Math Problem Statement

Suppose matrix A is m×n, and  T : Rn → Rm, where T(x) = Ax

If for every b ∈ Rm  in the range of T, there is at most one x ∈ Rn so that Ax = b, then T must be onto.

Solution

The statement is incorrect. Let's break it down:

Given:

  • Matrix AA is m×nm \times n,
  • T:RnRmT: \mathbb{R}^n \to \mathbb{R}^m, defined by T(x)=AxT(x) = A x.

The condition "for every bRmb \in \mathbb{R}^m in the range of TT, there is at most one xRnx \in \mathbb{R}^n such that Ax=bA x = b" is saying that the transformation is injective (one-to-one), meaning there are no two distinct xx's that map to the same bb.

Does injectivity imply surjectivity (onto)?

  • Injectivity means that each bb in the range of TT has at most one corresponding xx.
  • **Surject

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Matrix Transformations
Injective and Surjective Functions

Formulas

T(x) = Ax (Linear transformation formula)

Theorems

Injectivity does not imply Surjectivity
Rank-Nullity Theorem

Suitable Grade Level

College-level (Linear Algebra)