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Let $F$ be a field, and let $V_1,V_2$ be vector spaces over F. Assume that ${u_1 ,u_2 ,...,u_k }$ is a linearly independent subset of $V_1$ , and that f is an injective linear map from V1 to V2.

Show that $f(u_1), f(u_2), . . . , f(u_k)$ is a linearly independent subset of $V_2$.

I know how to prove the left side but I'm unsure how to prove this.

Let $x \in V$ and let $f(x)=0$. Since ${u_1 ,u_2 ,...,u_k }$ spans $V_1$ there exists $λ_1 ,λ_2 ,...,λ_k$ such that $λ_1u_1$+$λ_2u_2$+...$λ_ku_k$$=x$

this implies 0=f(x)=$λ_1f(u_1)$+$λ_2f(u_2)$+...$λ_kf(u_k)$

since $f(u_1)...f(u_k)$ is linearly independent, $λ_1=...=λ_k=0$ and finally $x=0$ therefore Ker $f$ is $0$ and $f$ is injective

How do I prove the right side?

Mathissohardlmao
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  • https://math.stackexchange.com/questions/70876/showing-that-a-one-to-one-linear-transformation-maps-a-linearly-independent-set – user744868 Mar 05 '20 at 05:14
  • $u_1,...,u_k$ is just an independent set, they do not neccessarily span $V_1$. You rather should assume that there are $\lambda_j$'s, not all 0, such that $\sum_{j=1}^k \lambda_jf(u_j)=0$, then due to linearity the vector $x=\sum_j \lambda_j u_j\neq 0$ maps to 0, hence $f$ is not injective, which is a contradiction. – Mick Mar 05 '20 at 05:20

1 Answers1

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Suppose the set $\{v_{1},v_{2},\ldots,v_{n}\}\subset V_{1}$ is linear independent and the linear mapping $f:V_{1}\rightarrow V_{2}$ in injective. Based on such assumptions, consider the linear combination: \begin{align*} & \alpha_{1}f(v_{1}) + \alpha_{2}f(v_{2}) + \ldots + \alpha_{n}f(v_{n}) = 0 \Longrightarrow f(\alpha_{1}v_{1} + \alpha_{2}v_{2} + \ldots + \alpha_{n}v_{n}) = 0\Longrightarrow\\\\ & \alpha_{1}v_{1} + \alpha_{2}v_{2} + \ldots + \alpha_{n}v_{n} = 0 \Longrightarrow \alpha_{1} = \alpha_{2} = \ldots = \alpha_{n} = 0 \end{align*} Consequently, the set $\{f(v_{1}),f(v_{2}),\ldots,f(v_{n})\}$ is linear independent.

user0102
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