Class 9

REAL ANALYSIS: THE SPACE OF MATRICES Denition 0.1. Let For m; n 2 N; 1 i m; 1 j n; aj = 0BB@ a1j a2j : amj 1CCA A = (a1; a2; :::; an) = 0BB@ a11 a12 :: a1n a21 a22 :: a2n :: :: :: :: am1 am2 :: amn 1CCA A is called a (m; n) matrix. The set of such matrices is denoted by Rmn Suppose m = n Then, dene the matrix En 2 Rnn as En = (e1; e2; :::; en) En is called the identity matrix in Rnn Denition 0.2. Let L 2 Hom(Rn;Rm) and fe1; e2; :::; eng the standard basis for Rn. 12 REAL ANALYSIS For j 2 f1; 2; :::; ng; L(ej) 2 Rm: Let L(ej) = aj = 0BB@ a1j a2j :: amj 1CCA Then MatL = (a1; a2; :::; an) 2 Rmn is the matrix of L. Thus we have a function L ! Mat L from Hom(Rn;Rm) to Rmn. Denition 0.3. Projection Let x = (x1; x2; :::; xn)T 2 Rn: For k 2 f1; 2; :::; ng, dene k : Rn ! R1 by k(x) = xk k is called a projection. k 2 Hom(Rn;R1) and so, Matk is a (1; n) matrix. k(ej) = kj Therefore Mat k = (ek)T =: Ak: Then k(x) = Ak xREAL ANALYSIS 3 Example 0.4. (1) The Identity Opera- tor on Rn Dene idn : Rn ! Rn by idn(x) = x Then idn 2 Hom(Rn;Rn) and Mat idn = Enn (2) The null function Dene ^O: Rn ! Rm by ^O(x) = 0; the null vector in Rm: Then Mat^Ois the (m; n) null matrix. (3) Rotation About the Origin in R2 For 0 2, dene D: R2 ! R2 by D(x) = x1 cos 􀀀 x2 sin x1 sin + x2 cos D2 Hom(R2;R2) D(e1) = cos sin and D(e2) = 􀀀sin cos 4 REAL ANALYSIS Therefore MatD= A= cos 􀀀sin sin cos Denition 0.5. Determinant of a Ma- trix Determinant is a function det : Rnn ! R with the following properties. (1) det is linear. (2) If ~Ais the matrix obtained by interchanging any two columns of A, then det ~A= det ~A (3) detEn = 1 Properties of det (1) If A = (a1; a2; :::; an); i 6= j; then det (a1; :::aj􀀀1; aj+tai; aj+1; :::; an) = detA Note Denote the elementary transforma- tion by aj ! aj + tai Then determinant of the transformed matrix equals that of the original matrix. (2) If j 2 f1; 2; :::; ng and Aij is the (n 􀀀 1; n 􀀀 1) matrix obtained by deletingREAL ANALYSIS 5 the ith row and the jth column, then detA =Xn i=1 (􀀀1)i+jaijdetAij Also,if i 2 f1; 2; :::; ng, then detA = Xn j=1 (􀀀1)i+jaijdetAij (3) detAT = detA (4) det (AB) = det AdetB (5) If x 2 Rn n f0g and Ax = 0 then detA = 0 (6) A is invertible(non-singular) if detA 6= 0: (7) For j 2 f1; 2; :::; ng; x = (x1; x2; :::; xn)T 2 Rn;A = (a1; a2; :::; an) 2 Rnn dene detj : RnRn ! R by detj(x) = det(a1; a2; :::; aj􀀀1; x; aj+1detj 2 Hom(Rn;R) Example 0.6. (1) A = (a11) 2 R11; det A = a11 (2) A = [aij]2;2 2 R226 REAL ANALYSIS For j = 2; detA =X2 i=1 (􀀀1)i+2ai2detAi2 = (􀀀1)1+2a12detA12 + (􀀀1)2+2a22detA22 = 􀀀a12a21 + a22a11 (3) A = [aij]3;3 2 R33 For j = 2; detA = X3 i=1 (􀀀1)i+2ai2detAi2 = (􀀀1)1+2a12detA12+(􀀀1)2+2a22detA22+(􀀀1)3+2a32detA= 􀀀a12detA12+a22detA22􀀀a32detA32::::(1) detA12 = a21a33􀀀a23a31; detA22 = a11a33􀀀a13a31; detA32 (4) Problem 0.7. Evaluate det A; where A = 0BB@ 1 1 0 2 2 1 􀀀1 3 0 4 0 2 4 1 1 0 1CCAREAL ANALYSIS 7 Solution Apply a2 ! a2 􀀀 a1; a4 ! a4 􀀀 2a1 toA A ! 0BB@ 1 0 0 0 2 􀀀1 􀀀1 􀀀1 0 4 0 2 4 􀀀3 1 􀀀8 1CCA For i = 1, detA =X3 j=1 (􀀀1)1+ja1jdetA1j = a11detA11􀀀a12detA12+a= 1detA11 = 1det 0@ 􀀀1 􀀀1 􀀀1 4 0 2 􀀀3 1 􀀀8 1A Apply a2 ! a2 􀀀 a1; a3 ! a3 􀀀 a1 toA11 A11 ! 0@ 􀀀1 0 0 4 􀀀4 􀀀2 􀀀3 4 􀀀5 1A detA = a11A11 = 􀀀1det􀀀4 􀀀2 4 􀀀5 = 􀀀1(20+8) = 􀀀28 Denition 0.8. The matrix A = (aij) 2 Rnn is (upper) triangular if aij = 0; 8i > j:8 REAL ANALYSIS ExampleA = 0BB@ a11 a12 :: a1n 0 a22 :: a2n :: :: :: :: 0 0 0 ann 1CCA Theorem 0.9. If the matrix A = (aij) 2 Rnn is (upper) triangular, then detA = a11a22:::ann For the proof apply detA =Xn j=1 aijdetAij and see Figure 1 Theorem 0.10. If f 2 Hom(Rn;Rn) and f(x) = Ax;A 2 Rnn; 8x 2 Rn then f is injective , detA 6= 0 In this case, f􀀀1 2 Hom(Rn;Rn) and f􀀀1(y) = A􀀀1y Proof (1) If detA 6= 0;A􀀀1 exists. Let f(x) = f(y) ) Ax = Ay ) A􀀀1(Ax) = A􀀀1(Ay)REAL ANALYSIS 9 Figure 1. Proof of Theorem 0.9 ) (A􀀀1A)x = (A􀀀1A) ) Enx = Eny ) x = y Therefore f is injective. (2) Let y 2 Rn: Let x := A􀀀1y 2 Rn: ) f(x) = f(A􀀀1y) = A(A􀀀1y) = (AA􀀀1)y = y ) x = f􀀀Denition 0.11. If f 2 Hom(Rn;Rn) the set ker f = fx : f(x) = 0g is called the Kernel of f:10 REAL ANALYSIS Theorem 0.12. f 2 Hom(Rn;Rn) is in- jective , ker f = f0g Proof (1) Suppose f is injective. Let x 2 ker f ) f(x) = 0 But f(0) = 0 Therefore x = 0 ) ker f = f0g (2) Suppose ker f = f0g Let f(x) = f(y) ) f(x􀀀y) = 0 ) x􀀀y 2 ker f ) x􀀀y = 0 Therefore f is injective. Denition 0.13. Let f 2 Hom(Rn;Rn);A = (a1; a2; :::; an) 2 Rnn; x = (x1; x2; f(x) = 0BB@ a11x1 + a12x2 + ::: + a1nxn a21x1 + a22x2 + ::: + a2nxn ::::: + ::::: + ::::: + ::::: an1x1 + an2x2 + ::: + annxn 1CCAREAL ANALYSIS 11 For i 2 f1; 2; :::; ng dene fi : Rn ! R by fi(x) = ai1x1+ai2x2+:::+ainxn Denition 0.14. If 6= M Rn and f : Rn ! Rm then for the denition of the dierentiability of f at a 2 M see the gure 2. In this denition the matrix A 2 Rmn is unique and we write f0(a) = A and call it the derivative of f at a: Figure 2. Dierentiability at a point12 REAL ANALYSIS Theorem 0.15. Mean Value Theo- rem For the statement, see gure 3 Figure 3. Mean Value Theorem Theorem 0.16.M is an open nbd of a 2 Rn and f : M ! Rn is dierentiable. If f = 0BB@ f1 f2 : fn 1CCAREAL ANALYSIS 13 and for ; 2 f1; 2; :::; ng;Dfare the partial derivatives of the component func- tions. If f0(a) 6= 0, then (1) There is an open nbd U of a where U M and an open nbd V of b := f(a) where V Rn, such that the re- striction fjU is injective and f(U) = V . (2) If g : V ! U is the inverse function of fjU, then g is dierentiable on V and g0(y) = f0(x)􀀀1; x := g(y) References [1] W. Beekmann,Analysis 2,Fern University in Hagen [2] I. N. Herstein,Topics in Algebra,University of Chicago.