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Network Flows
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1. the flow in an arc is only in one directions
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2. flow into a node = flow out of a node
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3. flow into the network = flow out of the network
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Balancing Chemical Equations
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1. add x's before each combo and both side
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2. carbo = x1 + 2(x3), set as system, solve
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Matrix
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augmented matrix
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variables and solution(rhs)
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coefficient matrix
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coefficients only, no rhs
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maginitude (norm) of vector v is ||v||; ||v|| ≥ 0
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if k>0, kv same direction as v
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magnitude = k||v||
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if k<0, kv opposite direction to v
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magnitude = |k| ||v||
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vectors in Rn (n = dimension)
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v = (v1, v2, ..., vn)
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v = P1P2 = OP2 - OP1
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displacement vector
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norm/magnitude of vector ||v||
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sqrt( (v1)2+(v2)2...)
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||v|| = 0 iff v =0
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||kv|| = |k| ||v||
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unit vector u in same direct as v
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u = (1/ ||v||) v
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e1 = (1,0...) ... en = (0,...1) in Rn
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standard unit vector
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d(u,v) = sqrt((u1-v1)2 + (u2-v2)2 ... (un-vn)2) = ||u-v||
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d(u,v) = 0 iff u = v
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u·v = u1v1 + u2v2 ...+unvn
||u|| ||v|| cos(θ) |
dot product
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u and v are orthogonal if u·v = 0 (cos(θ) = 0)
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a set of vectors is an orthogonal set iff vi·vj = 0,if i≠j
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a set of vectors is an orthonormal set iff vi·vj = 0,if i≠j, and ||vi|| = 1 for all i
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(u·v)2 ≤ ||u||2||v||2 or
|u·v| ≤ ||u|| ||v|| |
Cauchy-Schwarz Inequality
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d(uv) ≤ d(u,w) + d(w,v)
||u+v|| ≤ ||u|| + ||v|| |
Triangle Inequality
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||v1 + v2 ... + vk|| = ||v1|| + ||v2|| ... + ||vk||
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a vector equation with parameter t
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x = x0 + tv,
-∞ < t < +∞ |
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solutin set for 3 dimension linear equation is a plane
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if x is a point on this plane
(point-normal equation) |
n·(x-x0) = 0
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A(x-x0)+B(y-y0)+C(z-z0) = 0
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x0 = (x0,y0,z0),
n = (A, B, C) |
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general/algebraic equation
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Ax+By+Cz = D
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two planes are parallel if n1 = kn2,
orthogonal if n1·n2 = 0 |
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(A + B)ij = (A)ij + (B)ij
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(A - B)ij = (A)ij - (B)ij
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(cA)ij = c(A)ij
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(AT)ij = (A)ji
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(AB)ij = ai1b1j + ai2b2j + ... aikbkj
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Inner Product (number) is uTv = u·v, u and v same size
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Outer Product (matrix) is uvT, u and v can be any size
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(AT)T = A
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(kA)T = k(A)T
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(A+B)T = AT + BT
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(AB)T = BTAT
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tr(AT) = tr(A)
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tr(AB) = tr(BA)
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uTv = tr(uvT)
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tr(uvT) = tr(vuT)
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tr(A) = a11 + a22 ... + ann
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(AT)ij = Aji
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Identity matrix is square matrix with 1 along diagonals
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If A is m x n, AꞮn = A and ꞮmA = A
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a square matrix is
invertible(nonsingular) if: |
AB = Ɪ = BA
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B is the inverse of A
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B = A-1
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if A has no inverse, A is not invertible (singular)
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det(A) = ad - bc ≠ 0 is invertible
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if A is invertible:
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(AB)-1 = B-1A-1
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(An)-1 = A-n = (A-1)n
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(AT)-1 = (A-1)T
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(kA)-1
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1/k(A-1), k≠0
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elementary matrices are invertible
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A-1 = Ek Ek-1 ... E2 E1
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[ A | Ɪ ] -> [ Ɪ | A-1 ]
(how to find inverse of A) |
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Ax = b; x = A-1b
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- 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. |
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EAx = Eb -> Rx = b' , where b' = Eb
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(Ax=b) [ A | b ] -> [ EA | Eb ] (Rx = b')
(but treat b as unknown: b1, b2...) |
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For it to be consistent, if R has zero rows at the bottom, b' that row must equal to zero
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Linear Combination of the vectors:
v = c1v1 + c2v2 ... + cnvn (use matrix to find c) |
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Ax = 0
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Homogeneous
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Ax = b
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Non-homogenous
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x = x0 + t1v1 + t2v2 ... + tkvk
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Homogeneous
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x = t1v1 + t2v2 ... + tkvk
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Non-homogeneous
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xp is any solution of NH system
and xh is a solution of H system |
x = xp + xh
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IF: w1, w2 are within S
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then w1+w2 are within S
and kw1 is within S |
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- the zero vector 0 it self is a subspace
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- Rn is a subspace of all vectors
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- Lines and planes through the origin are subspaces
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- The set of all vectors b such that Ax = b is consistent, is a subspace
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- 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
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W = {c1v1 + c2v2 + ... ckvk}; c are within real numbers
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- the span of a set of vectors { v1, v2, ... vk} is the set of all linear combinations of these vectors
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span { v1, v2, ... vk} = { v11t, t2v2, ... , tkvk}
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If S = { v1, v2, ... vk}, then W = span(S) is a subspace
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Ax = b is consistent if and only if b is a linear combination of col(A)
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- if unique solution for a set of vectors, then it is linearly independent
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c1v1 + c2v2 ... + cnvn = 0; all the c = 0
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- for dependent, not all the c = 0
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Dependent if:
- a linear combination of the other vectors - a scalar multiple of the other - a set of more than n vectors in Rn |
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Independent if:
- the span of these two vectors form a plane |
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- list the vectors as the columns of a matrix, row reduce it, if many solution, then it is dependent
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- after RREF, the columns with leading 1's are a maxmially linearly independent subset according to Pivot Theorem
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Diagonal Matrices
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all zeros along the diagonal
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Lower Triangular
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zeros above diagonal
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Upper Triangular
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zeros below the diagonal
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Symmetric if:
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AT = A
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Skew-Symmetric if:
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AT = -A
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det(A) = a1jC1j + a2jC2j ... + anjCnj
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expansion along jth column
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det(A) = ai1Ci1 + ai2Ci2 ... + ainCin
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expansion along the ith row
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Cij = (-1)i+j Mij
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Mij = deleted ith row and jth column matrix
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- pick the one with most zeros to calculate easier
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det(AT) = det(A)
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det(A-1) = 1/det(A)
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det(AB) = det(A)det(B)
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det(kA) = kndet(A)
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- A is invertible iff det(A) not equal 0
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- det of triangular or diagonal matrix is the product of the diagonal entries
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det(A) for 2x2 matrix
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ad - bc
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adj(A) = CT
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CT = matrix confactor of A
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A-1 = (1/det(A)) adj(A)
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adj(A)A = det(A) I
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x1 = det(A1) / det(A)
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x2 = det(A2) / det(A)
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xn = det(An) / det(A)
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det(A) not equal 0
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An is the matrix when the nth column is replaced by b
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a hyperplane in Rn
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a1x1 + a2x2 ... + anxn = b
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- can also written as ax = b
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to find aperp
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ax = 0, find the span
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if A is 2x2 matrix:
- |det(A)| is the area of parallelogram |
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if A is 3x3 matrix:
- |det(A)| is the volume of parallelepiped |
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- subtract points to get three vectors, then make it to a matrix to find the area/volume
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u x v = (u2v3 - u3v2, u3v1 - u1v3, u1v2 - u2v1)
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u x v = -v x u
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k(u x v) = (ku) x v = u x (kv)
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u x u = 0
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parallel vectors has 0 for c.p.
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u (u x v) = 0
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v (u x v) = 0
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u x v is perpendicular to span {u, v}
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||u x v|| = ||u|| ||v|| sin(theta), where theta is the angle between vectors
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complex number
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a + ib
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(a + ib) + (c + id) = (a + c) + i(b + d)
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(a + ib) - (c + id) = (a - c) + i(b - d)
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(a + ib) (c + id) = (ac + bd) + i(ad + bc)
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(a + bx) (c + dx) = (ac + bdx2) + x(ad + bc)
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i2 = -1
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z = a + ib
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z bar = a - ib
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the length(magnitude) of vector z
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|z| = sqrt(z x z bar)
= sqrt(a2 + b2) |
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z-1 = 1/z = z bar / |z|2
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z1 / z2 = z1z2-1
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z = |z| (cos(θ) + i (sin(θ))
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polar form (r = |z|)
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z1z2 = |z1| |z2| (cos(θ1 + θ2) + i (sin(θ1 + θ2))
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z1/z2 = |z1| / |z2| (cos(θ1 - θ2) + i (sin(θ1 - θ2))
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zn = rn(cos(n θ) + i sin(n θ))
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r = |z|
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ei theta = cos(θ) + i sin(θ)
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ei pi = -1
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ei pi +1 = 0
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z1z2 = r1r2 ei (θ1 + θ2)
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zn = rn ei nθ
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z1 /z2 = r1 / r2 ei (θ1 - θ2)
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Ax= λx
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det(λI - A) = (-1)n det(A - λI)
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pa(λ) = 3x3: det(A - λI); 2x2: det(λI - A)
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- solve for (λI - A)x = 0 for eigenvectors
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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 |
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f: Rn -> Rm, n = domain, m = co-domain
f(x1, x2, ...xn) = (y1, ...ym) |
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T: Rn -> Rm is a linear transformatin if
1. T(cu) = cT(u) 2. T(u +v) = T(u）+ T(v) |
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for any linear transformation, T(0) = 0
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Rθ = [T(e1) T(e2)] = [cosθ −sinθ]
[sinθ cosθ] |
matrix for rotation
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reflection across y-axis: T(x, y) = (-x, y)
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reflection across x-axis: T(x, y) = (y, -x)
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reflection across diagonal y = x, T(x, y) = (y, x)
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orthogonal projection onto the x-axis: T(x, y) = (x, 0)
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orthogonal projection onto the y-axis: T(x, y) = (0, y)
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u = (1/ ||v||)v; express it vertically as u1 and u2
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A = [(u1)2 u2u1]
[u1u2 (u2)2] |
projection matrix
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contraction with 0 ≤ k < 1 (shrink), k > 1 (stretch)
[x, y] -> [kx, ky] |
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compression in x-direction [x, y] -> [kx, y]
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compression in y-direction [x, y] -> [x, ky]
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shear in x-direction T(x,y) = (x+ky, y);
[x+ky (1, k), y( 0, 1)] |
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shear in y-direction T(x,y) = (x, y+kx);
[x (1, 0), y (k, 1)] |
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orthogonal projection on the xy-plane: [x, y , 0]
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orthogonal projection on the xz-plane: [x, 0 , z]
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orthogonal projection on the yz-plane: [0, y , z]
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reflection about the xy-plane: [x, y, -z]
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reflection about the xz-plane: [x, -y, z]
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reflection about the yz-plane: [-x, y, z]
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an orthogonal transformation is a linear transformation T; Rn -> Rn that preserves lengths; ||T(u)|| = ||u||
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||T(u)|| = ||u|| <=> T(x)·T(y) = x·y for all x,y in Rn
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orthogonal matrix is square matrix A such that AT = A-1
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1. if A is orthogonal, then so is AT and A-1
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2. a product of orthonal matrices is orthogonal
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3. if A is orthogonal, then det(A) = 1 or -1
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4. if A is orthogonal, then rows and columns of A are each orthonormal sets of vectors
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ker(T) is the set of all vectors x such that T(x) = 0, RREF matrix, find the vector, ker(T) = span{(v)}
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the solution space of Ax = 0 is the null space;
null(A) = ker(A) |
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range of T, ran(T) is the set of vectors y such that
y = T(x) for some x |
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ran(T) = col([T]) = span{ [col1], [col2] ...}; Ax = b
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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} |
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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 |
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The composition of T2 with T1 is: T2 ◦ T1
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(T2 ◦ T1)(x) = T2(T1(x)); T2 ◦ T1: Rn -> Rm
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compostion of linear transformations corresponds to matrix application: [T2 ◦ T1] = [T1][T2]
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[T(θ1+θ2)] = [Tθ2] ◦ [Tθ1];
rotate then shear ≠ shear then rotate |
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linear trans T: Rn->Rm has an inverse iff T is one to one, T-1: Rm -> Rn, Tx = y <=> x = T-1y
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for Rn to Rn, [T-1] = [T]-1; [T]-1◦T = 1n <=> [T-1][T]=Ɪn
1n is identity transformation; Ɪn is identity matrix |
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S is a basis for the subspace V of Rn if:
S is linearly idenpendent and span(S) = V |
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dim(V) = k, k is the # of vectors
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row(A) = rows with leading ones after RREF
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col(A) = columns with leading ones from original A
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null(A) = free variable's vectors
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null(AT) = after transform, the free variable vector
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The Rank Theorem: rank(A) = rank(AT) for any matrix have the same dimension
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rank(A) = # of free vectors in span
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dim(row(A)) = dim(col(A)) = rank(A)
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dim(null(A)) = nullity(A)
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S⟂ = {v ∈ Rn | v · w = 0 for all w ∈ S}
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S⟂ is a subspace of Rn; S⟂ = span(S)⟂ = W⟂
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row(A)⟂ = null(A)
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null(A)⟂ = row(A)
((S⟂)⟂ = S iff S is subspace |
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col(A)⟂ = null(AT)
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null(AT)⟂ = col(A)
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The Dimension Theorem
A is m x n matrix |
rank(A) + nullity(A) = n
(k + (n-k) = n) |
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if W is a subspace of Rn
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dim(W) + dim(W⟂) = n
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