Math 110 Review List
... complement of the row space and that of the column space; using the Gram-‐Schmidt process to find an orthogonal and an orthonormal basis for a given space. d. Relevant Sections: 5.1, 5.2 and 5.3 ...
... complement of the row space and that of the column space; using the Gram-‐Schmidt process to find an orthogonal and an orthonormal basis for a given space. d. Relevant Sections: 5.1, 5.2 and 5.3 ...
Vectors and Vector Spaces
... that if V has a finite basis, then every linearly independent set having the same number of vectors is also a basis. This result is the content of the next section. However, to prove it we need the Extension theorem. Corollary 1.4.1. If S = {v1 , . . . , vk } is linearly dependent then the represent ...
... that if V has a finite basis, then every linearly independent set having the same number of vectors is also a basis. This result is the content of the next section. However, to prove it we need the Extension theorem. Corollary 1.4.1. If S = {v1 , . . . , vk } is linearly dependent then the represent ...
Definition of a Vector Space A collection of vectors: V , scalars for
... As conditions (1),(2),(3) are all valid, H is a subspace of V . The above proof allows obvious generalization to: Thm 1 (P.210): For any v1 , . . . , vp ∈ V , the collection of vectors H = Span {v1 , . . . , vp } is a subspace of V . Note: We will call H to be the subspace spanned (or generated) by ...
... As conditions (1),(2),(3) are all valid, H is a subspace of V . The above proof allows obvious generalization to: Thm 1 (P.210): For any v1 , . . . , vp ∈ V , the collection of vectors H = Span {v1 , . . . , vp } is a subspace of V . Note: We will call H to be the subspace spanned (or generated) by ...
MATH 51 MIDTERM 1 SOLUTIONS 1. Compute the following: (a). 1
... we have rank(A) ≤ k, therefore nullity(A) > 0. If b is in the column space of A we will have infinitely many solutions, otherwise we will have no solution. (e). Suppose A is a matrix with 5 rows and 4 columns. Suppose that the equation Ax = 0 has only one solution. What, if anything, can you conclud ...
... we have rank(A) ≤ k, therefore nullity(A) > 0. If b is in the column space of A we will have infinitely many solutions, otherwise we will have no solution. (e). Suppose A is a matrix with 5 rows and 4 columns. Suppose that the equation Ax = 0 has only one solution. What, if anything, can you conclud ...
1 DELFT UNIVERSITY OF TECHNOLOGY Faculty of Electrical
... a scalar, equal to the length of the first vector times the length of the projection of the second vector on the first vector. a vector with length equal to the area of the parallelogram spanned by the two vectors. a vector with length equal to the length of the first vector times the length of the ...
... a scalar, equal to the length of the first vector times the length of the projection of the second vector on the first vector. a vector with length equal to the area of the parallelogram spanned by the two vectors. a vector with length equal to the length of the first vector times the length of the ...
The Zero-Sum Tensor
... On the topic of rare matrices, some properties of a matrix, here defined as a zero-sum matrix, are analyzed and three rules are derived governing multiplication involving such matrices. The suggested category (the zero-sum matrix) does not seem to presently exist, and is as expected neither included ...
... On the topic of rare matrices, some properties of a matrix, here defined as a zero-sum matrix, are analyzed and three rules are derived governing multiplication involving such matrices. The suggested category (the zero-sum matrix) does not seem to presently exist, and is as expected neither included ...
Sample Problems for Midterm 2 1 True or False: 1.1 If V is a vector
... 1.13 If dim V = n, then any independent set in V has at least n elements. 1.14 If dim V = n, then any independent set in V with n elements is a basis for V. 1.15 If W is a subspace of V, then dim W ≤ dim V. 1.16 If W = spn S and S is linearly independent, then S is a basis for W. 1.17 [v − w] S = [ ...
... 1.13 If dim V = n, then any independent set in V has at least n elements. 1.14 If dim V = n, then any independent set in V with n elements is a basis for V. 1.15 If W is a subspace of V, then dim W ≤ dim V. 1.16 If W = spn S and S is linearly independent, then S is a basis for W. 1.17 [v − w] S = [ ...
= 0. = 0. ∈ R2, B = { B?
... Sn−1 to both sides. Since Sn = 0, all the terms except the first one vanish, and we have c1 Sn−1 v = 0, and hence c1 = 0 because Sn−1 v 6= 0. Now we can similarly apply Sn−2 to show that c2 = 0, and so on (this may again be formalized by induction if desired), and we conclude that all the c j are 0 ...
... Sn−1 to both sides. Since Sn = 0, all the terms except the first one vanish, and we have c1 Sn−1 v = 0, and hence c1 = 0 because Sn−1 v 6= 0. Now we can similarly apply Sn−2 to show that c2 = 0, and so on (this may again be formalized by induction if desired), and we conclude that all the c j are 0 ...
