Basis maps to Linearly Independent Set How do I prove that a linear map f will map a basis to a linearly independent set?
Supposedly this is examinable but damned if it's not on either set of notes.
Thanks!
 A: This is not true in general. For instance, any linear map from a vector space of dimension $> 1$ to a space of dimension 1 will not map to a linearly independent set.
Crucially, it must be an injective linear map.
Let $f: V \rightarrow V'$ be injective and $v_1, ..., v_n$ be a basis for $V$ (in fact, this works for any linearly independent family).
Suppose that $\lambda_1f(v_1) + ... + \lambda_nf(v_n) = \mathbf{0}.$
Then by linearity, $f(\lambda_1v_1 + ... + \lambda_nv_n) = \mathbf{0}.$
Then $\lambda_1v_1 + ... + \lambda_nv_n$ must lie in the kernel of $f$. But the kernel of an injective map contains only $\mathbf{0}$, and $v_1, ..., v_n$ are linearly independent. So $\lambda_1 = ... = \lambda_n = 0$, and therefore $f(v_1), ... ,f(v_n)$ must be linearly independent too.
A: Suppose that $T:V\rightarrow W$ is a injective linear map. Let $\left\{v_1,\dots ,v_n\right\}$ be a basis of $V$ (You can do a similar argument if $V$ is infinite-dimensional). Suppose that $\sum_{i=1}^n\lambda_iT(v_i)=0$, then $T(\sum_{i=1}^n\lambda_iv_i)=0$. Hence $\sum_{i=1}^n\lambda_iv_i\in \text{Ker}(T)$, but since $T$ is injective, $\text{Ker}(T)=\left\{0\right\}$. Hence $\sum_{i=1}^n\lambda_iv_i=0$. Since $\left\{v_1,\dots , v_n\right\}$ is a basis, we need that $\lambda_i=0$ for all $i$. Hence the $T(v_i)$'s are linearly independent.
