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Here is my question:

Let $L:\mathbb{R}^n\to \mathbb{R}^m$ be a linear mapping. Assume that $\{\vec{v}_1,\ldots,\vec{v}_n\}$ is a basis for $\mathbb{R}^n$ such that $\{\vec{v}_1,\ldots,\vec{v}_k\}$ is a basis for Ker$(L)$. Is $\{L(\vec{v}_{k+1}),\ldots,L(\vec{v}_n)\}$ is linearly independent?

I'm trying to write a proof but am having a bit of trouble ..

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Let $v = \alpha_{k+1} v_{k+1}+\cdots \alpha_n v_n$ and suppose $L(v) = 0$. Then you must have $v \in \ker L$. This means that $v = \alpha_1 v_1+\cdots \alpha_k v_k$, or, in other words, $\alpha_1 v_1+\cdots \alpha_k v_k - (\alpha_{k+1} v_{k+1}+\cdots \alpha_n v_n) = 0$. Since the vectors $v_k$ are linearly independent, this means $\alpha_k = 0$ for all $k$.

This means that if $\alpha_{k+1} L(v_{k+1})+\cdots \alpha_n L(v_n) = 0$, then we must have $\alpha_{k+1} = \cdots = \alpha_n = 0$. Hence $L(v_{k+1}), ..., L(v_n)$ are linearly independent.

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