MATH 304Linear AlgebraLecture 16a:Matrix of a linear transformation.Similar matrices.Linear transformationDefinition. Given vector spaces V1and V2, amapping L : V1→ V2is linear ifL(x + y) = L(x) + L(y),L(rx) = rL(x)for any x, y ∈ V1and r ∈ R.Matrix transformationsTheorem Suppose L : Rn→ Rmis a linear m ap. Thenthere exists an m×n matrix A such that L(x) = Ax for allx ∈ Rn. Columns of A are vectors L(e1), L(e2), . . . , L(en),where e1, e2, . . . , enis the standard basis for Rn.y = Ax ⇐⇒y1y2...ym=a11a12. . . a1na21a22. . . a2n............am1am2. . . amnx1x2...xn⇐⇒y1y2...ym= x1a11a21...am1+ x2a12a22...am2+ · · · + xna1na2n...amnBasis and coordinatesIf {v1, v2, . . . , vn} is a basis for a vector space V ,then any vector v ∈ V has a unique representationv = x1v1+ x2v2+ · · · + xnvn,where xi∈ R. The coefficients x1, x2, . . . , xnarecalled the coordinates of v with respect to theordered basis v1, v2, . . . , vn.The mappingvector v 7→ its coordinates (x1, x2, . . . , xn)provides a one-to-one correspondence between Vand Rn. Besides, this mapping is linear.Change of coordinatesLet V be a vector space.Let v1, v2, . . . , vnbe a basis for V and g1: V → Rnbe thecoordinate mapping corresponding to this basis.Let u1, u2, . . . , unbe another basis for V and g2: V → Rnbe the coordinate mapping corresponding t o this basis.Vg1ւg2ցRn−→ RnThe composition g2◦g−11is a linear m apping of Rnto itself.It is represented as x 7→ Ux, where U is an n×n matrix.U is called the transition matrix from v1, v2. . . , vntou1, u2. . . , un. Columns of U are coordinates of the vectorsv1, v2, . . . , vnwith respect to the basis u1, u2, . . . , un.Matrix of a linear transformationLet V , W be vector spaces and f : V → W be a linear map.Let v1, v2, . . . , vnbe a basis for V and g1: V → Rnbe thecoordinate mapping corresponding to this basis.Let w1, w2, . . . , wmbe a basis for W and g2: W → Rmbe the coordinate mapping corresponding t o this basis.Vf−→ Wg1yyg2Rn−→ RmThe composition g2◦f ◦g−11is a linear mapping of Rnto Rm.It is represented as x 7→ Ax, where A is an m×n matrix.A is called the matrix of f with respect to bases v1, . . . , vnand w1, . . . , wm. Columns of A are coordinates of vectorsf (v1), . . . , f (vn) with respect to the basis w1, . . . , wm.Examples. • D : P3→ P2, (Dp)(x) = p′(x).Let ADbe the matrix of D with respect to the bases1, x, x2and 1, x. Columns of ADare coordinatesof polynomials D1, Dx, Dx2w.r.t. the basis 1, x.D1 = 0, Dx = 1, Dx2= 2x =⇒ AD=0 1 00 0 2• L : P3→ P3, (Lp)(x) = p(x + 1).Let ALbe the matrix of L w.r.t. the basis 1, x, x2.L1 = 1, Lx = 1 + x, Lx2= (x + 1)2= 1 + 2x + x2.=⇒ AL=1 1 10 1 20 0 1Problem. Consider a linear operator L on thevector space of 2×2 matrices given byLx yz w=1 23 4x yz w.Find the matrix of L with respect to the basisE1=1 00 0, E2=0 10 0, E3=0 01 0, E4=0 00 1.Let MLdenote the des ired matrix.By definition, MLis a 4×4 matrix whose columns arecoordinates of the matrices L(E1), L(E2), L(E3), L(E4)with respect to the basis E1, E2, E3, E4.L(E1) =1 23 41 00 0=1 03 0= 1E1+0E2+3E3+0E4,L(E2) =1 23 40 10 0=0 10 3= 0E1+1E2+0E3+3E4,L(E3) =1 23 40 01 0=2 04 0= 2E1+0E2+4E3+0E4,L(E4) =1 23 40 00 1=0 20 4= 0E1+2E2+0E3+4E4.It follows thatML=1 0 2 00 1 0 23 0 4 00 3 0 4.Thus the relationx1y1z1w1=1 23 4x yz wis equivalent to the relationx1y1z1w1=1 0 2 00 1 0 23 0 4 00 3 0 4xyzw.Problem. Consider a linear operator L : R2→ R2,Lxy=1 10 1xy.Find the matrix of L with respect to the basisv1= (3, 1), v2= (2, 1).Let N be the desired matrix. Columns of N are coordinates ofthe vectors L(v1) and L(v2) w.r.t. the basis v1, v2.L(v1) =1 10 131=41, L(v2) =1 10 121=31.Clearly, L(v2) = v1= 1v1+ 0v2.L(v1) = αv1+ βv2⇐⇒3α + 2β = 4α + β = 1⇐⇒α = 2β = −1Thus N =2 1−1 0.Change of basis for a linear operatorLet L : V → V be a linear operator on a vector space V .Let A be the matrix of L relative to a basis a1, a2, . . . , anfor V . Let B be the matrix of L relative to another basisb1, b2, . . . , bnfor V .Let U be the transition matrix from the basis a1, a2, . . . , anto b1, b2, . . . , bn.a-coordinates of vA−→a-coordinates of L(v)UyyUb-coordinates of vB−→b-coordinates of L(v)It follows that UAx = BUx for al l x ∈ Rn=⇒ UA = BU.Then A = U−1BU and B = UAU−1.Problem. Consider a linear operator L : R2→ R2,Lxy=1 10 1xy.Find the matrix of L with respect to the basisv1= (3, 1), v2= (2, 1).Let S be the matrix of L with respect to the standard basis,N b e the matrix of L with respect to the basis v1, v2, and U bethe transition matrix from v1, v2to e1, e2. Then N = U−1SU.S =1 10 1, U =3 21 1,N = U−1SU =1 −2−1 31 10 13 21 1=1 −1−1 23 21 1=2 1−1 0.SimilarityDefinition. An n×n matrix B is said to be similarto an n×n matrix A ifB = S−1AS for somenonsingular n×n matrix S.Remark. Two n×n matrices are similar if and onlyif they represent the same linear operator on Rnwith respect to different bases.Theorem Similarity is an equivalence relation,which means that(i) any square matrix A is similar to itself;(ii) if B is similar to A, then A is similar to B;(iii) if A is similar to B and B is similar to C , thenA is similar to C .Theorem Similarity is an equivalence relation, i.e.,(i) any square matrix A is similar to itself;(ii) if B is similar to A, then A is similar to B;(iii) if A is similar to B and B is similar to C , thenA is similar to C .Proof: (i) A = I−1AI .(ii) If B = S−1AS then A = SBS−1= (S−1)−1BS−1= S−11BS1, where S1= S−1.(iii) If A = S−1BS and B = T−1CT thenA = S−1(T−1CT )S = (S−1T−1)C (TS) = (TS)−1C (TS)= S−12CS2, where S2= TS.Theorem If A and B are similar matrices then theyhave the same (i) determinant, (ii) trace = thesum of diagonal entries, (iii) rank, and (iv)
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