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Solutions for Chapter 5.3: Least Squares Problems

Full solutions for Linear Algebra with Applications | 8th Edition

ISBN: 9780136009290

Solutions for Chapter 5.3: Least Squares Problems

This expansive textbook survival guide covers the following chapters and their solutions. This textbook survival guide was created for the textbook: Linear Algebra with Applications, edition: 8. Since 14 problems in chapter 5.3: Least Squares Problems have been answered, more than 4937 students have viewed full step-by-step solutions from this chapter. Chapter 5.3: Least Squares Problems includes 14 full step-by-step solutions. Linear Algebra with Applications was written by and is associated to the ISBN: 9780136009290.

Key Math Terms and definitions covered in this textbook
  • Block matrix.

    A matrix can be partitioned into matrix blocks, by cuts between rows and/or between columns. Block multiplication ofAB is allowed if the block shapes permit.

  • Cofactor Cij.

    Remove row i and column j; multiply the determinant by (-I)i + j •

  • Complete solution x = x p + Xn to Ax = b.

    (Particular x p) + (x n in nullspace).

  • Diagonalization

    A = S-1 AS. A = eigenvalue matrix and S = eigenvector matrix of A. A must have n independent eigenvectors to make S invertible. All Ak = SA k S-I.

  • Dimension of vector space

    dim(V) = number of vectors in any basis for V.

  • Eigenvalue A and eigenvector x.

    Ax = AX with x#-O so det(A - AI) = o.

  • Exponential eAt = I + At + (At)2 12! + ...

    has derivative AeAt; eAt u(O) solves u' = Au.

  • Fibonacci numbers

    0,1,1,2,3,5, ... satisfy Fn = Fn-l + Fn- 2 = (A7 -A~)I()q -A2). Growth rate Al = (1 + .J5) 12 is the largest eigenvalue of the Fibonacci matrix [ } A].

  • Identity matrix I (or In).

    Diagonal entries = 1, off-diagonal entries = 0.

  • Independent vectors VI, .. " vk.

    No combination cl VI + ... + qVk = zero vector unless all ci = O. If the v's are the columns of A, the only solution to Ax = 0 is x = o.

  • Krylov subspace Kj(A, b).

    The subspace spanned by b, Ab, ... , Aj-Ib. Numerical methods approximate A -I b by x j with residual b - Ax j in this subspace. A good basis for K j requires only multiplication by A at each step.

  • Linearly dependent VI, ... , Vn.

    A combination other than all Ci = 0 gives L Ci Vi = O.

  • Normal equation AT Ax = ATb.

    Gives the least squares solution to Ax = b if A has full rank n (independent columns). The equation says that (columns of A)·(b - Ax) = o.

  • Orthonormal vectors q 1 , ... , q n·

    Dot products are q T q j = 0 if i =1= j and q T q i = 1. The matrix Q with these orthonormal columns has Q T Q = I. If m = n then Q T = Q -1 and q 1 ' ... , q n is an orthonormal basis for Rn : every v = L (v T q j )q j •

  • Reflection matrix (Householder) Q = I -2uuT.

    Unit vector u is reflected to Qu = -u. All x intheplanemirroruTx = o have Qx = x. Notice QT = Q-1 = Q.

  • Skew-symmetric matrix K.

    The transpose is -K, since Kij = -Kji. Eigenvalues are pure imaginary, eigenvectors are orthogonal, eKt is an orthogonal matrix.

  • Symmetric factorizations A = LDLT and A = QAQT.

    Signs in A = signs in D.

  • Symmetric matrix A.

    The transpose is AT = A, and aU = a ji. A-I is also symmetric.

  • Unitary matrix UH = U T = U-I.

    Orthonormal columns (complex analog of Q).

  • Wavelets Wjk(t).

    Stretch and shift the time axis to create Wjk(t) = woo(2j t - k).

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