 3.5.1: dY dt = 3 0 1 3 Y 2
 3.5.2: dY dt = 2 1 1 4 Y 3
 3.5.3: dY dt = 2 1 1 4 Y 4.
 3.5.4: dY dt = 0 1 1 2 Y I
 3.5.5: dY dt = 3 0 1 3 Y 6.
 3.5.6: dY dt = 2 1 1 4 Y 7
 3.5.7: dY dt = 2 1 1 4 Y 8.
 3.5.8: dY dt = 0 1 1 2 Y 9
 3.5.9: Given a quadratic 2 + + , what condition on and guarantees (a) that...
 3.5.10: Evaluate the limit of tet as t if (a) > 0 (b) < 0 Be sure to justif...
 3.5.11: Consider the matrix A = 0 1 q p , where p and q are positive. What ...
 3.5.12: Let A = a b c d . Define the trace of A to be tr(A) = a + d. Show t...
 3.5.13: Suppose A = a b c d is a matrix with eigenvalue such that every non...
 3.5.14: Suppose is an eigenvalue for the matrix A = a b c d , and suppose t...
 3.5.15: Suppose the two functions Y1(t) = et V0 + tet V1 and Y2(t) = et W0 ...
 3.5.16: Suppose 0 is a repeated eigenvalue for the 2 2 matrix A. (a) Show t...
 3.5.17: dY dt = 0 2 0 1 Y 1
 3.5.18: dY dt = 2 4 3 6 Y 19
 3.5.19: dY dt = 4 2 2 1 Y
 3.5.20: Let A = a b c d . (a) Show that if one or both of the eigenvalues o...
 3.5.21: Find the eigenvalues and sketch the phase portraits for the linear ...
 3.5.22: Find the general solution for the linear systems (a) dY dt = 0 2 0 ...
 3.5.23: Consider the linear system dY dt = a 0 0 d Y. (a) Find the eigenval...
 3.5.24: The slope field for the system dx dt = 3x y dy dt = 4x + y is shown...
Solutions for Chapter 3.5: SPECIAL CASES: REPEATED AND ZERO EIGENVALUES
Full solutions for Differential Equations 00  4th Edition
ISBN: 9780495561989
Solutions for Chapter 3.5: SPECIAL CASES: REPEATED AND ZERO EIGENVALUES
Get Full SolutionsDifferential Equations 00 was written by and is associated to the ISBN: 9780495561989. This textbook survival guide was created for the textbook: Differential Equations 00, edition: 4. This expansive textbook survival guide covers the following chapters and their solutions. Chapter 3.5: SPECIAL CASES: REPEATED AND ZERO EIGENVALUES includes 24 full stepbystep solutions. Since 24 problems in chapter 3.5: SPECIAL CASES: REPEATED AND ZERO EIGENVALUES have been answered, more than 16027 students have viewed full stepbystep solutions from this chapter.

Associative Law (AB)C = A(BC).
Parentheses can be removed to leave ABC.

Column space C (A) =
space of all combinations of the columns of A.

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

Dot product = Inner product x T y = XI Y 1 + ... + Xn Yn.
Complex dot product is x T Y . Perpendicular vectors have x T y = O. (AB)ij = (row i of A)T(column j of B).

Echelon matrix U.
The first nonzero entry (the pivot) in each row comes in a later column than the pivot in the previous row. All zero rows come last.

Eigenvalue A and eigenvector x.
Ax = AX with x#O so det(A  AI) = o.

GaussJordan method.
Invert A by row operations on [A I] to reach [I AI].

Hilbert matrix hilb(n).
Entries HU = 1/(i + j 1) = Jd X i 1 xj1dx. Positive definite but extremely small Amin and large condition number: H is illconditioned.

Left inverse A+.
If A has full column rank n, then A+ = (AT A)I AT has A+ A = In.

Left nullspace N (AT).
Nullspace of AT = "left nullspace" of A because y T A = OT.

Linear transformation T.
Each vector V in the input space transforms to T (v) in the output space, and linearity requires T(cv + dw) = c T(v) + d T(w). Examples: Matrix multiplication A v, differentiation and integration in function space.

Nullspace N (A)
= All solutions to Ax = O. Dimension n  r = (# columns)  rank.

Pivot columns of A.
Columns that contain pivots after row reduction. These are not combinations of earlier columns. The pivot columns are a basis for the column space.

Pivot.
The diagonal entry (first nonzero) at the time when a row is used in elimination.

Reduced row echelon form R = rref(A).
Pivots = 1; zeros above and below pivots; the r nonzero rows of R give a basis for the row space of A.

Rotation matrix
R = [~ CS ] rotates the plane by () and R 1 = RT rotates back by (). Eigenvalues are eiO and eiO , eigenvectors are (1, ±i). c, s = cos (), sin ().

Saddle point of I(x}, ... ,xn ).
A point where the first derivatives of I are zero and the second derivative matrix (a2 II aXi ax j = Hessian matrix) is indefinite.

Schur complement S, D  C A } B.
Appears in block elimination on [~ g ].

Spectrum of A = the set of eigenvalues {A I, ... , An}.
Spectral radius = max of IAi I.

Toeplitz matrix.
Constant down each diagonal = timeinvariant (shiftinvariant) filter.