Let A : ℝ^m → ℝⁿ be a non-zero linear transformation. Which of the following statements is true? 

Concept:

Linear Transformation: A function \( A\)  between two vector spaces that preserves the operations of

vector addition and scalar multiplication. In this case, \( A\) is a linear transformation from \(\mathbb{R}^m\) 

(an  m -dimensional space) to \( \mathbb{R}^n\) (an n -dimensional space).

One-to-One (Injective): A linear transformation is injective if distinct vectors in \(\mathbb{R}^m\) are mapped to

distinct vectors in \(\mathbb{R}^n\) . In terms of matrices, \( A\) is injective if its null space only contains the zero vector.

Onto (Surjective): A linear transformation is surjective if for every vector in \(\mathbb{R}^n\), there is at least one vector

in \(\mathbb{R}^m\) that maps to it. In terms of matrices, \( A\) is surjective if its image spans the entire space \(\mathbb{R}^n\) (i.e., if \(\text{im}(A) = \mathbb{R}^n\)).

Bijective: A linear transformation is bijective if it is both injective (one-to-one) and surjective (onto).

A bijection implies that the linear transformation has an inverse, meaning \( A\) can map \(\mathbb{R}^m\) to \(\mathbb{R}^n\) perfectly

without losing or repeating information.

Explanation:

Option 1: 

Consider the set X = {1, 2, 3} so m = 3 and set Y = {a, b, c, d} so n = 4.

Define the function \(A: X \to\) Y by A(1) = a, A(2) = b and A(3) = c

This function is one-to-one (no two elements in X map to the same element in Y)

 but not onto (the element d \in Y is not mapped by any element of X).

In this case, m = 3 and n = 4, but m < n. Thus, the function is injective but not surjective,

providing a counterexample to the condition m > n.

Option 2:

Consider the set X = {1, 2, 3, 4} (so m = 4) and set Y = {a, b, c} (so n = 3).

Define the function \(A: X \to\) Y by
A(1) = a, A(2) = b, A(3) = c and A(4) = c

This function is onto because every element in Y is mapped by some element in X.

However, it is not one-to-one because two elements in X (3 and 4) map to the same element c in Y.

In this case, m = 4 and n = 3, but m > n. Therefore, the function is surjective but not injective,

providing a counterexample to the condition m < n.

Option 3: 

A transformation is bijective if it is both one-to-one and onto, meaning every element of \( \mathbb{R}^n\) has

a unique preimage in \( \mathbb{R}^m \). This can only happen if m = n, so this is a correct statement.

Option 4:

Consider the set X = {1, 2, 3} (so m = 3) and the set Y = {a, b, c, d} (so n = 4).

Define the function \(A: X \to\) Y by A(1) = a, A(2) = b and A(3) = c

This function is one-to-one (injective) because no two elements of X map to the same element of Y.

However, m \(\neq\) n, as m = 3 and n = 4.

Thus, the function is injective but m \(\neq\) n, providing a valid counter example.

Thus, the correct statements is option 3).