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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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    No, that is of course true. I have gotten to used to finding direct-sum decompositions of vector spaces.2012-09-17

2 Answers 2

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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 $V \supseteq W + \mathbb{F}{\bf v}_1 + \ldots + \mathbb{F}{\bf v}_m$.

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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.........