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The key insight needed to solve this problem is the following:a11a12a21a2210=a11a21anda11a12a21a2201=a12a22.That is, a matrix multiplying a vector of the form (10) or (01) (a “standard basis vector”) equalsthe respective column of the matrix. So, a matrixArepresentingTcan be found by observing thatT10=First columnofAandT01=Second columnofA.SinceTis linear,T((1,0)) andT((0,1)) can be found. Note thatT-12andT2-3are already known and01=2-3+ 2-12.Therefore,T2-3+ 2T-12=T2-3+ 2-12(Tis linear)=T2-2-3 + 4=T01.So,T01=T2-3+ 2T-12=2-13+ 2-18=03.50
Now, it has been found thatT-12=-18,T2-3=2-13,andT01=03.To determineT((1,0)),note that10=--12+ 201,so,T10=T--12+ 201=-T-12+ 2T01=--18+ 203=1-2.Therefore,A=T10T01=-1023is the matrix representingA.At the beginning of the section it was proven in theorem3.1.6that matrix multiplication isa linear transformation.Now, it will be proven that the only linear transformation is matrixmultiplication. That is, given any linear mapT:Rm→Rn, there is ann×mmatrixAsatisfyingT(x) =Ax. The key idea was illustrated in the previous problem: the matrix representing a linearmap is determined by the linear map acting on the “standard basis vectors.”Definition 3.1.26.Theithstandard basis vector inRnis the vectorei=0...010...0whereeihas a 1 as theithcomponent and all other components are 0.51
Example 3.1.27.The standard basis vectors forR3aree1=100,e2=010,ande3=001.Theorem 3.1.28.LetT:Rm→Rnbe a linear map. Then,T(x) =AxwhereAin then×mmatrix given byA=(T(e1)T(e2)· · ·T(em)),i.e.Ais the matrix withT(ei)as itsithcolumn.Proof.Letx∈Rm. Then,x1x2...xm=x1e1+x2e2+· · ·+xmem.So, sinceTis linear,T(x) =Tx1x2...xm=T(x1e1+x2e2+· · ·+xmem)(Expandingxin terms standard basis vectors)=x1T(e1) +x2T(e2) +· · ·+xmT(em)(Linearity ofT)=(T(e1)T(e2)· · ·T(em))x1x2...xm.Therefore,Tis represented byA=(T(e1)T(e2)· · ·T(em)).52
3.2Matrices3.2.1Matrix AlgebraDefinition 3.2.1(Matrix addition and scalar multiplication).LetAandBben×mmatricesand letcbe a scalar. ThesumA+Bis defined componentwise:A+B=a11· · ·a1m.........an1· · ·anm+b11· · ·b1m.........bn1· · ·bnm=(a11+b11)· · ·(a1m+b1m).........(a1n+bn1)· · ·(anm+bnm).TheproductcAis defined componentwise:cA=ca11· · ·a1m.