If $\{v_1, v_2, ..., v_n\}$ is a basis and $f$ is an injective morphism, show that $\{f(v_1), f(v_2), ..., f(v_n)\}$ is linearly independent.

Question

Let $V_1$, $V_2$ be two $K$-vector spaces with $dim_K V_1 = dim_K V_2 = n$, $f:V_1 \rightarrow V_2$ a morphism and $B = \{v_1, v_2, ..., v_n\}$ a basis for $V_1$.

Now consider the set $T = \{f(v_1), f(v_2), ..., f(v_n)\}$.

Is there a way to show that if $f$ is injective, then $T$ is linearly independent?

Answer 1

Hint 1: Since $f$ is injective, the kernel of $f$ is $\{0\}$ (i.e., the only vector from $V_{1}$ being mapped to the $0$ vector in $V_{2}$ is the $0$ vector -- can you prove this?).

Hint 2: $a_{1}f(v_{1}) + \dots + a_{n}f(v_{n}) = f(a_{1}v_{1} + \dots + a_{n}v_{n})$.

From hint 2, if the LHS $= 0$, what does that tell us about $a_{1}v_{1} + \dots + a_{n}v_{n}$? What does this imply about $a_{1}, \dots, a_{n}$?