has a spectral decomposition $ \ Bex H = V \ diag (\ lm_1, \ cdots, \ lm_n) V ^ *, \ EEx $ \ Bex w \ diag (\ lm_1, \ cdots, \ lm_n) = \ diag (\ lm_1, \ cdots, \ lm_n) W, \ quad W = V ^ * U. \ EEx $ therefore, $ \ Bex W = \ diag (W_1, \ cdots, w_s ), \ EEx $

Label: Style Color SP on bs ad Size nbsp C 1. Set $ A \ In M_n $. Prove that if $ AA ^ * = a ^ 2 $, $ A ^ * = A $. Proof: By Schur ry triangle theorem, $ U $ exists, making $ \ Bex a = u ^ * Bu, \ EEx $ where $ B = (B _ {IJ}) $ is the upper triangle array. so $ \ Bex U ^ * BB ^ * u = AA ^ * = a ^ 2 = u ^ * B ^ 2u \ rA BB ^ * = B ^ 2. \ EEx $ compare the diagonal yuan between two ends $ \ Bex | B _ {II} | ^ 2 + \ cdots + | B _ {In} | ^ 2 = B _{ II} ^ 2, \ Quad 1 \ Leq I \ Leq n. \ EEx $ whi

":Return ' death 'elif action = = "Slowly Place the bomb":Return ' Escape_pod 'ElseReturn ' The_bridge 'def escape_pod (self):Good_pod = Randint (1,5)Guess = Raw_input ("[Pod #]>>>")if int (guess)! = good_pod:Return ' death 'ElseExit (0)Class Engine (MAP):def __init__ (Self,start):Self.start = Startdef play (self):Next = Self.startWhile True:print "\ n-----------"GetAttr (Self,next)Next = Guest ()A_game = Engine ("Central_corridor")A_game.play () Remark: Beginner young novice, Welcome to the gre