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![Determinant and Area • |■8(a_1&a_2@b_1&b_2 )| = area of parallelogram with sides a=(█(a_1@a_2 )), b=(█(b_1@b_2 )) • Proof by graph • Proof ○ Area(A_1,A_2 )=signed area of parallelogram spanned by A_1, A_2 ○ If A_1→A_2 is counter-clockwise = area ○ If A_1→A_2 is clockwise =−area ○ Then Area(A_1,A_2 )=det(A_1,A_2 ), because ○ Alternating § Area(A_1,A_2 )=−Area(A_2,A_1 ) § (by definition, same area, but different orientation) ○ Linearity(Homogeneous) § Area(t⋅A_1,A_2 )=t⋅Area(A_1,A_2 ) § (Easy to prove from picture) ○ Linearity(Additive) § Area(A+B,C)=Area(A,C)+Area(B,C) § If A,C is parallel, then □ Area(A,C)=0 § If A,C is independent , then □ Area(A+sC,C)=Area(A,C), ∀A,C § Let B=t⋅A+s⋅C, then □ Area(A+B,C) □ =Area(A+t⋅A+s⋅C,C) □ =Area(A+t⋅A,C) □ =(1+t)Area(A,C) □ =Area(A,C)+t⋅Area(A,C) □ =Area(A,C)+Area(t⋅A,C) □ =Area(A,C)+Area(t⋅A+s⋅C,C) □ =Area(A,C)+Area(B,C) § Therefore Area(A+B,C)=Area(A,C)+Area(B,C) ○ Uniqueness Theorem § Area(A,B) § =det〖(A,B)⋅Area(I_1,I_2 )〗 § =det〖(A,B)⋅Area(unit square)〗 § =det(A,B) Determinant and Volume • det〖(A,B,C)=signed volume〗 of parallelepiped spanned by A,B,C Inverse of a Matrix • Setup ○ T:Rn→Rn linear ○ T has a matrix m(T)=[■8(T_11&⋯&T_1n@⋮&⋱&⋮@T_n1&⋯&T_nn )] • The following statements are equivalent ○ N(T)={0} ○ T is injective ○ T is one\-to\-one ○ T is bijective § because T:Rn→Rn § dimN(T)+dimrange(T)=dim〖Rn 〗 § ⇒dimrange(T)=n § ⇒R(T)=Rn ○ There is a map S:Rn→Rn with ST=TS=I • Find the inverse of 2×2 matrix ○ T=[■8(1&3@2&5)] ○ Find T^(−1), i.e. solve Tx=y ○ Note: Tx=y⟺x=T^(−1) y ○ Normal version § [■8(1&3@2&5)][█(x_1@x_2 )]=[█(y_1@y_2 )] § {█(x_1+3x_2=1⋅y_1+0⋅y_2@2x_2+5x_2=0⋅y_1+1⋅y_2 )┤ § ⇒{█(x_1=−5y_1+3y_2@x_2=2y_1−y_2 )┤ § ⇒x=T^(−1) y § where T^(−1)=[■8(−5&3@2&−1)] ○ Shorthand § [T│I] § ~[■8(1&3@2&5) │ ■8(1&0@0&1)] § ~[■8(1&3@0&−1) │ ■8(1&0@−2&1)] § ~[■8(1&0@0&−1) │ ■8(−5&3@−2&1)] § ~[■8(1&0@0&1) │ ■8(−5&3@2&−1)] § ~[I│T^(−1) ] § Therefore T^(−1)=[■8(−5&3@2&−1)] Minors and Cofactors • Theorem ○ |■8(a_11&a_12&⋯&a_1n@⋮&⋮&⋮&⋮@a_k1&a_k2&…&a_kn@⋮&⋮&⋮&⋮@a_n1&a_n2&⋯&a_nn )|=a_k1 C_k1+a_k2 C_k2+…+a_kn C_kn ○ C_kl=cofactor matrix • Cofactor Matrix C_kl=(−1)^(k+l) {█((n−1)×(n−1) determinant obtained@by deleting row k and column l @from the original determinant)} • Example ○ |■8(1&7&2@4&π&−1@3&ln2&2)| ○ =3×|■8(7&2@π&−1)|−ln2 |■8(1&2@4&−1)|+2|■8(1&7@4&π)| ○ =3×(−7−2π)−ln2×(−9)+2×(π−28) ○ =−77−4π+9ln2 • Matrix Multiplication ○ Let P=[■8(a_11&a_12&⋯&a_1n@a_21&a_22&…&a_2n@⋮&⋮&⋱&⋮@a_n1&a_n2&⋯&a_nn )][■8(C_11&C_21&⋯&C_n1@C_12&C_22&…&C_n2@⋮&⋮&⋱&⋮@C_1n&C_2n&⋯&C_nn )] ○ P_11=a_11 C_11+a_12 C_12+…+a_1n C_1n=detA ○ P_21=a_21 C_11+a_21 C_12+…++a_21 C_1n=0 ○ Because we have two equal row ○ Therefore P=detA [■(1&&@&⋱&@&&1)] •](https://shawnzhong.com/wp-content/uploads/2017/11/img_5a020d4a91900.png)
![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)
