I am trying to solve the exercise
Let $W$ be a subspace of $V$ and ${\bf v}_1,\ldots,{\bf v}_m \in V$ be linearly independent modulo $W$. Show that $m \leq \dim V - \dim W$.
So far, I have used a theorem:
Theorem
Let $W \leq V$. We say ${\bf v}_1,\ldots,{\bf v}_m \in V$ are linearly independent modulo $W$ if any of the following equivalent conditions hold:
- The sum $W + \mathbb{F}{\bf v}_1 + \ldots + \mathbb{F}{\bf v}_m$ is direct
- $\{{\bf v}_1,\ldots,{\bf v}_m\}$ is linearly independent and spans a vector space complement to $W$ in $W + \mathbb{F}{\bf v}_1 + \ldots > + \mathbb{F}{\bf v}_m$
from our lecture notes, to conclude that the sum
$V = W + \mathbb{F}{\bf v}_1 + \ldots + \mathbb{F}{\bf v}_m$
is direct. Now I can get that
$\dim V = \dim( W + \mathbb{F}{\bf v}_1 + \ldots + \mathbb{F}{\bf v}_m)$
and since the sum is direct, I think I should be able to break that sum into
$\dim V = \dim W + \dim(\mathbb{F}{\bf v}_1 + \ldots + \mathbb{F}{\bf v}_m)$
which consequently gives that $m = \dim V - \dim W$, but that is not what I should show, so there must be an error somewhere.