Solutions to Homework Set 6
... order 21. To this end, let P be a Sylow 3-subgoup and consider P L. It is easy to show that this is a subgroup of order 21. To show that its normal, argue that its normalizer cannot have order 21 and then it must be all of G. 3) Prove that for every even integer p ≥ 2, there exists a constant c(p) s ...
... order 21. To this end, let P be a Sylow 3-subgoup and consider P L. It is easy to show that this is a subgroup of order 21. To show that its normal, argue that its normalizer cannot have order 21 and then it must be all of G. 3) Prove that for every even integer p ≥ 2, there exists a constant c(p) s ...
TEST I Name___________________________________ Show
... one batch. Matrix B gives the costs of ingredients from suppliers X and Y. What is the cost of 100 batches of each candy using ingredients from supplier X? ...
... one batch. Matrix B gives the costs of ingredients from suppliers X and Y. What is the cost of 100 batches of each candy using ingredients from supplier X? ...
Applications
... At each transition period, each member must either leave or stay. This means that the sum of the entries in any column in the above matrix must equal 1. Thus, for instance, we must have p 11 + p 21 + ⋯ + p n1 = 1 Such a matrix (whose entries are nonnegative and whose individual column entries sum to ...
... At each transition period, each member must either leave or stay. This means that the sum of the entries in any column in the above matrix must equal 1. Thus, for instance, we must have p 11 + p 21 + ⋯ + p n1 = 1 Such a matrix (whose entries are nonnegative and whose individual column entries sum to ...
Self Study : Matrices
... Two matrices are equal only if they fulfil both of the following conditions: a. they have the same order; b. all corresponding elements are equal. ...
... Two matrices are equal only if they fulfil both of the following conditions: a. they have the same order; b. all corresponding elements are equal. ...
Reformulated as: either all Mx = b are solvable, or Mx = 0 has
... 2.9.2. The inverse of a linear transformation. Let T : U ! V be a linear transformation between two vector spaces U and V . If T is onto to one and onto, then the function T has an inverse T 1 , T 1 : V ! U . Exercise. Show that the inverse of a linear transformation is also a linear transformation. ...
... 2.9.2. The inverse of a linear transformation. Let T : U ! V be a linear transformation between two vector spaces U and V . If T is onto to one and onto, then the function T has an inverse T 1 , T 1 : V ! U . Exercise. Show that the inverse of a linear transformation is also a linear transformation. ...
Linear models 2
... Proof. If x ∈ S1 ⊕ S2 then by definition there exist s1 ∈ S1 , s2 ∈ S2 such that x = s1 + s2 . Assume x = s01 + s02 , s01 ∈ S1 , s02 ∈ S2 , then s1 − s01 = s2 − s02 . This implies that s1 − s01 and s2 − s02 are in S1 and also in S2 . However, S1 ∩ S2 = {0}, so we conclude s1 = s01 and s2 = s02 . We ...
... Proof. If x ∈ S1 ⊕ S2 then by definition there exist s1 ∈ S1 , s2 ∈ S2 such that x = s1 + s2 . Assume x = s01 + s02 , s01 ∈ S1 , s02 ∈ S2 , then s1 − s01 = s2 − s02 . This implies that s1 − s01 and s2 − s02 are in S1 and also in S2 . However, S1 ∩ S2 = {0}, so we conclude s1 = s01 and s2 = s02 . We ...
form Given matrix The determinant is indicated by
... together… this is your “DOWN” total. 4) Draw “Up” diagonals under each of the three 3-term “Up” diagonals. ...
... together… this is your “DOWN” total. 4) Draw “Up” diagonals under each of the three 3-term “Up” diagonals. ...
![λ1 [ v1 v2 ] and A [ w1 w2 ] = λ2](http://s1.studyres.com/store/data/020256186_1-44523acdcc73497aa300703df377fe57-300x300.png)
λ1 [ v1 v2 ] and A [ w1 w2 ] = λ2
... 1.) Find the eigenvalues of A. Solve det(A − λI) = 0 for λ. 2.) Find a basis for each of the eigenspaces. Solve (A − λj I)x = 0 for x. 3.) Use the Gram-Schmidt process to find an orthonormal basis for each eigenspace. That is for each λj use Gram-Schmidt to find an orthonormal basis for N ul(A − λj ...
... 1.) Find the eigenvalues of A. Solve det(A − λI) = 0 for λ. 2.) Find a basis for each of the eigenspaces. Solve (A − λj I)x = 0 for x. 3.) Use the Gram-Schmidt process to find an orthonormal basis for each eigenspace. That is for each λj use Gram-Schmidt to find an orthonormal basis for N ul(A − λj ...
Subspace sampling and relative
... columns (C) and a few rows (R) of A, we can compute U and “reconstruct” A as CUR. If th e sam pling prob ab ilities are not “too b ad ”, w e get provab ly good accuracy. ...
... columns (C) and a few rows (R) of A, we can compute U and “reconstruct” A as CUR. If th e sam pling prob ab ilities are not “too b ad ”, w e get provab ly good accuracy. ...
Matrix Worksheet 7
... 1. An ice-cream stall sells both green tea and mocha ice cream. A small portion of either costs $0.75 and a large portion costs $1.25. During a short period of time, the number of ice creams sold is own in the table below. ...
... 1. An ice-cream stall sells both green tea and mocha ice cream. A small portion of either costs $0.75 and a large portion costs $1.25. During a short period of time, the number of ice creams sold is own in the table below. ...
Lecture 2 Matrix Operations
... Properties of matrix multiplication • 0A = 0, A0 = 0 (here 0 can be scalar, or a compatible matrix) • IA = A, AI = A • (AB)C = A(BC), so we can write as ABC • α(AB) = (αA)B, where α is a scalar • A(B + C) = AB + AC, (A + B)C = AC + BC • (AB)T = B T AT ...
... Properties of matrix multiplication • 0A = 0, A0 = 0 (here 0 can be scalar, or a compatible matrix) • IA = A, AI = A • (AB)C = A(BC), so we can write as ABC • α(AB) = (αA)B, where α is a scalar • A(B + C) = AB + AC, (A + B)C = AC + BC • (AB)T = B T AT ...