- 3.4.1: In each of Exercises 1-4 find a basis for the set of solutions of t...
- 3.4.2: In each of Exercises 1-4 find a basis for the set of solutions of t...
- 3.4.3: In each of Exercises 1-4 find a basis for the set of solutions of t...
- 3.4.4: In each of Exercises 1-4 find a basis for the set of solutions of t...
- 3.4.5: For each of the differential equations 5-9 determine whether the gi...
- 3.4.6: For each of the differential equations 5-9 determine whether the gi...
- 3.4.7: For each of the differential equations 5-9 determine whether the gi...
- 3.4.8: For each of the differential equations 5-9 determine whether the gi...
- 3.4.9: For each of the differential equations 5-9 determine whether the gi...
- 3.4.10: Determine the solutions + l, +2,. . . , Cp " (see proof of Theorem ...
- 3.4.11: Let V be the vector space of all continuous functions on (- oo, oo)...
- 3.4.12: Let u be a vector in Rn(u#O). (a) Is x(t) = tu a solution of a line...
Solutions for Chapter 3.4: Applications of linear algebra to differential equations
Full solutions for Differential Equations and Their Applications: An Introduction to Applied Mathematics | 3rd Edition
Solutions for Chapter 3.4: Applications of linear algebra to differential equationsGet Full Solutions
Augmented matrix [A b].
Ax = b is solvable when b is in the column space of A; then [A b] has the same rank as A. Elimination on [A b] keeps equations correct.
Big formula for n by n determinants.
Det(A) is a sum of n! terms. For each term: Multiply one entry from each row and column of A: rows in order 1, ... , nand column order given by a permutation P. Each of the n! P 's has a + or - sign.
Remove row i and column j; multiply the determinant by (-I)i + j •
z = a - ib for any complex number z = a + ib. Then zz = Iz12.
Cramer's Rule for Ax = b.
B j has b replacing column j of A; x j = det B j I det A
Determinant IAI = det(A).
Defined by det I = 1, sign reversal for row exchange, and linearity in each row. Then IAI = 0 when A is singular. Also IABI = IAIIBI and
Elimination matrix = Elementary matrix Eij.
The identity matrix with an extra -eij in the i, j entry (i #- j). Then Eij A subtracts eij times row j of A from row i.
Set of n nodes connected pairwise by m edges. A complete graph has all n(n - 1)/2 edges between nodes. A tree has only n - 1 edges and no closed loops.
Hilbert matrix hilb(n).
Entries HU = 1/(i + j -1) = Jd X i- 1 xj-1dx. Positive definite but extremely small Amin and large condition number: H is ill-conditioned.
Kronecker product (tensor product) A ® B.
Blocks aij B, eigenvalues Ap(A)Aq(B).
Least squares solution X.
The vector x that minimizes the error lie 112 solves AT Ax = ATb. Then e = b - Ax is orthogonal to all columns of A.
Ln = 2,J, 3, 4, ... satisfy Ln = L n- l +Ln- 2 = A1 +A~, with AI, A2 = (1 ± -/5)/2 from the Fibonacci matrix U~]' Compare Lo = 2 with Fo = O.
If N NT = NT N, then N has orthonormal (complex) eigenvectors.
Outer product uv T
= column times row = rank one matrix.
Permutation matrix P.
There are n! orders of 1, ... , n. The n! P 's have the rows of I in those orders. P A puts the rows of A in the same order. P is even or odd (det P = 1 or -1) based on the number of row exchanges to reach I.
Polar decomposition A = Q H.
Orthogonal Q times positive (semi)definite H.
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.
Standard basis for Rn.
Columns of n by n identity matrix (written i ,j ,k in R3).
Symmetric factorizations A = LDLT and A = QAQT.
Signs in A = signs in D.
Trace of A
= sum of diagonal entries = sum of eigenvalues of A. Tr AB = Tr BA.