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 = 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)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 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 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
(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 Theoremelementary 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. |
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 | 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 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 | - 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 MatricesDiagonal 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 |
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(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 Ruleadj(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/Volumea 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 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 + 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 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: 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 Transformationan 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, 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: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, RankS 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 TheoremS⟂ = {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 |
|
Created By
Metadata
Favourited By
Comments
No comments yet. Add yours below!
Add a Comment
Related Cheat Sheets