Show that a linear map is one-to-one if and only if it preserves linear independence.

Question

From Jim Heffron's free textbook Linear Algebra: 3rd Edition, question 3.2.27 states: "Show that a linear map is one-to-one if and only if it preserves linear independence."

The answer key, also given for free, hasn't been much help. Any ideas?

Answer 1

Assume $\sum_{k=1}^n a_kx_k \neq 0$ whenever at least one $a_k \neq 0$. If $f$ is $1-1$ and $f(\sum_{k=1}^n a_kx_k)=\sum_{k=1}^n a_kf(x_k)= 0$, since $f$ is linear we know that $f(0)=0$ so by the definition of $1-1$, we know that $\sum_{k=1}^n a_kx_k=0$, and since the $x_k$ are linearly independent, we also know that $a_k=0$ for all $k$. Thus, if $f$ is $1-1$, it preserves linear independence.

Conversely, assume $f$ preserves linear independence. If $f(x)=f(y)$, then because $f$ is linear, we know $f(x)-f(y)=f(x-y)=0$. But any nonzero vector is a linearly independent set, so if $\{f(x-y)\}=\{0\}$ is not linearly independent and $f$ preserves linear independence, then $\{x-y\}$ can't be a linearly independent set, so the only possibility is that $x-y=0,$ so $x=y$. Thus, $f(x)=f(y) \Rightarrow x=y$ so $f$ is $1-1$.