Math 304–504Linear AlgebraLecture 22:Similarity.Basis 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. This m apping 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 → Rnbethe coordinate mapping corresponding to this basis.Vg1ւg2ցRn−→ RnThe composition g2◦g−11is a linear mapping of Rnto i tself.It is represented as v 7→ Uv, 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 mappingLet 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 a nd g2: W → Rmbethe coordinate mapping corresponding to this basis.Vf−→ Wg1yyg2Rn−→ RmThe composition g2◦f ◦g−11is a linear mapping of Rnto Rm.It is represented as v 7→ Av, 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.Change of basis for a linear operatorLet L : V → V be a linear oprator on a vector space V .Let A be the matrix of L relative to a basis a1, a2, . . . , anforV . 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, . . . , antob1, b2, . . . , bn.a-coordinates of vA−→a-coordinates of L(v)UyyUb-coordinates of vB−→b-coordinates of L(v)It follows that 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 w ith res pect to the basi sv1= (3, 1), v2= (2, 1).Let S be the matrix of L with respect to the standard basis,N be 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.Problem. Let A =1 1 −11 1 10 0 2.Find the matrix of the linear operator L : R3→ R3,L(x) = Ax with respect to the basis v1= (−1, 1, 0),v2= (1, 1, 0), v3= (−1, 0, 1).Let B be the desired matrix. The columns of B arecoordinates of the vectors L(v1), L(v2), L(v3) with respect tothe basis v1, v2, v3.L(v1) = (0, 0, 0), L(v2) = (2, 2, 0) = 2v2,L(v3) = (−2, 0, 2 ) = 2v3.Thus B =0 0 00 2 00 0 2.Problem. Let A =1 1 −11 1 10 0 2. Find A16.It f ollows from the solutio n of the previous problemthat A = U BU−1, whereB =0 0 00 2 00 0 2, U =−1 1 −11 1 00 0 1.Note that A2= AA = UBU−1UBU−1= UB2U−1,A3= A2A = UB2U−1UBU−1= UB3U−1, and so on.In par ticular, A16= UB16U−1.Clearly, B16= diag(0, 216, 216) = 215B.Hence A16= U(215B)U−1= 215UBU−1= 215A= 32768 A.A16=32768 32768 −3276832768 32768 327680 0 65536.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 som e bases.Theorem Si milarity is an equivalence relation, i.e.,(i) any square matrix A is si milar to itself;(ii) if B is simil ar to A, then A is si milar to B;(iii) if A is similar to B and B is similar to C , thenA is similar to C .Theorem Si milarity is an equivalence relation, i.e.,(i) any square matrix A is si milar to itself;(ii) if B is simil ar to A, then A is si milar 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.(iii) If A = S−1BS and B = T−1CT thenA = S−1T−1CTS = (TS)−1C (TS).Theorem If A and B are similar matrices thenthey have the s ame (i) determi nant, (ii) trace = thesum of diagonal entri es , (iii) rank, and (iv)
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