Cheatography

# Linear Algebra - MT2 Cheat Sheet by fionaw

This is for math 232 - linear algebra, midterm 2

### Matrix Algebra

 AT = 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) (AT)ij = Aji tr(A) = a11 + a22 ... + ann

### Elementary Matrix

 elementary matrices are invertible - A -> RREF = I - 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 If not, then all no. A-1 = Ek Ek-1 ...  E2 E1 [ A | I ] -> [ I | A-1 ] (how to find inverse of A) Ax = b; x = A-1b

### 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 systemand xh is a solution of H system x = xp + xh

### 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

### 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

 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