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![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)
![Question 1 • T:R3→R2 with T defined as ○ T(i)=(0,0) ○ T(j)=(1,1) ○ T(k)=(1,−1) • Find the matrix for normal basis ○ M(T,{i,j,k},{i,j})=[■8(0&1&1@0&1&−1)] • Find the matrix using (█(1@1)),(█(1@2)) as the basis for R2 ○ M(T,{i,j,k},{(█(1@1)),(█(1@2))})=[■8(1&1@1&2)]^(−1) [■8(0&1&1@0&1&−1)]=[■8(0&1&3@0&0&−2)] • Find bases for R3 and R2 so that the matrix is diagonal ○ M(T,{(█(0@1/2@1/2)),(█(0@1/2@−1/2)),(█(1@0@0))},{i,j})=[■8(1&0&0@0&1&0)] ○ M(T,{j,k,i},{T(i),T(k)})=[■8(1&0&0@0&1&0)] Question 2 • Let T: R2→R2 be an abitrary linear map. Can one choose a basis (v_1,v_2) on the domain and a basis (w_1,w_2) on the codomain such that the matrix of T with respect to these bases is diagonal? ○ Yes ○ Rank 0: [■8(0&0@0&0)] ○ Rank 1: [■8(1&0@0&0)] ○ Rank 2: [■8(1&0@0&1)] • Can one choose a basis (v_1,v_2) on both the domain and codomain -- the same basis on both -- such that the matrix of T is diagonal? ○ No ○ T(x,y)=(y,0) cannot be diagonal ○ M(T)=[■8(0&1@0&0)]](https://shawnzhong.com/wp-content/uploads/2017/10/img_59f6b63497baf.png)
![Matrix Representation of Linear Transformation • Definition ○ T:V→W ○ {e_1,e_2…e_n }:basis for V ○ {f_1,f_2…f_n }:basis for W ○ matrix(T,{e_k },{f_l })=m(T)=[■8(T_11&⋯&T_1n@⋮&⋱&⋮@T_m1&⋯&T_mn )] • Example ○ {■(Te_1=T_11 f_1+…+T_m1 f_m@Te_2=T_12 f_1+…+T_m2 f_m@⋮@Te_n=T_1n f_1+…+T_mn f_m ) ┤a Algebra of Linear Transformations vs. Algebra of Matrices • Comparison Linear Transformations Matrices T+S m(T+s)=m(T)+m(S) c⋅T m(cT)=c⋅m(T) S∘T m(S∘T)=m(S)⋅m(T) • Proof: m(S∘T)=m(S)m(T) ○ Setup § T:U→V, S:V→W § {e_1…e_n }: basis of U § {f_1…f_m }: basis of V § {g_1…g_k }: basis of W ○ Let m(R)=m(S∘T), where R=S∘T ○ m(T) is defined by § {■(Te_1=T_11 f_1+…+T_m1 f_m@Te_2=T_12 f_1+…+T_m2 f_m@⋮@Te_n=T_1n f_1+…+T_mn f_m ) ┤ ○ m(S) is defined by § {■(Sf_1=S_11 g_1+…+S_k1 g_k@Sf_2=S_12 g_1+…+S_k2 g_k@⋮@Sf_m=S_1m g_m+…+S_km g_k ) ┤ ○ m(R) is defined by § {■(Re_1=R_11 e_1+…+R_k1 g_k@Re_2=R_12 e_1+…+R_k2 g_k@⋮@Re_n=R_1n e_1+…+R_kn g_k ) ┤ ○ R_ij=Coefficient of g_i in 〖Re〗_j=Coefficient of g_i in (S∘T) e_j ○ Expanding (S∘T) e_j, we have § (S∘T) e_j=S(〖Te〗_j ) § =S(T_1j f_1+T_2j f_2+…+T_mj f_m ) § =T_1j⋅〖Sf〗_1+T_2j⋅〖Sf〗_2+…+T_mj⋅〖Sf〗_m § =T_1j (S_11 g_1+…+S_k1 g_k )+…+T_mj (S_1m g_1+…+S_km g_k ) ○ Terms containing g_i § T_1j S_i1 g_i+T_2j S_i2 g_i+…+T_mj S_im g_i § =(S_i1 T_1j+S_i2 T_2j+…+S_im T_mj ) g_i ○ Therefore § R=(R_ij )_(i,j=1)^(n,k)=(S_i1 T_1j+S_i2 T_2j+…+S_im T_mj )_(i,j=1)^(n,k) § m(S)m(T)=[■8(S_11&⋯&S_1m@⋮&⋱&⋮@S_k1&⋯&S_km )]×[■8(T_11&⋯&T_1n@⋮&⋱&⋮@T_m1&⋯&T_mn )] § =(S_i1 T_1j+S_i2 T_2j+…+S_im T_mj )_(i,j=1)^(n,k) § ⇒m(S∘T)=m(S)m(T) Matrix Multiplication • Example ○ V=W=R2 with standard basis ○ T=rotation by θ § m(T)=[■8(cos〖θ 〗&−sinθ@sinθ&cos〖θ 〗 )] ○ S=rotation by φ § m(S)=[■8(cos〖φ 〗&−sinφ@sinφ&cos〖φ 〗 )] ○ S=rotation by θ+φ § m(S)m(T) § =[■8(cos〖φ 〗 cosθ−sinφ sinθ&〖−sin〗θ cosφ−sinφ cosθ@sinθ cosφ+sinφ cosθ&cos〖φ 〗 cosθ−sinφ sinθ )] § =[■8(cos〖(θ+φ) 〗&−sin(θ+φ)@sin(θ+φ)&cos〖(θ+φ) 〗 )] § =m(ST) § Therefore m(ST)=m(S)m(T) • Example: T≠0, but T^2=0 ○ T=[■8(0&1@0&0)], T(x,y)=(0,x) ○ ⇒T^2=T×T=[■8(0&1@0&0)]×[■8(0&1@0&0)]=[■8(0&0×1+1×0@0&0)]=[■8(0&0@0&0)] ○ Note: T≠0, but T^2=0 • Example: ST≠TS ○ T=[■8(0&1@0&0)], S=[■8(0&0@1&0)] ○ TS=[■8(0&1@0&0)][■8(0&0@1&0)]=[■8(1&0@0&0)] ○ ST=[■8(0&0@1&0)][■8(0&1@0&0)]=[■8(0&0@0&1)] ○ Note:ST≠TS ○ Therefore matrix multiplication is not commutative • Example ○ S,T:V→V, (or S,T are square matrice) ○ (S+T)^2=(S+T)(S+T)=S^2+ST+TS+T^2 ○ Note: (S+T)^2≠S^2+2TS+T^2≠S^2+2ST+T^2 Solving Linear Equations using Matrix • Matrix representation of Linear Equations ○ {█(a_11 x_1+a_12 x_2+…+a_1n x_n=y_1@a_21 x_1+a_22 x_2+…+a_2n x_n=y_2@⋮@a_m1 x_1+a_m2 x_2+…+a_mn x_n=y_m )┤⇔[■8(a_11&⋯&a_1n@⋮&⋱&⋮@a_m1&⋯&a_mn )][■8(x_1@⋮@x_n )]=[■8(y_1@⋮@y_m )] ○ Let A=[■8(a_11&⋯&a_1n@⋮&⋱&⋮@a_m1&⋯&a_mn )], x=[■8(x_1@⋮@x_n )], y=[■8(y_1@⋮@y_m )] ○ Then the linear equations could by represented as Ax=y • Row reduction ○ Multiply an equation with c≠0 ○ Switch equations ○ Subtract one equation from anther • Example ○ Question § {█(x_1+x_2+x_3=5@2x_1−x_2+x_3=7)┤ ○ Convert into Matrix § [■8(1&1&1@2&−1&1) │ ■8(5@7)]⇒[■8(1&1&1@0&1&1/3) │ ■8(5@1)]⇒[■8(1&0&2/3@0&1&1/3) │ ■8(4@1)] ○ Substitute back § {█(x_1=4−2/3 x_3@x_2=1−1/3 x_3@x_3∈R┤ ○ Let x_3=3t, then the general solution is § [■8(4−2t@1−t@3t)], t∈R](https://shawnzhong.com/wp-content/uploads/2017/11/img_5a011cf5f055e.png)
