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![Uniqueness Theorem • Theorem ○ Suppose f(A_1,…,A_n ) is a function of A_1,…,A_n∈Rn ○ That satisfies Linearity and Alternating § f(B+C,A_2,…,A_n )=f(B,A_2,…,A_n )+f(C,A_2,…,A_n ) § f(t⋅A_1,A_2,…,A_n )=t⋅f(A_1,A_2,…,A_n ) § f(A_1,A_2,…,A_i,..,A_j,…A_n )=−f(A_1,A_2,…,A_j,..,A_i,…A_n ) ○ Then f(A_1,…,A_n )=det(A_1,…,A_n )⋅f(I_1,…,I_n ) where § I_1=[1,0,0,…,0] § I_2=[0,1,0,…,0] § ⋮ § I_n=[0,0,0,…,1] • Proof ○ f(A_1,…,A_n ) ○ =f(a_11 I_1+a_12 I_2+…+a_1n I_n,…,a_n1 I_1+a_n2 I_2+…+a_nn I_n ) ○ =∑_█(1≤i_1,i_2,…,i_n≤n@all different)^n▒a_(1i_1 ) a_(2i_2 )…a_(ni_n )⋅f(I_(i_1 ),I_(i_2 ),…,I_(i_n ) ) ○ =∑_█(1≤i_1,i_2,…,i_n≤n@all different)^n▒a_(1i_1 ) a_(2i_2 )…a_(ni_n )⋅sign(i_1,…,i_n )⋅f(I_1,I_2,…,I_n ) ○ =f(I_1,I_2,…,I_n )⋅∑_█(1≤i_1,i_2,…,i_n≤n@all different)^n▒a_(1i_1 ) a_(2i_2 )…a_(ni_n )⋅sign(i_1,…,i_n ) ○ =f(I_1,I_2,…,I_n )⋅det(A_1,…,A_n ) • Example ○ |■8(A_(k×k)&0@C_(l×k)&B_(l×l) )|=|■(a_11&…&a_1k&&&@⋮&⋱&⋮&&&@a_k1&…&a_kk&&&@c_11&…&c_1k&b_11&…&b_1l@⋮&⋱&⋮&⋮&⋱&⋮@c_l1&…&c_lk&b_l1&…&b_ll )|=detA⋅detB ○ Consider a function f that satisfies the Uniqueness Theorem § f((A_1 ) ̅+(A_1 ) ̅ ̅,A_2,…,A_n )=f((A_1 ) ̅,A_2,…,A_n )+d((A_1 ) ̅ ̅,A_2,…,A_n ) § f(tA_1,A_2,…,A_n )=f(A_1,A_2,…,A_n ) § f(A_1,A_2,…,A_i,..,A_j,…A_n )=f(A_1,A_2,…,A_j,..,A_i,…A_n ) ○ Let f_BC (A_1,…,A_k )=|■8(A_(k×k)&0@C_(l×k)&B_(l×l) )| with B,C fixed, and A as variable § f_BC (A_1,…,A_k ) § =det(A_1,…,A_k ) f_BC (I_1,…,I_k ) § =detA⋅|■(1&&&&&@&⋱&&&&@&&1&&&@c_11&…&c_1k&b_11&…&b_1l@⋮&⋱&⋮&⋮&⋱&⋮@c_l1&…&c_lk&b_l1&…&b_ll )| § =detA⋅|■(1&&&&&@&⋱&&&&@&&1&&&@&&&b_11&…&b_1l@&&&⋮&⋱&⋮@&&&b_l1&…&b_ll )| § =detA⋅|■(I&@&B)| ○ Let g(B)=|■(I&@&B)| that satisfies the Uniqueness Theorem § g(B)=detB⋅g(I)=detB⋅|■(1&&@&⋱&@&&1)|=detB ○ Therefore |■8(A_(k×k)&0@0&B_(l×l) )|=detA⋅detB Properties of Determinant • det〖(AB)=detA⋅detB 〗 (where A_(n×n), B_(n×n)) ○ detA⋅detB ○ =|■8(A&0@I&B)| ○ =|■8(0&−AB@I&B)| ○ =(−1)^(n^2 ) |■8(I&B@0&−AB)| ○ =(−1)^(n^2 ) detI⋅det(−AB) ○ =(−1)^(n^2 )⋅det(−AB) ○ =(−1)^(n^2 ) (−1)^n det(AB) ○ =(−1)^(n^2+n) det(AB) ○ =det(AB) • Power of Determinants ○ det(A^n )=det(A⋅A…A)=det〖(A)⋅〗 det(A)…det(A)=(detA )^n • Determinant of Inverse ○ If A has an inverse(A^(−1)), and detA≠0, then ○ A^(−1) A=I ○ ⇒det〖A^(−1) 〗⋅detA=detI=1 ○ ⇒det〖A^(−1)=1/detA 〗 • Matrix Product and Determinant ○ |■8(A_(n×n)&0@I&B_(n×n) )| ○ =|■(a_11&…&a_1n&&&@⋮&⋱&⋮&&&@a_n1&…&a_nn&&&@1&&&b_11&…&b_1n@&⋱&&⋮&⋱&⋮@&&1&b_n1&…&b_nn )| ○ =|■(0&…&a_1n&−a_11 b_11&…&−a_11 b_1n@⋮&⋱&⋮&⋮&⋱&⋮@0&…&a_nn&−a_n1 b_11&…&−a_n1 b_1n@1&…&0&b_11&…&b_1n@⋮&⋱&⋮&⋮&⋱&⋮@0&…&1&b_n1&…&b_n )|=… ○ =|■(0&…&0&−∑_(i=1)^n▒〖a_1i b_i1 〗&…&−∑_(i=1)^n▒〖a_1i b_in 〗@⋮&⋱&⋮&⋮&⋱&⋮@0&…&0&−∑_(i=1)^n▒〖a_ni b_i1 〗&…&−∑_(i=1)^n▒〖a_ni b_in 〗@1&…&0&b_11&…&b_1n@⋮&⋱&⋮&⋮&⋱&⋮@0&…&1&b_n1&…&b_nn )| ○ =|■8(0&−AB@I&B)|](https://shawnzhong.com/wp-content/uploads/2017/11/img_5a0cf537a2c42.png)
![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)
