# Dimension of subspace $W$ and linearly independent vectors modulo $W$

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.

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What is the exact theorem you used to conclude that the sum was direct? – user38268 Sep 17 '12 at 4:50
I have included the theorem in the question now. – utdiscant Sep 17 '12 at 4:59
Saying that the sum $W + v_1 +\ldots v_m$ is direct is not the same as saying that it is equal to $V$. – user38268 Sep 17 '12 at 5:02
No, that is of course true. I have gotten to used to finding direct-sum decompositions of vector spaces. – utdiscant Sep 17 '12 at 5:04

The error is in the equality $V = W + \mathbb{F}{\bf v}_1 + \ldots + \mathbb{F}{\bf v}_m$. Since the ${\bf v}_i$ are merely independent modulo $W$ but are not necessarily a basis for $V$ modulo $W$, it should read $V \supseteq W + \mathbb{F}{\bf v}_1 + \ldots + \mathbb{F}{\bf v}_m$.
I would think about your problem like this: To say that the $v_i's$ are linearly independent mod $W$ is the same as saying that their residues in the quotient $V/W$ are linearly independent. This does not mean that the residues span the whole of the quotient. From linear algebra we know that the cardinality of any linearly independent subset is less than or equal to the dimension of the space. The dimension of the quotient $V/W$ is $\dim V - \dim W$ and consequently.........