Quiz 1 Solutions, Math 309 (Vinroot) (1): The set of integers Z, with
... adding b to both sides and using properties of fields we would have a = b. So a + (−b) has some multiplicative inverse in F , say c. Multiplying the left side by c gives, using properties of vector spaces and fields, c(a + (−b))x = (c(a + (−b)))x = 1x = x, while the right side becomes c0 = 0. We thu ...
... adding b to both sides and using properties of fields we would have a = b. So a + (−b) has some multiplicative inverse in F , say c. Multiplying the left side by c gives, using properties of vector spaces and fields, c(a + (−b))x = (c(a + (−b)))x = 1x = x, while the right side becomes c0 = 0. We thu ...
3.7.5 Multiplying Vectors and Matrices
... It is important to realize that you can use \dot" for both left- and rightmultiplication of vectors by matrices. Mathematica makes no distinction between \row" and \column" vectors. Dot carries out whatever operation is possible. (In formal terms, a.b contracts the last index of the tensor a with th ...
... It is important to realize that you can use \dot" for both left- and rightmultiplication of vectors by matrices. Mathematica makes no distinction between \row" and \column" vectors. Dot carries out whatever operation is possible. (In formal terms, a.b contracts the last index of the tensor a with th ...
Dot Product, Cross Product, Determinants
... Note that for two nonzero, non-parallel vectors ~a,~b there are two vectors ~x which satisfy these two conditions where one is the negative of the other. We will see below that the three vectors ~a,~b,~x should have “positive orientation” in the sense of the “right hand rule”: If the thumb of the ri ...
... Note that for two nonzero, non-parallel vectors ~a,~b there are two vectors ~x which satisfy these two conditions where one is the negative of the other. We will see below that the three vectors ~a,~b,~x should have “positive orientation” in the sense of the “right hand rule”: If the thumb of the ri ...
Introductory Notes on Vector Spaces
... Note that {(1, 0), (0,1)} is a basis for R2. It is easy to prove this (see class notes). Note that {(1,0,0), (0,1,0), (0,0,1)} is a basis for R3. It is easy to prove this. In R3, names for the above basis vectors are: i = (1, 0, 0) j = (0, 1, 0) k = (0, 0, 1) ...
... Note that {(1, 0), (0,1)} is a basis for R2. It is easy to prove this (see class notes). Note that {(1,0,0), (0,1,0), (0,0,1)} is a basis for R3. It is easy to prove this. In R3, names for the above basis vectors are: i = (1, 0, 0) j = (0, 1, 0) k = (0, 0, 1) ...
Motion in an Inverse-Square Central Force Field
... where r̂ is the unit vector in the direction of ~r. In the case of planetary motion, K = GM , where G is the universal gravitational constant and M is the mass of the sun. In the case of motion in the electric field of a static charged particle, K = Cq, where C is a universal constant and q is the p ...
... where r̂ is the unit vector in the direction of ~r. In the case of planetary motion, K = GM , where G is the universal gravitational constant and M is the mass of the sun. In the case of motion in the electric field of a static charged particle, K = Cq, where C is a universal constant and q is the p ...
