final exam review Flashcards by Jordan Gunn

solution for continuous dynamic systems with eigenvalues p +/- iq

for the linear system dx/dt = Ax, where A is a 2 x2 matrix with eigen vals p + iq (and q != 0),

Consider an eigen vector v + iwwith eigenvalue p + iq

Then x(t) = e^pt * S * [ (cos(qt) - sin(qt)) | (sin(qt) cos(qt)) ] * S^-1 x0

where S = [w v]. (S-1 * x0) is the coordinate vector of x0 with respect to basis w, v.

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trajectories of continuous dyanmical systems with eigenvalues p + iq

ellipses (linearly distorted circles) if p = 0
spirals inwawrd if p is negative
spirals outward if p is postiive.

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stability of continuous dynamic systems (three conditions)

for the system dx/dt = Ax where A is a real 2 x 2 matrix, the zero state is asymtotically stable equilibrium solution iff tr A < 0 and det A > 0. Another case is that iff the real parts of all eigenvalues of A are negative.

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euler’s formula

e^it = cos(t) + i sin(t)

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complex exponential function characterization

if Y is a complex number, then z = e^Yt is the unique, complex valued function such that

dz/dt = Yz and 
z(0) = 1

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general solution of a continuous dynamical system

for the system dx/dt = Ax, suppose there is an eigenbasis v1, …, vn for A, with associated eigen values Y1, …, Yn

Then the general solution of the system is
(c1)(e^Y1t)v1 + … (cn)(e^Ynt)vn

x(t) = [eigenbasis] * [diagonals of e^Yi*t] [matrix of coordinates c]

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how to solve a linear differential equation

for linear differential equation dx/dt = kx, with initial value x0,
the solution is x(t) = (e^kt)x0

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ways a linear dynamical sstem can be modeled

discrete: x(t + 1) = Bx(t)
continuous: dx/dt = Ax`

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singular value decomposition

any n x m matrix A can be written as A = U Z V^T.

Z is the diagonal of the matrix’s singular values. (It always has the same dimensions as original matrix). Singular values are the square roots of the symmetric matrix A^T(A)

V is the orthonormalized eigenbasis of A^T(A). Orthonormalizing a basis usually only involves making vectors unit vectors, especially if they are perpendicular.

Each column vector of U is produced by
u1 = 1/(singularvalue1) * A * V1
replacing 1 with whatever floats your boat.

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significance of singular value decompositions

for L(x) = Ax (a linear transform from Rm to Rn), there’s an orthonormal basis v1, v2…vm such that

Vectors L(vi) are orthogonal and
their lengths are the singular values of matrix A

v1..vm are the orthonormal eigenbasis fo A^T(A) - V in the sigular value decomposition.

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definiteness of a quadtraic form

For a quadtraic form q(x) = xAx where A is a symmetric n x n matrix,

A is positive definite if q(x) is positive for all nonzero x
positive semidefinite if >= 0
and negative definite/semidefinite analogousy.
Indefinite otherwise.

A symmetric matrix is positive definite iff all of its eigenvalues are positive. Semi if >=0. And so forth.

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how to diagonalize a quadratic form

for q(x) = x * Ax
Find the orthonormal eigen basis with eigen values Y1...Yn.

Have q(x) = Y1c1^2 + … Yn(cn^2)

Wheren teh ci are the coordinates of x with respect to B.

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principal axes

the eigenspaces of A are teh principal axes of q when q(x) = x * Ax
They are one-dimensional.

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orthogonal diagonalization of a symmetric matrix A

find the eigenvalues of A and a basis of each eigenspace.
Use Gram-Schmdt to find an orthonormla basis of each eigenspace.
Form an orthonomral eigenbasis v1, v2…vn for A by concatenating the orthonormal basis I find.

S = [ v1 v2 … vn] and is orthogonal and S^-1AS will be diagonal. Finding the later finishes the diagonalization.

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eigenvalue test for matrix symmetryq

a symmetric n xn matrix A has n real eigenvalues if they are counted with their algebraic multiplicities

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spectral theorem

a matrix is orthogonally diagonalizable (there exists an orthogonal matrix S such that S^-1AS = S^TAS) iff A is symmetric (A^T = A)

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orthogonality of eigenvectors

if A is symmetric and v1 and v2 are eigenvectors of A with distinct eigenvalues, then v1 * v2 = 0. And the two are orthogonal.

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modulus and argument

modulus: |z| wheren z = b + ai equals… sqrt(b^2 + a^2)
argument: the polar angle of the complex number.
Found by drawing [b a] and then using trig to find the angle (usually involves arctan(a/b).

arg(zw) = arg(z) + arg(w)

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polar form

z + r(cosZ + isinZ)

where Z is the polar angle and r is the magnitude of the compex number

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de moivre’s formula

(cosZ + isinZ)^n = cos(nZ) + isin(nZ)

trace, determininat, complex eigenvalues

tr A = Y1 + .. Yn
det A + Y1 * .. Yn
when the eigenvalues are all complex.

stability and eigenvalues

a dynamical system x(t + 1) = Ax(t): its zero state is asymtotically stable iff the modulus of all the complex eigenvalues of A is less than 1.

discrete model with complex eigen vales - solution

for x(t + 1) = Ax(t), with eigen values p+/- iq = r(cosZ +/i sin(Z))

let v + iw be an eigen vector of A with eigenvalue p + iq,

x(t) = r^t S[cos(Zt) -sin(Zt) | sinZt cosZt ] [coordinates]

where S = [w v]

how to determine if a matrix is diagonalizable

we wish tifnd an invertible matrix S such that S^-1 A S = B is diagonal.

a. find the eigenvalues of A
b. for eigen eigenvalue find the eigenbasis v1..vn
c A is diagonalizable iff the dimensions of the eigenspaces add up to n.
S = [v1 …vn] and B = [diagonals of eigenvalues]

eigenvalues of similar matrices

if A is similar to B, A and B hav the same characteristic polynomial, rank A = rank B and nullity A = nullity B A and B have the same eigenvalues, etc, same determinant, same trace

characterizations of invertible

``` Ax = b has a unique solution for all b in Rn rref A = In rank A = n im A = Rn ker A = {0} columns of A form a a basis of Rn columns of A span Rn columns of A are linearly independent det A != 0 0 fails to be an eigenvalye of A ```

volume of a parallelpiped

for a matrix [v1 v2 v3] and the parallpiped defined by its vectors, its volumn is |detA| For n x m matrices its sqrt(det(A^T * A))

expansion factor

area of T(space)/area of space

determinant of the transpose

if A is square, det(A^T) = det A

elementary row operations and determinants

if b comes from dividng a row of A by a scalar k, det B = (1/k) det A row swap: detB = -A adding a multple of A to another row: detB = detA

determinants of powers and products

det(AB) = det A * detB | powers is now obvious

determinants of similar matrices

det A = detB

determinants of inversion of a matrix

det(A^-1) = 1/detA

imA and A^T

imA^perpendicular = ker(A^T) ^perpendicular is the orthogonal complement, the st of vecotrs x in Rn orthogonal to all vectos in A

kernels and transpose

``` kerA = ker(A^T A) if ker(A) = {0)} then A^T A is iinvertible. Duh. ```

least squares solution

for Ax = b, a least square soslution is he one that minimizes the difference between b and Ax for all x in R^m. the least square solution is the exact solution of the consistent system A^TA x = A^T b. That's the normla equation of Ax = b

the matrix of the orthogonal project using the transpose

A (A^T A)-1 A^T is that matrix.

matrix of orthogonal projection found by orthonomrla basis

for a subapce V with orthonomral basis u1..vm, the matrix P of the orthogonal project onto V is P = QQ^T where Q is thr matrix of u1..m

properties of the transpose

``` (A + B) ^ T = A^T + B^T (kA)^T = k(A^T) (AB)^T = B^T A^T rank(A^T) = rankA (A^T)^-1 = (A^-1)^T ```

characteristics of orthogonal matrices

A is orthogongal. L(x) = Ax preserves length. columns of A forma an orthonormal basis of Rn A^T A = In A^-1 = A^T A preserves the dot projet for all x and y. (Ax) * (Ay) = x * y

angle between two vectors

arrcos [ x * y / ||x||||y||] also the correlation coefficient

formlar for the orthogonal project

different from gram schmit )u1 * x)u .. (um * x) um

coordinates in a subspace of Rn

[x]b is the coordinates of the funciton x cv1 + c2v2 + ... cmvm where the coordinates express relations between the values of x and the vectors of the basis of V, b. They have linearity properties,

rank-nulity theorem

dim(kerA) + dim(imA) = m for any n * m matrix.

matrix for orthogonal project

[u^2 u1u2 u1u2 u2^2] where [u1 u2] is a unit vector parallel to L, the line being projected onto also (ui * x)ui ...

reflection

[a -b b -a] find with 2projL(x) - x for a specific line. a^2 + b^2 = 1

rotation

cos Z -sinZ sinZ cos Z or [a -b b a] where a^2 + b^2 = 1 Z is the angle. Multiply by r to also scale.

horizontal shear

[1 k 0 1] Vertical is by swap)