Such that if $W$ is a subspace of $ℝ^n$ and $V=${$v∈ ℝ^n | v ∉ W$}, is $V$ also a subspace of $ℝ^n$?

I think that it is, considering that if all elements of $V$ are in the set $ℝ^n$ that would mean that $V$ has to be as well, but another student disagrees. Why is it/is it not?

  • $\begingroup$ You state your reason for thinking that $V$ is a subspace as “if all elements of $V$ are in $\mathbb{R}^n$ that would mean that $V$ has to be as well”. That sounds to me like you are confusing subspace with subset. You are correct that $V$ is a subset of $\mathbb{R}^n$. A subspace is a subset that satisfies certain properties. To check if a subset is a subspace, you have to check if it satisfies those properties. $\endgroup$ – Joe Oct 1 '19 at 2:49

$W$ is a subspace, so it contains the zero vector $\mathbf{0}$. Hence $V$ cannot contain $\mathbf{0}$, so it cannot be a vector space.

Here's a slightly different take, if this helps. Take $W$ to be the $x$-axis in $\mathbb{R}^2$. Then your $V$ consists of all vectors off the $x$-axis. Now, $(1,1)$ and $(-1,-1)$ are both vectors in $V$. But their sum is $\mathbf{0}$, which isn't in $V$ (because it is in $W$). So, $V$ isn't closed under addition.

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  • $\begingroup$ Why can it not? Because it cannot contain the vectors in W? $\endgroup$ – tsall98 Oct 1 '19 at 0:42
  • $\begingroup$ Your very definition of $V$ tells me that $\mathbf{0} \notin V$. If $\mathbf{0} \in V$ then this means $\mathbf{0} \notin W$, which is silly. Have you looked at any concrete examples? Consider $W$ as the $x$-axis inside $\mathbb{R}^2$. $\endgroup$ – Randall Oct 1 '19 at 0:42
  • $\begingroup$ @tsall98: Yes, precisely, that's it. $\endgroup$ – Ted Shifrin Oct 1 '19 at 0:50

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