MA 237-102 Linear Algebra I Homework 5 Solutions 3/3/10 1. Which
... (ii) Show that the set of vectors X in V with the property that T (X) = 0 is a subspace of V (this subspace is called the kernel of T ); If T (X) = 0 and T (Y ) = 0, then for any real numbers s and t we have T (sX + tY ) = sT (X) + tT (Y ) = s0 + t0 = 0. Thus the kernel of T is a subspace of V. (ii ...
... (ii) Show that the set of vectors X in V with the property that T (X) = 0 is a subspace of V (this subspace is called the kernel of T ); If T (X) = 0 and T (Y ) = 0, then for any real numbers s and t we have T (sX + tY ) = sT (X) + tT (Y ) = s0 + t0 = 0. Thus the kernel of T is a subspace of V. (ii ...
Solution - Math-UMN
... Solution: The cross product of the vectors is orthogonal to both of them. h0, 1, 2i × h1, −1, 2i = h0, 2, −1i ...
... Solution: The cross product of the vectors is orthogonal to both of them. h0, 1, 2i × h1, −1, 2i = h0, 2, −1i ...
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... List the free variables for the system Ax = b and find a basis for the vector space null(A). Find the rank(A). 3. Explain why the rows of a 3 × 5 matrix have to be linearly dependent. 4. Let A be a matrix wich is not the identity and assume that A2 = A. By contradiction show that A is not invertible ...
... List the free variables for the system Ax = b and find a basis for the vector space null(A). Find the rank(A). 3. Explain why the rows of a 3 × 5 matrix have to be linearly dependent. 4. Let A be a matrix wich is not the identity and assume that A2 = A. By contradiction show that A is not invertible ...
LINEAR ALGEBRA (1) True or False? (No explanation required
... Every nonzero matrix A has an inverse A−1 The inverse of a product AB of square matrices A, B is equal to A−1 B −1 Homogeneous linear systems of equations always have a solution The rank of an m × n-matrix is always ≤ n The set of polynomials of degree = 2 is a vector space The product of an m × n- ...
... Every nonzero matrix A has an inverse A−1 The inverse of a product AB of square matrices A, B is equal to A−1 B −1 Homogeneous linear systems of equations always have a solution The rank of an m × n-matrix is always ≤ n The set of polynomials of degree = 2 is a vector space The product of an m × n- ...
Linear Algebra, Norms and Inner Products I. Preliminaries A. Definition
... space V iff for every x ∈ V , there exist scalars α1 , . . . , αn , such that x = i=1 αi xi • Fact: all bases of a space have the same number of vectors (the dimension of the space). • Fact: every vector x has a unique representation in a given basis. Thus we can represent a vector by its coefficien ...
... space V iff for every x ∈ V , there exist scalars α1 , . . . , αn , such that x = i=1 αi xi • Fact: all bases of a space have the same number of vectors (the dimension of the space). • Fact: every vector x has a unique representation in a given basis. Thus we can represent a vector by its coefficien ...
Basis (linear algebra)
Basis vector redirects here. For basis vector in the context of crystals, see crystal structure. For a more general concept in physics, see frame of reference.A set of vectors in a vector space V is called a basis, or a set of basis vectors, if the vectors are linearly independent and every vector in the vector space is a linear combination of this set. In more general terms, a basis is a linearly independent spanning set.Given a basis of a vector space V, every element of V can be expressed uniquely as a linear combination of basis vectors, whose coefficients are referred to as vector coordinates or components. A vector space can have several distinct sets of basis vectors; however each such set has the same number of elements, with this number being the dimension of the vector space.