If the image of a set is linearly independent, is the set linearly independent?

Question

Suppose that $T: \mathbb{R}^m \to \mathbb{R}^n$ is a linear transformation and that $v_1, v_2, \ldots, v_p$ are vectors in $\mathbb{R}^m$.

If $W = \{T(v_1), T(v_2), \ldots, T(v_p)\}$ is linearly independent in $\mathbb{R}^n$, does it follow that $S = \{v_1, v_2, \ldots, v_p\}$ is linearly independent in $\mathbb{R}^m$? Justify your answer with either a proof or a counterexample.

Answer 1

Assuming $T: \mathbb R^m \to \mathbb R^n$ is linear and $v_1, v_2, \dots, v_p$ are vectors in $\mathbb R^m$ and the vectors in $W = \{ T(v_1), T(v_2), \dots, T(v_p)\}$ are linearly independent in $\mathbb R^m$. Showing that $S = \{v_1, v_2, \dots, v_p\}$ is linearly independent in $\mathbb R^n$.

Assume that the vectors in $S$ are linearly dependent, say

$$\alpha_1 v_1 + \alpha_2 v_2 + \dots + \alpha_p v_p = 0$$

where at least one $\alpha_i \neq 0$. Then we get:

$$T(\alpha_1 v_1 + \alpha_2 v_2 + \dots + \alpha_p v_p) = \alpha_1 T(v_1) + \alpha_2 T(v_2) + \dots + \alpha_p T(v_p) = 0$$

and the vectors in $W$ are thus linearly dependent, a contradiction. Therefore the vectors in $S$ have to be linearly independent.

Answer 2

You need not argue to contraction. Pick vectors $v_1, \cdots , v_n$ and constants $\alpha_1, \cdots ,\alpha_n$ are constants with $\sum_k\alpha_kv_k = 0.$ Now apply $T$; $0 = T(\sum_k \alpha_k v_k) = \sum_k \alpha_k T(v_k)$. By the linear independence of the $T(v_k)$, $\alpha_k = 0$, $1\le k \le n$.