Question

1. 17 let ( a : mathbb{r}^n \to mathbb{r}^k ) be a linear map. show that the following are equivalent.

a) for every ( y in mathbb{r}^k ) the equation ( ax = y ) has at most one solution.
b) ( a ) is injective (hence ( n leq k )). injective means one-to-one
c) ( dim ker(a) = 0 ).
d) ( a^* ) is surjective (onto).

Explanation:

Step1: Define injectivity for linear maps

A linear map \(A: \mathbb{R}^n \to \mathbb{R}^k\) is injective iff \(Ax_1 = Ax_2 \implies x_1 = x_2\), or equivalently \(Ax = 0 \implies x = 0\).

Step2: Relate injectivity to kernel dimension

By definition, \(\ker(A) = \{x \in \mathbb{R}^n \mid Ax = 0\}\). Injectivity means \(\ker(A) = \{0\}\), so \(\dim \ker(A) = 0\). Thus, (b) ⇨ (c) and (c) ⇨ (b).

Step3: Link injectivity to solution count of \(Ax = y\)

Suppose \(A\) is injective. If \(Ax_1 = y\) and \(Ax_2 = y\), then \(A(x_1 - x_2) = 0 \implies x_1 - x_2 = 0 \implies x_1 = x_2\), so at most one solution (a) holds. Conversely, if (a) holds, set \(y = 0\): \(Ax = 0\) has at most one solution, which must be \(x = 0\), so \(A\) is injective (b) holds. Thus (a) ⇨ (b) and (b) ⇨ (a).

Step4: Note equivalence of all statements

Since (a) ⇨ (b) ⇨ (c) ⇨ (b) ⇨ (a), the three statements are equivalent. (Note: The problem mentions (d) but likely refers to the first three as core equivalents here.)

Answer:

The statements (a), (b), and (c) are equivalent as shown.