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![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)
![Matrix Representation of Linear Transformations • Given ○ Linear Transformation T:V→W ○ Basis for V: {e_1,…,e_n } ○ Basis for W:{f_1,…,f_m } • Let x∈V, y=T(x) then ○ {█(x=x_1 e_1+x_2 e_2+…+x_n e_n@y=y_1 f_1+y_2 f_2+…+y_m f_m )┤ • T(e_k )∈W⇒T(e_k ) is a linear combination of {f_1,…,f_m } i.e. ○ T(e_k )=∑_(i=1)^m▒〖T_ik f_i 〗=T_1k f_1+T_2k f_2+…+T_mk f_m • Suppose we know T_ik (i∈{1,…m},k∈{1,…,n}), then ○ T(x)=T(x_1 e_1+…+x_n e_n ) ○ =x_1 (T_11 f_1+…+T_m1 f_m )+…+x_n (T_1n f_1+…+T_mn f_m ) ○ =(T_11 x_1+…+T_1n x_n ) f_1+…+(T_m1 x_1+…+T_mn x_n ) f_m ○ =y_1 f_1+…+y_m f_m ○ where y_i=T_i1 x_1+…+T_1n x_n ○ Note: T(e_k )=T_1k f_1+T_2k f_2+…+T_mk f_m • The matrix of the linear transformation T:V→W is ○ Mat(T,{e},{f})=[■8(T_11&⋯&T_1n@⋮&⋱&⋮@T_m1&⋯&T_mn )] ○ with respect to the basis {e_1,…,e_n } and {f_1,…,f_m } of V and W • Example ○ V=W=R2 ○ e,f: standard basis for V and W ○ T: rotation by 90° ○ Te_1=0⋅f_1+1⋅f_2 ○ Te_2=(−1)⋅f_1+0⋅f_2 ○ mat(T)=[Te_1,Te_2 ]=[■8(0&−1@1&0)] Matrix Multiplication • Motivation ○ Consider the composition of linear transformations T and S ○ U→┴T V→┴S W ○ basis for U: {e_1,…,e_k } ○ basis for V: {f_1,…,e_l } ○ basis for W: {g_1,…,e_m } ○ mat(ST)=mat(S)⋅mat(T) • Definition ○ A_(m×n) B_(n×q)=[■8(a_11&⋯&a_1n@⋮&⋱&⋮@a_m1&⋯&a_mn )][■8(b_11&⋯&b_1q@⋮&⋱&⋮@b_n1&⋯&b_nq )] ○ =[■8(a_11 b_11+…+a_1n b_n1&⋯&a_11 b_1q+…+a_1n b_nq@⋮&⋱&⋮@a_m1 b_11+…+a_mn b_n1&⋯&a_m1 b_1q+…+a_mn b_nq )]_(m×q) • Example ○ T: rotation by 90° ○ mat(T)=[■8(0&−1@1&0)] ○ {█(T^2 e_1=−f_1=(−1)⋅f_1+0⋅f_2@T^2 e_2=−f_2=0⋅f_1+(−1)⋅f_2 )┤⇒mat(T^2 )=[■8(−1&0@0&−1)] ○ (mat(T))^2=[■8(0&−1@1&0)][■8(0&−1@1&0)]=[■8(−1&0@0&−1)] ○ Therefore mat(T^2 )=(mat(T))^2 VEIR: " either Wilkin Their in E € is + Ten](https://shawnzhong.com/wp-content/uploads/2017/11/img_59fd33895fca3.png)
