2.3 Characterizations of Invertible Matrices Theorem 8 (The
... a. A is an invertible matrix. b. A is row equivalent to I n . c. A has n pivot positions. d. The equation Ax = 0 has only the trivial solution. e. The columns of A form a linearly independent set. f. The linear transformation x →Ax is one-to-one. g. The equation Ax = b has at least one solution for ...
... a. A is an invertible matrix. b. A is row equivalent to I n . c. A has n pivot positions. d. The equation Ax = 0 has only the trivial solution. e. The columns of A form a linearly independent set. f. The linear transformation x →Ax is one-to-one. g. The equation Ax = b has at least one solution for ...
1 Linear Transformations
... 1. Theorem 11: Suppose T : Rn → Rm is a linear transformation. Then T is one-to-one if and only if the equation T (x) = 0 has only the trivial solution. 2. Proof: First suppose that T is one-to-one. Then the transformation T maps at most one input vector in Rn to the output vector 0. Thus the equati ...
... 1. Theorem 11: Suppose T : Rn → Rm is a linear transformation. Then T is one-to-one if and only if the equation T (x) = 0 has only the trivial solution. 2. Proof: First suppose that T is one-to-one. Then the transformation T maps at most one input vector in Rn to the output vector 0. Thus the equati ...
Sheet 14 - TCD Maths home
... Now, for any choice of a, b, c, we can find a solution (k1 , k2 , k3 , k4 ). Hence any v can be written as a linear combination k1 v1 + k2 v2 + k3 v3 + k4 v4 and therefore the vectors span IR3 . ...
... Now, for any choice of a, b, c, we can find a solution (k1 , k2 , k3 , k4 ). Hence any v can be written as a linear combination k1 v1 + k2 v2 + k3 v3 + k4 v4 and therefore the vectors span IR3 . ...
