Does the image of a linearly independent set remain linearly independent under canonical projection quotient map?

Question

Let $V$ be a vector space and $V_0$ be a proper subspace of $V$. Then we have the canonical projection map $\pi: V \to V/V_0$. Now consider a linearly independent set $S = \{ v_1,\cdots,v_n \} \subset V\setminus V_0 $.

Question: is the image of $S$ linearly independent in $V/V_0$ ? does this hold for infinite an infinite set $S = \{ v_1,v_2,\cdots \}$

My Proof:

Since the map $\pi$ is linear then for any linear combination of the images of $\{ v_1,\cdots,v_n \}$ $$ \sum_{j=1}^n a_j \pi\left(v_j \right) =\pi\left( \sum_{j=1}^n a_j v_j \right) \overset{!}{=}0 \in V/V_0 \quad\implies \quad \sum_{j=1}^n a_j v_j \in V_0 $$ But the set $\{ v_1,\cdots,v_n \} \notin V_0$ is linearly independent, then it follows that $$ \sum_{j=1}^n a_j v_j = 0 $$ and hence $a_j = 0$ for all $j \in \{1,\cdots,n\}$.

I would like to know if there is a mistake in the proof, and whether it is true in general. Thanks in advance!

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Addition

I realized when it was too late that I did not state the question as I had in mind. Instead of "$S = \{ v_1,\cdots,v_n \} \subset V\setminus V_0 $" I should have written "$\text{span} \{ v_1,\cdots,v_n \} \cap V_0 = \{0\}$ where $\{ v_1,\cdots,v_n \}$ is linearly independent".

Answer 1

Sorry, it’s incorrect.

Consider $V$ with basis $\{v_1,v_2\}$ and $V_0$ spanned by $v_1+v_2$. All the hypotheses are satisfied, but no two-element set can be linearly independent in $V/V_0$, which has dimension $1$.

This can be extended to any space which has dimension greater than $2$, even infinite. Suppose $B$ is a basis of $V$, with $v_1,v_2\in B$ distinct (they exist by assumption). Then with $S=B$ and $V_0$ spanned by $v_1+v_2$, you have $\pi(v_1+v_2)\in V_0$, so the set $\pi(S)$ is not linearly independent.

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Addition

The statement is true with an additional assumption, namely that $$ \operatorname{span}\{v_1,v_2,\dots,v_n\}\cap V_0=\{0\} $$ which is much stronger than $v_1,\dots,v_n\notin V_0$. When you have $$ \sum_{i=1}^n a_iv_i\in V_0 $$ you can in fact conclude that the linear combination is zero and so, by linear independence, $a_1=a_2=\dots=a_n=0$.

Answer 2

Consider the vectors $(1,1)$ and $(1,2)$ in $\mathbb{R}^2$. The are linearly independent but they have the same projection onto $(1,0)$, which is $(1,0)$ and so their projection is not linearly independent.