Math 375 – 11/1 Nov 02, 2017 Shawn Math 375 No comments yet Math 375 – 10/31 Nov 01, 2017 Shawn Math 375 No comments yet Math 375 – 10/30 Nov 01, 2017 Shawn Math 375 No comments yet Math 431 – 9/11 Nov 21, 2017 Shawn Math 375 No comments yet 1 … 3 4 5
![Question 1 • Let A be an n×n square matrix which has a row or column of all zeros • Prove: A is singular (i.e. not invertible) • Proof: Column of all zeros ○ Ae_i=(■8(∗&…&0&…&∗@⋮&…&⋮&…&⋮@∗&…&0&…&∗))┬█(⏟@i-th) (█(0@⋮@1@⋮@0))}├ i-┤th=0 ○ Ae_i=0⇒A is not injective⇒A is not invertable • Proof: Rows of all zeros ○ ∀v∈V⇒Av=(■(∗&…&∗@0&…&0@∗&…&∗))v=(█(⋮@0@⋮)) ○ Av=0⇒A is not surjective⇒A is not invertable Question 2 • Let T:R2→R2 be a linear map. • Computer the area of the image of the unit square [0,1]^2 • i.e. the set T([0,1]^2 )={T(x,y):x,y∈[0,1]}⊆R2 • Answer ○ Area of image = det(T) • Proof Question 3 • Let V be a finite-dimensional vector space • Let T:V→V be a linear map such that TS=ST for all linear maps S:V→V • Prove that there exists c∈R such that for all v∈V, we have Tv=cv • Prove (Version 1) ○ Let E_ij=(⇳112 [■(0&&0&&0@&⋱&⋮&⋰&@0&…&1&…&0@&⋰&⋮&⋱&@0&&0&&0)] ⇳12)┬█(⏟@j-th)}├ i-th┤, where i≠j § TE_ij=[■8(a_11&⋯&a_1n@⋮&⋱&⋮@a_n1&⋯&a_n1 )][■(0&&0&&0@&⋱&⋮&⋰&@0&…&1&…&0@&⋰&⋮&⋱&@0&&0&&0)]=[■8(0&…&a_1j&…&0@⋮&…&⋮&…&⋮@0&…&a_jj&…&0@⋮&…&⋮&…&⋮@0&…&a_nj&…&0)] § E_ij T=[■(0&&0&&0@&⋱&⋮&⋰&@0&…&1&…&0@&⋰&⋮&⋱&@0&&0&&0)][■8(a_11&⋯&a_1n@⋮&⋱&⋮@a_n1&⋯&a_n1 )]=[■8(0&…&0&…&0@…&…&…&…&…@a_i1&…&a_ii&…&a_in@…&…&…&…&…@0&…&0&…&0)] ○ Because TS=ST for all linear maps S:V→V § TE_ij=E_ij T § [■8(0&…&a_1j&…&0@⋮&…&⋮&…&⋮@0&…&a_jj&…&0@⋮&…&⋮&…&⋮@0&…&a_nj&…&0)]=[■8(0&…&0&…&0@…&…&…&…&…@a_i1&…&a_ii&…&a_in@…&…&…&…&…@0&…&0&…&0)] § ⇒{■8(a_ii=a_jj&∀i,j∈{1,2,…,n},i≠j@a_kl=0&∀k,l∈{1,2,…,n}, k≠l)┤ § Let a_11=a_22=…a_nn=c ○ Therefore T=[■(c&&@&⋱&@&&c)] is a scalar matrix i.e. Tv=cv ○ Also, T satisfied the following property for all linear maps S:V→V § TSv=T(Sv)=c⋅Sv=S(cv)=STv • Proof (Version 2) ○ Assume Tv and v is linearly independent § i.e. Tv≠cv ○ Then the following is a basis for V § {v,Tv,e_1,e_2,…} ○ Define S to be § S≝{█(S(v)=v@S(Tv)=v@S(e_1 )=0@S(e_2 )=0@⋮)┤ ○ Then § T(v)=T(S(v))=TS(v)=ST(v)=S(Tv)=v ○ Which makes a contradiction ○ Therefore Tv and v is linearly dependent i.e. Tv=cv](https://shawnzhong.com/wp-content/uploads/2017/11/img_59f8d90c487b6.png)
![Probability Space • Ω: sample space (list of all outcomes) • F: collection of events (subsets of Ω) • P: probability measure ○ P(A)∈[0,1] ○ P(∅)=0 ○ P(Ω)=1 ○ For disjoint A_1,A_2,…:P(⋃24_(i=1)^∞▒A_i )=∑_(i=1)^∞▒PA_i ) Equally Likely Outcome • P(ω)=1/(#Ω),∀ω∈Ω • P(A)=(#A)/(#Ω) • Example: 431 game with full deck ○ Ω={(c_1,c_2,c_3 )│■8(c_1 is my card@c_2 is your first card@c_3 is your second card@and they are all distinct)} ○ P(A)=(#A)/(#Ω) ○ W_7={(c_1,c_2,c_3 )∈Ω| (c_2≥7 and c_2c_1 ) or (c_27 and c_3c_1 )} ○ #Ω=52×51×50=(52)_3 ○ Note: (n)_k=n!/(n−k)! • Example: 431 game with replacement ○ Ω={(c_1,c_2,c_3 )│■8(c_1 is my card@c_2 is your first card@c_3 is your second card)} ○ W_7={(c_1,c_2,c_3 )∈Ω| (c_2≥7 and c_2c_1 ) or (c_27 and c_3c_1 )} ○ #Ω=〖52〗^3 Different Types of Random Experiments • S={1,…,n} • Sampling with replacement where order matters ○ Ω=S^k={(s_1,…,s_k )|s_i∈S} ○ #Ω=n^k • Sampling without replacement where order matters ○ Ω={(s_1,…,s_k )|s_i∈S and ∀i≠j:s_i≠s_j } ○ #Ω=n(n−1)⋯(n−k+1)=n!/(n−k)!=(n)_k • Sampling without replacement where order is irrelevant ○ Ω={A⊆S|#A=k}](https://shawnzhong.com/wp-content/uploads/2017/11/img_5a13a28c17868.png)
