Let $V$ be a vector space, let $\lambda_i \in V^*$ for $1 \leq i \leq n$ be distinct. Is there a $v \in V$ such that $\lambda_i(v)$ are all distinct? $V$ is a vector space over $\mathbb{C}$, and we have distinct $\lambda_i \in V^*$ for $1 \leq i \leq n$. I'm trying to find (or just prove the existence of) a $v \in V$ such that $\lambda_i(v)$ are all distinct.
For the vector space in question, I don't think there are any restrictions on the dimension. I've thought about this for a while and can't think of a reason why this is true. If $V$ has dimension $n$, then if the $\lambda_i$'s are linearly independent, we should be able to find a vector $v$ such that $\lambda_i(v) = i$ as this becomes an invertible matrix equation.
I have some intuition for why it is true for other cases. If the $\lambda_i$'s are not linearly independent, we can set a maximal linearly independent subset to be equal to a bunch of distinct numbers such that somehow none of the linear dependence equations give something that is equal. I kind of see why we can do this, but not fully and definitely can't prove it.
Can someone help prove this fact? Thanks!
 A: Your observation is correct!
To prove it, we first note that we can reframe the question as follows. Let $T:V \to \Bbb C^n$ denote the linear map given by
$$
T(v) = (\lambda_1(v),\dots,\lambda_n(v)).
$$
Let $T(V)$ denote the range of $T$. Let $U_{ij} \subset \Bbb C^n$ denote the subspace
$$
U_{ij} = \{(x_1,\dots,x_n) \in \Bbb C^n : x_i = x_j\}.
$$
Note that $T(V)$ is a subspace of $U_{ij}$ for some $i,j$ if and only if $\lambda_i = \lambda_j$. On the other hand, the statement that there exists a $v$ such that all values of $\lambda_i(V)$ are distinct is equivalent to the statement that $T(V)$ is not a subset of the union $U = \bigcup_{1 \leq i,j \leq n} U_{ij}$ of these subspaces. With that in mind, it suffices to use/show the following statement

Claim: Given (finitely many) subspaces $U_1,\dots,U_k \subset \Bbb C^n$ and a subset $W \subset \Bbb C^n$, $W \subset \bigcup_{i=1}^n U_i$ holds if and only if $W \subset U_i$ for some $i$.

To prove this, we use the following lemma.

Lemma: Given a complex vector space $V$ and (finitely many) subspaces $U_1,\dots,U_k \subset V$, if $\bigcup_{i} U_i = V$, then it there exists a $j$ such that $U_j = V$.

This lemma is proved here, for instance.
Proof of claim: If $W \subset \bigcup_{i=1}^n U_i$, then it follows that
$$
W = \bigcup_{i=1}^k (W \cap U_i).
$$
However, $W \cap U_i$ is a subspace of $W$. Thus, by the lemma, there exists a $j$ such that $W \cap U_j = W$, which is to say that $W \subset U_j$, which was what we wanted.
