Basic EquationsNetwork 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 Productmaginitude (norm) of vector v is v; v ≥ 0  if k>0, kv same direction as v  magnitude = kv  if k<0, kv opposite direction to v  magnitude = k v  vectors in R^{n} (n = dimension)  v = (v1, v2, ..., vn)  v = P1P2 = OP2  OP1  displacement vector  norm/magnitude of vector v  sqrt( (v1)^{2}+(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 R^{n}  standard unit vector  d(u,v) = sqrt((u1v1)^{2} + (u2v2)^{2} ... (unvn)^{2}) = uv  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  CauchySchwarz Inequality  d(uv) ≤ d(u,w) + d(w,v) u+v ≤ u + v  Triangle Inequality  v1 + v2 ... + vk = v1 + v2 ... + vk 
Lines and Planesa 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 (pointnormal equation)  n·(xx0) = 0  A(xx0)+B(yy0)+C(zz0) = 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  (A^{T})ij = (A)ji  (AB)ij = ai1b1j + ai2b2j + ... aikbkj  Inner Product (number) is u^{T}v = u·v, u and v same size  Outer Product (matrix) is uv^{T}, u and v can be any size  (A^{T})^{T} = A  (kA)^{T} = k(A)^{T}  (A+B)^{T} = A^{T} + B^{T}  (AB)^{T} = B^{T}A^{T}  tr(A^{T}) = tr(A)  tr(AB) = tr(BA)  u^{T}v = tr(uv^{T})  tr(uv^{T}) = tr(vu^{T})  tr(A) = a11 + a22 ... + ann  (A^{T})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^{1}A^{1}  (A^{n})^{1} = A^{n} = (A^{1})^{n}  (A^{T})^{1} = (A^{1})^{T}  (kA)^{1}  1/k(A^{1}), k≠0 
Elementary Matrix and Unifying Theoremelementary matrices are invertible  A^{1} = Ek Ek1 ... E2 E1  [ A  Ɪ ] > [ Ɪ  A^{1} ] (how to find inverse of A)  Ax = b; x = A^{1}b   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 R^{n}  Ax = b has eactly 1 solution for every b in R^{n}  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. 
ConsistencyEAx = 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 SystemsLinear Combination of the vectors: v = c1v1 + c2v2 ... + cnvn (use matrix to find c)  Ax = 0  Homogeneous  Ax = b  Nonhomogenous  x = x0 + t1v1 + t2v2 ... + tkvk  Homogeneous  x = t1v1 + t2v2 ... + tkvk  Nonhomogeneous  xp is any solution of NH system and xh is a solution of H system  x = xp + xh 
  Examples of SubspacesIF: 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   R^{n} 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 R^{n}, 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 R^{n}  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 MatricesDiagonal Matrices  all zeros along the diagonal  Lower Triangular  zeros above diagonal  Upper Triangular  zeros below the diagonal  Symmetric if:  A^{T} = A  SkewSymmetric if:  A^{T} = A 
Determinantsdet(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(A^{T}) = det(A)  det(A^{1}) = 1/det(A)  det(AB) = det(A)det(B)  det(kA) = k^{n}det(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 Ruleadj(A) = C^{T}  C^{T} = 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/Volumea hyperplane in R^{n}  a1x1 + a2x2 ... + anxn = b   can also written as ax = b  to find a^{perp}  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 Productu 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 Numbercomplex 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 + bdx^{2}) + x(ad + bc)  i^{2} = 1  z = a + ib  z bar = a  ib  the length(magnitude) of vector z  z = sqrt(z x z bar) = sqrt(a^{2} + b^{2})  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))  z^{n} = r^{n}(cos(n θ) + i sin(n θ))  r = z  e^{i theta} = cos(θ) + i sin(θ)  e^{i pi} = 1  e^{i pi} +1 = 0  z1z2 = r1r2 e^{i (θ1 + θ2)}  z^{n} = r^{n} e^{i nθ}  z1 /z2 = r1 / r2 e^{i (θ1  θ2)} 
  Eigenvalues and EigenvectorsAx= 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 Transformationf: R^{n} > R^{m}, n = domain, m = codomain f(x1, x2, ...xn) = (y1, ...ym)  T: R^{n} > R^{m} 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 yaxis: T(x, y) = (x, y)  reflection across xaxis: T(x, y) = (y, x)  reflection across diagonal y = x, T(x, y) = (y, x)  orthogonal projection onto the xaxis: T(x, y) = (x, 0)  orthogonal projection onto the yaxis: 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 xdirection [x, y] > [kx, y]  compression in ydirection [x, y] > [x, ky]  shear in xdirection T(x,y) = (x+ky, y); [x+ky (1, k), y( 0, 1)]  shear in ydirection T(x,y) = (x, y+kx); [x (1, 0), y (k, 1)]  orthogonal projection on the xyplane: [x, y , 0]  orthogonal projection on the xzplane: [x, 0 , z]  orthogonal projection on the yzplane: [0, y , z]  reflection about the xyplane: [x, y, z]  reflection about the xzplane: [x, y, z]  reflection about the yzplane: [x, y, z] 
Orthogonal Transformationan orthogonal transformation is a linear transformation T; R^{n} > R^{n} that preserves lengths; T(u) = u  T(u) = u <=> T(x)·T(y) = x·y for all x,y in R^{n}  orthogonal matrix is square matrix A such that A^{T} = A^{1}  1. if A is orthogonal, then so is A^{T} 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, Compositionker(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:R^{n} > R^{m} is onto iff the system Tx = y has a solution x in R^{n} for every y in R^{m} 2. Ax = b is consistent for every b in R^{m}(A is onto) iff col(A) = R^{m}  The composition of T2 with T1 is: T2 ◦ T1  (T2 ◦ T1)(x) = T2(T1(x)); T2 ◦ T1: R^{n} > R^{m}  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: R^{n}>R^{m} has an inverse iff T is one to one, T^{1}: R^{m} > R^{n}, Tx = y <=> x = T^{1}y  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, RankS is a basis for the subspace V of R^{n} 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(A^{T}) = after transform, the free variable vector  The Rank Theorem: rank(A) = rank(A^{T}) 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 TheoremS^{⟂} = {v ∈ R^{n}  v · w = 0 for all w ∈ S}  S^{⟂} is a subspace of R^{n}; S^{⟂} = span(S)^{⟂} = W^{⟂}  row(A)^{⟂} = null(A)  null(A)^{⟂} = row(A) ((S^{⟂})^{⟂} = S iff S is subspace  col(A)^{⟂} = null(A^{T})  null(A^{T})^{⟂} = col(A)  The Dimension Theorem A is m x n matrix  rank(A) + nullity(A) = n (k + (nk) = n)  if W is a subspace of R^{n}  dim(W) + dim(W^{⟂}) = n 

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