How can two vectors be dependent in one field and independent in another field?

I have some basic thing I don't understand.

How can two vectors be dependent in one field and independent in another one?

For example, assume I have the following two vectors: $u = (1-2i, 3+i)$, $v = (2-5i, 7-i)$. Is it possible that they are dependent above $\mathbb{C}$ and independent over $\mathbb{R}$?

What I had in mind is that if two vectors are dependent, that means that there exists a scalar $\alpha$, such that $u = \alpha v$ and if $\alpha \in \mathbb{C}$ then they are dependent only in $\mathbb{C}$. Is that correct?

Let $V$ be a vector space over the field $\mathbb{F}$ and suppose $\mathbb{E}$ is a subfield of $\mathbb{F}$. Then we can also regard $V$ as a vector space over $\mathbb{E}$; for $\alpha \in \mathbb{E}$ and $v \in V$, $\alpha v$ is defined as $\alpha \in \mathbb{F}$. Now let $v_1, v_2 \in V$. We say $\{v_1, v_2\}$ is linearly independent over $\mathbb{F}$ if $\alpha_1v_1 + \alpha_2v_2 = 0$ only has the trivial solution (where $\alpha_1, \alpha_2 \in \mathbb{F}$). We make the same definition for $\mathbb{E}$ by making the adjustment $\alpha_1, \alpha_2 \in \mathbb{E}$.

Proposition: If $\{v_1, v_2\}$ is linearly independent over $\mathbb{F}$, then $\{v_1, v_2\}$ is linearly independent over $\mathbb{E}$.

Proof: Suppose not, then there are $\alpha_1, \alpha_2 \in \mathbb{E}$, at least one of which is non-zero, such that $\alpha_1v_1 + \alpha_2v_2 = 0$. But as $\mathbb{E} \subseteq \mathbb{F}$, $\alpha_1, \alpha_2 \in \mathbb{F}$ so $\alpha_1v_1 + \alpha_2v_2 = 0$ with $\alpha_1, \alpha_2 \in \mathbb{F}$, at least one of which is non-zero. This contradicts the fact that $\{v_1, v_2\}$ is linearly independent over $\mathbb{F}$.

The other direction is what fails. That is, if $\{v_1, v_2\}$ is linearly independent over $\mathbb{E}$, $\{v_1, v_2\}$ may or may not be linearly independent over $\mathbb{F}$. This is because there may exist $\alpha_1, \alpha_2 \in \mathbb{F}\setminus\mathbb{E}$ such that $\alpha_1v_1 + \alpha_2v_2 = 0$ in which case $\{v_1, v_2\}$ would be linearly independent over $\mathbb{E}$ but not $\mathbb{F}$.

The intuition here is that by looking at a subfield, you are considering less possibilities for the scalars. So, if you've shown something to hold for all pairs of scalars in this smaller collection, it may fail when considering pairs of scalars in the larger collection. Conversely, if you have shown something holds for all pairs of scalars in the large set of scalars, then it is automatically true for all pairs of scalars in any smaller subcollection.

Consider $\mathbb R$ as a vector space over $\mathbb Q$. Then $1$ and $\sqrt2$ are linearly independent. But they are linearly dependent when you consider $\mathbb R$ as a vector space over $\mathbb R$.