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Linear Algebra - MATH 232 Cheat Sheet by

This is for math 232 - linear algebra, midterm 2

Basic Equations

Network Flows
1. the flow in an arc is only in one directions
2. flow into a node = flow out of a node
3. flow into the network = flow out of the network
Bala­ncing Chemical Equati­ons
1. add x's before each combo and both side
2. carbo = x1 + 2(x3), set as system, solve
Matrix
augmented matrix
variables and soluti­on(rhs)
coeffi­cient matrix
coeffi­cients only, no rhs

Vectors, Norm, Dot Product

maginitude (norm) of vector v is ||v||; ||v|| ≥ 0
if k>0, kv same direction as v
magnitude = k||v||
if k<0, kv opposite direction to v
magnitude = |k| ||v||
vectors in Rn (n = dimension)
v = (v1, v2, ..., vn)
v = P1P2 = OP2 - OP1
displa­cement vector
norm/m­agn­itude of vector ||v||
sqrt( (v1)+(­v2)­2...)
||v|| = 0 iff v =0
||kv|| = |k| ||v||
unit vector u in same direct as v
u = (1/ ||v||) v
e1 = (1,0...) ... en = (0,...1) in Rn
standard unit vector
d(u,v) = sqrt((­u1-­v1)2 + (u2-v2)2 ... (un-vn­)2) = ||u-v||
d(u,v) = 0 iff u = v
u·v = u1v1 + u2v2 ...+unvn
||u|| ||v|| cos(θ)
dot product
u and v are orthogonal if u·v = 0 (cos(θ) = 0)
a set of vectors is an orthogonal set iff vi·vj = 0,if i≠j
a set of vectors is an orthon­ormal set iff vi·vj = 0,if i≠j, and ||vi|| = 1 for all i
(u·v)2 ≤ ||u||­2|­|v|­|2 or
|u·v| ≤ ||u|| ||v||
Cauchy­-Sc­hwarz Inequality
d(uv) ≤ d(u,w) + d(w,v)
||u+v|| ≤ ||u|| + ||v||
Triangle Inequality
||v1 + v2 ... + vk|| = ||v1|| + ||v2|| ... + ||vk||

Lines and Planes

a vector equation with parameter t
x = x0 + tv,
-∞ < t < +∞
solutin set for 3 dimension linear equation is a plane
if x is a point on this plane
(point-normal equati­on)
n·(x-x0) = 0
A(x-­x0)­+B(­y-y­0)+­C(z-z0) = 0
x0 = (x0,y0­,z0),
n = (A, B, C)
genera­l/a­lge­braic equation
Ax+By+Cz = D
two planes are parallel if n1 = kn2,
orthogonal if n1·n2 = 0

Matrix Algebra, Identity and Inverse Matrix

(A + B)ij = (A)ij + (B)ij
(A - B)ij = (A)ij - (B)ij
(cA)ij = c(A)ij
(AT)ij = (A)ji
(AB)ij = ai1b1j + ai2b2j + ... aikbkj
Inner Product (number) is uTv = u·v, u and v same size
Outer Product (matrix) is uvT, u and v can be any size
(AT)T = A
(kA)T = k(A)T
(A+B)T = AT + BT
(AB)T = BTAT
tr(AT) = tr(A)
tr(AB) = tr(BA)
uTv = tr(uvT)
tr(uvT) = tr(vuT)
tr(A) = a11 + a22 ... + ann
(AT)ij = Aji
Identity matrix is square matrix with 1 along diagonals
If A is m x n, AꞮn = A and ꞮmA = A
a square matrix is
invertible(nonsingular) if:
AB = Ɪ = BA
B is the inverse of A
B = A-1
if A has no inverse, A is not invertible (singular)
det(A) = ad - bc ≠ 0 is invert­ible
if A is invert­ible:
(AB)-1 = B-1A-1
(An)-1 = A-n = (A-1)n
(AT)-1 = (A-1)T
(kA)-1
1/k(A­-1), k≠0

Elementary Matrix and Unifying Theorem

elementary matrices are invertible
A-1 = Ek Ek-1 ...  E2 E1
[ A | Ɪ ] -> [ Ɪ | A-1 ]
(how to find inverse of A)
Ax = b; x = A-1b
- A -> RREF = Ɪ
- A can be express as a product of E
- A is invertible
- Ax = 0 has only the trivial solution
- Ax = b is consistent for every vector b in Rn
- Ax = b has eactly 1 solution for every b in Rn
- colum and rowvectors of A are linealy indepe­ndent
- det(A) ≠ 0
- λ = 0 is not an eigenvalue of A
- TA is one to one and onto
If not, then all no.

Consis­tency

EAx = Eb -> Rx = b' , where b' = Eb
(Ax=b) [ A | b ] -> [ EA | Eb ] (Rx = b')
(but treat b as unknown: b1, b2...)
For it to be consis­tent, if R has zero rows at the bottom, b' that row must equal to zero

Homoge­neous Systems

Linear Combin­ation of the vectors:
v = c1v1 + c2v2 ... + cnvn
(use matrix to find c)
Ax = 0
Homoge­neous
Ax = b
Non-ho­mog­enous
x = x0 + t1v1 + t2v2 ... + tkvk
Homoge­neous
x = t1v1 + t2v2 ... + tkvk
Non-ho­mog­eneous
xp is any solution of NH system
and xh is a solution of H system
x = xp + xh
 

Examples of Subspaces

IF: w1, w2 are within S
then w1+w2 are within S
and kw1 is within S
- the zero vector 0 it self is a subspace
- Rn is a subspace of all vectors
- Lines and planes through the origin are subspaces
- The set of all vectors b such that Ax = b is consis­tent, is a subspace
- If {v1, v2, ...vk} is any set of vectors in Rn, then the set W of all linear combin­ations of these vector is a subspace
W = {c1v1 + c2v2 + ... ckvk}; c are within real numbers

Span

- the span of a set of vectors { v1, v2, ... vk} is the set of all linear combin­ations of these vectors
span { v1, v2, ... vk} = { v11t, t2v2, ... , tkvk}
If S = { v1, v2, ... vk}, then W = span(S) is a subspace
Ax = b is consistent if and only if b is a linear combin­ation of col(A)

Linear Indepe­ndent

- if unique solution for a set of vectors, then it is linearly indepe­ndent
c1v1 + c2v2 ... + cnvn = 0; all the c = 0
- for dependent, not all the c = 0
Dependent if:
- a linear combin­ation of the other vectors
- a scalar multiple of the other
- a set of more than n vectors in Rn
Indepe­ndent if:
- the span of these two vectors form a plane
- list the vectors as the columns of a matrix, row reduce it, if many solution, then it is dependent
- after RREF, the columns with leading 1's are a maxmially linearly indepe­ndent subset according to Pivot Theorem

Diagonal, Triang­ular, Symmetric Matrices

Diagonal Matrices
all zeros along the diagonal
Lower Triangular
zeros above diagonal
Upper Triangular
zeros below the diagonal
Symmetric if:
AT = A
Skew-S­ymm­etric if:
AT = -A

Determ­inants

det(A) = a1jC1j + a2jC2j ... + anjCnj
expansion along jth column
det(A) = ai1Ci1 + ai2Ci2 ... + ainCin
expansion along the ith row
Cij = (-1)i+j Mij
Mij = deleted ith row and jth column matrix
- pick the one with most zeros to calculate easier
det(AT) = det(A)
det(A-1) = 1/det(A)
det(AB) = det(A)­det(B)
det(kA) = kndet(A)
- A is invertible iff det(A) not equal 0
- det of triangular or diagonal matrix is the product of the diagonal entries
det(A) for 2x2 matrix
ad - bc

Adjoint and Cramer's Rule

adj(A) = CT
CT = matrix confactor of A
A-1 = (1/det(A)) adj(A)
adj(A)A = det(A) I
x1 = det(A1) / det(A)
x2 = det(A2) / det(A)
xn = det(An) / det(A)
det(A) not equal 0
An is the matrix when the nth column is replaced by b

Hyperp­lane, Area/V­olume

a hyperplane in Rn
a1x1 + a2x2 ... + anxn = b
- can also written as ax = b
to find aperp
ax = 0, find the span
if A is 2x2 matrix:
- |det(A)| is the area of parall­elogram
if A is 3x3 matrix:
- |det(A)| is the volume of parall­ele­piped
- subtract points to get three vectors, then make it to a matrix to find the area/v­olume

Cross Product

u x v = (u2v3 - u3v2, u3v1 - u1v3, u1v2 - u2v1)
u x v = -v x u
k(u x v) = (ku) x v = u x (kv)
u x u = 0
parallel vectors has 0 for c.p.
u (u x v) = 0
v (u x v) = 0
u x v is perpen­dicular to span {u, v}
||u x v|| = ||u|| ||v|| sin(th­eta), where theta is the angle between vectors

Complex Number

complex number
a + ib
(a + ib) + (c + id) = (a + c) + i(b + d)
(a + ib) - (c + id) = (a - c) + i(b - d)
(a + ib) (c + id) = (ac + bd) + i(ad + bc)
(a + bx) (c + dx) = (ac + bdx2) + x(ad + bc)
i2 = -1
z = a + ib
z bar = a - ib
the length­(ma­gni­tude) of vector z
|z| = sqrt(z x z bar)
= sqrt­(a2 + b2)
z-1 = 1/z = z bar / |z|2
z1 / z2 = z1z2-1
z = |z| (cos(θ) + i (sin(θ))
polar form (r = |z|)
z1z2 = |z1| |z2| (cos(θ1 + θ2) + i (sin(θ1 + θ2))
z1/z2 = |z1| / |z2| (cos(θ1 - θ2) + i (sin(θ1 - θ2))
zn = rn(cos(n θ) + i sin(n θ))
r = |z|
ei theta = cos(θ) + i sin(θ)
ei pi = -1
ei pi +1 = 0
z1z2 = r1r2 ei (θ1 + θ2)
zn = rn ei nθ
z1 /z2 = r1 / r2 ei (θ1 - θ2)
 

Eigenv­alues and Eigenv­ectors

Ax= λx
det(λI - A) = (-1)n det(A - λI)
pa(λ) = 3x3: det(A - λI); 2x2: det(λI - A)
- solve for (λI - A)x = 0 for eigenv­ectors
Work Flow:
- form matrix
- compute pa(λ) = det(λI - A)
- find roots of pa(λ) -> eigenv­alues of A
- plug in roots then solve for the equation

Linear Transf­orm­ation

f: Rn -> Rm, n = domain, m = co-dom­ain
f(x1, x2, ...xn) = (y1, ...ym)
T: Rn -> Rm is a linear transf­ormatin if
1. T(cu) = cT(u)
2. T(u +v) = T(u)+ T(v)
for any linear transf­orm­ation, T(0) = 0
Rθ = [T(e1) T(e2)] = [cosθ −sinθ]
                                 ­[sinθ   cosθ]
matrix for rotation
reflection across y-axis: T(x, y) = (-x, y)
reflection across x-axis: T(x, y) = (y, -x)
reflection across diagonal y = x, T(x, y) = (y, x)
orthogonal projection onto the x-axis: T(x, y) = (x, 0)
orthogonal projection onto the y-axis: T(x, y) = (0, y)
u = (1/ ||v||)v; express it vertically as u1 and u2
A = [(u1)2 u2u1]
       ­[u1u2 (u2)2]
projection matrix
contra­ction with 0 ≤ k < 1 (shrink), k > 1 (stretch)
[x, y] -> [kx, ky]
compre­ssion in x-dire­ction [x, y] -> [kx, y]
compre­ssion in y-dire­ction [x, y] -> [x, ky]
shear in x-dire­ction T(x,y) = (x+ky, y);
[x+ky (1, k), y( 0, 1)]
shear in y-dire­ction T(x,y) = (x, y+kx);
[x (1, 0), y (k, 1)]
orthogonal projection on the xy-plane: [x, y , 0]
orthogonal projection on the xz-plane: [x, 0 , z]
orthogonal projection on the yz-plane: [0, y , z]
reflection about the xy-plane: [x, y, -z]
reflection about the xz-plane: [x, -y, z]
reflection about the yz-plane: [-x, y, z]

Orthogonal Transf­orm­ation

an orthogonal transf­orm­ation is a linear transf­orm­ation T; Rn -> Rn that preserves lengths; ||T(u)|| = ||u||
||T(u)|| = ||u|| <=> T(x)·T(y) = x·y for all x,y in Rn
orthogonal matrix is square matrix A such that AT = A-1
1. if A is orthog­onal, then so is AT and A-1
2. a product of orthonal matrices is orthogonal
3. if A is orthog­onal, then det(A) = 1 or -1
4. if A is orthog­onal, then rows and columns of A are each orthon­ormal sets of vectors

Kernel, Range, Compos­ition

ker(T) is the set of all vectors x such that T(x) = 0, RREF matrix, find the vector, ker(T) = span{(­v)}
the solution space of Ax = 0 is the null space;
null(A) = ker(A)
range of T, ran(T) is the set of vectors y such that
y = T(x) for some x
ran(T) = col([T]) = span{ [col1], [col2] ...}; Ax = b
Important Facts:
1. T is one to one iff ker(T) = {0}
2. Ax = b, if consis­tent, has a unique solution
iff null(A) = {0};   Ax = 0 has only the trivial solution iff null(A) = {0}
Important facts 2:
1.T:Rn -> Rm is onto iff the system Tx = y has a solution x in Rn for every y in Rm
2. Ax = b is consistent for every b in Rm(A is onto) iff col(A) = Rm
The compos­ition of T2 with T1 is: T2 ◦ T1
(T2 ◦ T1)(x) = T2(T1(x)); T2 ◦ T1: Rn -> Rm
compostion of linear transf­orm­ations corres­ponds to matrix applic­ation: [T2 ◦ T1] = [T1][T2]
[T(θ1+θ2)] = [Tθ2] ◦ [Tθ1];
rotate then shear ≠ shear then rotate
linear trans T: Rn->Rm has an inverse iff T is one to one, T-1: Rm -> Rn, Tx = y <=> x = T-1y
for Rn to Rn, [T-1] = [T]-1; [T]-1◦T = 1n <=> [T-1­][T]=Ɪn
1n is identity transf­orm­ation; Ɪn is identity matrix

Basis, Dimension, Rank

S is a basis for the subspace V of Rn if:
S is linearly idenpe­ndent and span(S) = V
dim(V) = k, k is the # of vectors
row(A) = rows with leading ones after RREF
col(A) = columns with leading ones from original A
null(A) = free variable's vectors
null(AT) = after transform, the free variable vector
The Rank Theorem: rank(A) = rank(AT) for any matrix have the same dimension
rank(A) = # of free vectors in span
dim(ro­w(A)) = dim(co­l(A)) = rank(A)
dim(nu­ll(A)) = nullity(A)

Orthogonal Compli­ment, DImention Theorem

S = {v ∈ Rn | v · w = 0 for all w ∈ S}
S is a subspace of Rn; S = span(S) = W
row(A) = null(A)
null(A) = row(A)
((S) = S iff S is subspace
col(A) = null(AT)
null(A­T) = col(A)
The Dimension Theorem
A is m x n matrix
rank(A) + nullity(A) = n
(k + (n-k) = n)
if W is a subspace of Rn
dim(W) + dim(W) = n

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