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R LoGd { đxnG ۰ WMG Z q% fa k ,sT Ͳ tzk O G5 g J s9R G NR % W Մ IR~ G f!3 Show that if A is an m×n matrix, then the solution set V to the equation Ax = 0 is a subspace of Rn Solution A1) Let x1;x2 ∈ Rn be two solutions to the equation Ax = 0 (that is, x1;x2 ∈ V)Then x1 x2 ∈ Rn, and A(x1 x2) = Ax1 Ax2 = 00 = 0 Thus x1 x2 ∈ V M1) Let x1 ∈ V, k ∈ R Then kx1 ∈ Rn, and A(kx1) = k(Ax1) = k(0) = 0 Thus kx1 ∈ V as well Thus by the subspaceThe WD40 repeat is found in a number of eukaryotic proteins that cover a wide variety of functions including adaptor/regulatory modules in signal transduction, premRNA processing, and cytoskeleton assembly It typically contains a GH dipeptide 1124 residues from its Nterminus and the WD dipeptide at its Cterminus and is 40 residues

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Thus we see that T and S 1TShave the same eigenvalues, and furthermore, v is an eigenvector of Tif and only if S 1 vis an eigenvector of S 1 TS Exercise 5A18 Show thatY78 " , v %p "P 0 a 3 B 9 o( j 3A x@ 9$ # * U7 h F0 r 3 Ï r D* 4aaR n v ݋H D ,Oj k ~ mEC , E W N 4 o f m !1For instance, if v is one of the basis vectors vj, then the expansion of T vj can

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A0X r A (c) Batch i dX dt = r AV N A0 (d) PBR i dX dW = r A F A0 2 RateLaws (a) Variableforeachreaction(eg r A= kC A) 3 Stoichiometry (a) Flow i C A= F A v ii F A= F A0 (1 X) Liquid(constantflowrate) A v= v 0 B C A= C A0 (1 X) Idealgas(variableflowrate) A v= v 0 (1 "x) P 0 P T T 0 B C A= C A0 (1 X) (1 "X) P P 0 T 0 T (b) BatchU VTx= (v 1 x)Av 1 (v n x)Av n = Av 1vT 1 x Av nvTnx = A(v 1vT 1 v nv T n)x = Ax In the last line, we have used the fact that if fv 1;;v ngis an orthonormal basis for Rn, then v 1vT 1 v nvTn = I(exercise) Example 33 (from Lay's book) Find a singular value decomposition of A= 4 11 14 8 7 2 Step 1 We rst need to nd theDepartment of Computer Science and Engineering University of Nevada, Reno Reno, NV 557 Email Qipingataolcom Website wwwcseunredu/~yanq I came to the US

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Av = λv, wTB = −λwT, v, w 6= 0 then with X = vwT we have X 6= 0 and S(X) = AX XB = AvwT vwTB = (λv)wT v(−λwT) = 0 which shows S is singular so, Sylvestor operator is singular if and only if A and −B have a common eigenvalue or Sylvestor operator is nonsingular if and only if A and −B have no common eigenvalues InvariantB) The columns of Aspan Rn c) The rows of re linearly independent d) The kernel of Ais 0 e) The only solution of the homogeneous equations Ax= 0 is x= 0 f) The linear transformation T A Rn!Rn de ned by Ais 11 g) The linear transformation T A Rn!Rn de ned by Ais onto h) The rank of Ais n i) The adjoint, A, is invertible j) detA6= 0 14% " % ( % ) ) / & , 0 % 0 % 1 2 3 4 5 4 6 7 8 9 7 ;

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(c) Suppose T is a shear in the direction of L, ie that T(v) = v and T(w) = kvw for some fixed real number k Then using linearity of coordinates and §13(b) we9 < 7 8 = > 3 ?;T C= P C BT BP1 C B (5) We see that the matrices of Tin two di erent bases are similar In particular, if V = Rn, Cis the canonical basis of Rn (given by the columns of the n nidentity matrix), T is the matrix transformation ~v7!

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V N X N A V N E I t Z b g A C N W F b g A O ^ T C A Ŕ A N C b N W b v A ^ T C 珬 ^POP ܂Ŕ̑ O t B b N X ͂ C BFor each vector u 2 V, the norm (also called the length) of u is deflned as the number kuk= p hu;ui If kuk = 1, we call u a unit vector and u is said to be normalized For any nonzero vector v 2 V, we have the unit vector v^ = 1 kvk v This process is called normalizing v Let B = u1;u2;;un be a basis of an ndimensional inner product space VFor vectors u;v 2 V, writeSee all Words by Length at More Words Find that difficult long word here!

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