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I have been trying to think of an alternate way to construct this proof because I think this is a bit ugly, but I've decided to settle on this way. I'm just not sure if it's a valid way though.

Background

Let $W$ be a subspace of a vector space $V$ over $\mathbb{F}$. For some fixed $\mathbf{v} \in V$, a coset of $W$ is defined to be the set

$$ \{\mathbf{v}\} + W = \{ \mathbf{v} + \mathbf{w} \,\,|\,\, \mathbf{w} \in W \}. $$ We usually denote this as $\mathbf{v} + W$, though.

Problem

I am trying to show that, for fixed $\mathbf{v_1}, \mathbf{v_2} \in V$, if $\mathbf{v_1} + W = \mathbf{v_2} + W$ then $\mathbf{v_1} - \mathbf{v_2} \in W$.

My Proof

Suppose $W = \{\mathbf{w_1}, \mathbf{w_2}, \ldots, \mathbf{w_n}\}$. Then,

$$ \mathbf{v_1} + W = \mathbf{v_2} + W \implies \{ \mathbf{v_1} + \mathbf{w_1}, \ldots, \mathbf{v_1} + \mathbf{w_n} \} = \{ \mathbf{v_2} + \mathbf{w_1}, \ldots, \mathbf{v_2} + \mathbf{w_n} \}. $$

Let $\{x_1, \ldots, x_n \,\,|\,\, 1 \leq x_i \leq n, \, x_i \in \mathbb{N} \}$ be an arbitrary ordering of the numbers $1, 2, \ldots, n$.

Then suppose, without loss of generality, that, $$ \begin{align*} \mathbf{v_1} + \mathbf{w_1} &= \mathbf{v_2} + \mathbf{w_{x_1}},\\ \mathbf{v_1} + \mathbf{w_2} &= \mathbf{v_2} + \mathbf{w_{x_2}},\\ &\ldots\\ \mathbf{v_1} + \mathbf{w_n} &= \mathbf{v_2} + \mathbf{w_{x_n}}. \end{align*} $$ More succinctly, $\mathbf{v_1} + \mathbf{w_i} = \mathbf{v_2} + \mathbf{w_{x_i}}$ $\forall i = 1, 2, ... n$.

Rearranging, $\mathbf{v_1} - \mathbf{v_2} = \mathbf{w_{x_i}} - \mathbf{w_i} \in W$ because $W$ is a subspace of $V$.

Therefore, $\mathbf{v_1} - \mathbf{v_2} \in W$.

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    $\begingroup$ The problem here is that you are assuming that $W$ is a finite set. A typical vector space is rarely a finite set. $\endgroup$ – Lord Shark the Unknown Apr 25 '17 at 5:26
  • $\begingroup$ The proof would be easier if you use the fact that $0\in W$. $\endgroup$ – Alan Wang Apr 25 '17 at 5:27
  • $\begingroup$ @LordSharktheUnknown Yes, you are absolutely right. I will edit the question to clarify, but my course only deals with finite vector spaces so it was assumed. Thanks for pointing that out. $\endgroup$ – Tristan Batchler Apr 25 '17 at 5:29
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    $\begingroup$ I'm very surprised that you course only deals with finite vector spaces. Many courses deal with finite-dimensional vector spaces, but non-trivial finite vector spaces only exist when the scalar field is a finite field. $\endgroup$ – Lord Shark the Unknown Apr 25 '17 at 5:31
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    $\begingroup$ $\mathbf v_2 \in \mathbf v_2 + \mathbf W$. If $\mathbf v_1 + \mathbf W = \mathbf v_2 + \mathbf W$, then $\mathbf v_2 \in \mathbf v_1 + \mathbf W$. So, for some $\mathbf w \in \mathbf W, \mathbf v_1 = \mathbf v_2 + w$. etc $\endgroup$ – steven gregory Apr 25 '17 at 5:47
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You don't need to suppose $W= \{ w_1, \ldots, w_n \}$. This will assume that $M$ is finite set. But vector spaces are usually infinite set. So you shouldn't write $W= \{ w_1, \ldots, w_n \}$.

Instead, you can just write: for some $w_1 \in W$, since $v_1+w_1 \in v_2+W$ so there exists $w_2 \in W$ so $v_1+w_1=v_2+w_2$ or $v_1-v_2=w_2-w_1 \in W$.

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