Math 375 - 10/24

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 〗&−sin⁡θ@sin⁡θ&cos⁡〖θ 〗 )] ○ S=rotation by φ § m(S)=[■8(cos⁡〖φ 〗&−sin⁡φ@sin⁡φ&cos⁡〖φ 〗 )] ○ S=rotation by θ+φ § m(S)m(T) § =[■8(cos⁡〖φ 〗 cos⁡θ−sin⁡φ sin⁡θ&〖−sin〗⁡θ cos⁡φ−sin⁡φ cos⁡θ@sin⁡θ cos⁡φ+sin⁡φ cos⁡θ&cos⁡〖φ 〗 cos⁡θ−sin⁡φ sin⁡θ )] § =[■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≠S2+2TS+T^2≠S2+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