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![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)
![Solving Linear Equations • Trying to solve the equation ○ Ax=y ○ where x∈V is sought, y∈W is given ○ V,W vector spaces ○ T:V→W linear transformation • Example 1 ○ {█(a_11 x_1+a_12 x_2+…+a_1n x_n=y_1@⋮@a_m1 x_1+a_m2 x_2+…+a_mn x_n=y_m )┤ ○ Let § x=(█(x_1@⋮@x_n ))∈Rn § y=(█(y_1@⋮@y_n ))∈Rn § A: Rn→Rn § A(█(x_1@⋮@x_n ))=(█(a_11 x_1+…+a_1n x_n@⋮@a_m1 x_1+…+a_mn x_n )) § A(█(x_1@⋮@x_n ))=(■8(a_11&⋯&a_1n@⋮&⋱&⋮@a_m1&⋯&a_mn ))(█(x_1@⋮@x_n )) § A with respect to standard bases of Rn, Rm ○ Then the linear equations could be represented as § Ax=y • Theorem 1 ○ Statement § If A:V→W is linear § and if u,v∈V are solutions to Ax=y § (i.e. if Au=y, and Av=y) § Then u−v∈N(A) ○ Proof § A(u−v)=Au−Av=y−y=0 ○ Text version § If 〖Ax〗_p=y then for all x∈V with Ax=y § There is an x_ℎ∈N(A) with x=x_p+x_ℎ • Theorem 2 ○ Statement § If u is a solution to Ax=y § and if w∈N(A) § then u+w is also a solution of Ax=y ○ Proof § A(u+w)=Au+Aw=y+0=y ○ Text version § For all x_p with 〖Ax〗_p=y and for all x_ℎ∈N(A) § A(x_p+x_h)=y • General solution ○ Homogeneous equation Ax=0 ○ Inhomogeneous equation § Ax=y, where y≠0 ○ The general solution to Ax=y is of the form § x_gen=x_p+x_ℎ, where § x_p is a particular solution § x_h is the general solution to the homogeneous equation ○ Set of all solutions § {x∈V│Ax=y}={x_p+x_h■8(Ax_p=y@x_hN(A) )} ○ Proof § We are given one solution x_p of Ax=y § If x_ℎ∈N(A) § then by definition 〖Ax〗_ℎ=0 § and hence A(x_p+x_h)=y § ⇒x_p+x_ℎ∈{x∈V│Ax=y} § Conversely if Ax=y then § A(x−x_p )=Ax−Ax_p=y−y=0 § So x_ℎ≝x−x_p∈N(A) • Example 2 ○ Solve the linear equation{█(x_1+2x_2−x_3=7@2x_1−x_2+x_3=4)┤ ○ Setup § V=R3⇒x=(█(x_1@x_2@x_3 )) § W=R2⇒y=(█(7@4)) § A: R3→R2 is matrix multiplication with [■8(1&2&−1@2&−1&1)] ○ Range(A) § ={Ax│x∈R3 } § ={all possible y∈R2 for which Ax=y has a solution} ○ By Rank–nullity theorem § dim〖N(A)+dim〖Range(A)〗=dim〖R3 〗=3〗 dim〖Range(A)〗 dimN(A) 0 3 1 2 2 1 ○ Solving the equation by Gaussian Elimination § [■8(1&2&−1@2&−1&1) │ ■8(7@4)]→[■8(1&0&1/5@0&1&−3/5) │ ■8(3@2)] § {█(x_1+1/5 x_3=3@x_2−3/5 x_3=2)┤ § Let x_3=5c § Then{█(x_2=2+3/5 x_3=2+3c@x_2=3−1/5 x_3=3−c)┤ § Therefore the general solution is § x=[█(3−c@2+3c@5c)]=⏟([█(3@2@0)] )┬(x_p )+c⏟([█(−1@3@5)] )┬(x_h) • Example 3 ○ Given § V=W={functions y:[a,b]→R § A:V→W where Af=f^′+xf ○ Question § Solve dy/dx+xy=x ○ The general solution is in form of § x+x_p+x_ℎ ○ It](https://shawnzhong.com/wp-content/uploads/2017/10/img_59f257143642c.png)
