(Curt) Answer: False

P124: The column space consists of all linear combinations of the columns. The combinations are all possible vectors $\mathbf{Ax}$. They fill the column space $C(A)$.

I remembered that a column space is a subspace, but don't grasp how to proceed. In general, can a certain set of vectors which doesn't belong to a particular subspace, form another subspace?

This question precedes nullspace, rank, REF, $\mathbf{Ax = b}$, linear independence, span, basis, dimension, dimensions/theorems of the 4 subspaces, Orthogonality, Determinants, eigenvalues and eigenvectors, and linear transformations. Please kindly omit them.

Supplementary on Chris Taylor's Counterexample $2$ (in his First Edit):

All nonzero vectors in the complement of a column space don't form a subspace.

$\Large{{1.}}$ What's the intuition behind this? How and why would you foreknow/envisage this, prior to contriving any counterexamples?

$\Large{{2.}}$ I'm trying to intuit and deduce whether $\mathbf{(u + w)} \in W \text{ or } W^C$ ?

Since $U = W^C \cup \{\mathbf{0}\}$, thus $\mathbf{u} \in U \equiv \mathbf{u} \in W^C \; \vee \; \mathbf{u = 0}$. Again, we are considering nonzero vectors for this 2nd counterexample so in actual fact, $\mathbf{u \neq 0}$. So $\mathbf{u}∈W^C$ (is the only possible disjunct). But how does this divulge whether $(\underbrace{\mathbf{u}}_{\large{\in W^C}} + \underbrace{\mathbf{w}}_{\large{\in W}}) \in W \text{ or } W^C$ ?

$\Large{{3.}}$ Can question $2$ be answered without rewriting $\color{ #B22222}{u = u_W + u_{W^C}}$ in Chris Taylor's latest comment? I wouldn't have perceived this critical step, so how and why can this rewrite be performed?

Also, what motivates the proof by contradiction? For want of a contradiction, it supposes $\color{ #B22222}{\mathbf{u}} + \color{green}{\mathbf{w}} = (\color{ #B22222}{\mathbf{u_W}} + \color{green}{\mathbf{w}}) + \color{ #B22222}{\mathbf{u_{W^C}}} \in W$. Then this $\iff \mathbf{u_{W^C} = 0} \iff \mathbf{u} = \mathbf{u_W} \in W. \Lightning$


Counterexample $1$: $\mathbf{0}$ isn't in the complement of the column space.

A set of vectors $U$ forms a subspace of a vector space $V$ if $U\subseteq V$ and if $U$ is itself a vector space.

Let $U$ be the set of vectors not in the column space of a particular linear map $A$. Trivially $0$ is in the column space, and so $0\notin U$. A subset that doesn't contain zero cannot be a vector space.

First Edit:

Counterexample $2$: All nonzero vectors not in a column space don't form a subspace.

Let $W\subset V$ be the column space, let $U$ be the complement of $W$ with zero attached. Pick a particular $w\in W$ and $u\in U$ with $u,w\neq 0$. Then $u+w\in U$. Now if $U$ is a subspace, then if two vectors are in $U$, their sum is also in $U$. But $(u+w)+(-u) = w \notin U$. Hence $U$ is not a subspace.

Second Edit:

You asked how I came up with this proof that $U$ wasn't a subspace. To do that you need to show that one of the vector space axioms doesn't hold. In the first case it was straightforward - vector spaces must contain $0$, and we could show that $U$ did not contain $0$.

In the second case we adjoined $0$ to $U$. Therefore we had to pick a different vector space axiom to violate. The simplest one seemed to be closure - the requirement that two vectors in a vector space must sum to give a third vector in the same vector space.

So all I had to do was pick two vectors in $U$ that summed to give a vector not in $U$. The only vectors not in $U$ are the vectors $w\in W$ with $w\neq 0$, ie: $[ w \neq \mathbf{0} ] \in W$. From that point, it was obvious that I should pick $u+w$ and $-u$ as the two vectors in $U$ that summed to give $[ w \neq \mathbf{0} ] \in W$.
All that remained was to show that $u+w$ and $-u$ were in $U$ (which is the same as saying that they are not in $W$), and that these two summed to give an arbitrary $[ w \neq \mathbf{0} ] \in W$.

  • $\begingroup$ Yes but even if you add zero to the complement it is still not a subspace. $\endgroup$ – dezign Nov 25 '13 at 9:44
  • $\begingroup$ @dezign That is true, but irrelevant to this question. $\endgroup$ – Chris Taylor Nov 25 '13 at 10:18
  • $\begingroup$ @dezign: Could you please enlarge on your comment? Provided that I add zero to $U$, how and why is $U$ still not a subspace? I'd appreciate your answer in an answer; they're easier to read than comments. $\endgroup$ – Greek - Area 51 Proposal Nov 26 '13 at 4:34
  • $\begingroup$ @LePressentiment I edited the answer for you. $\endgroup$ – Chris Taylor Nov 26 '13 at 8:58
  • $\begingroup$ @LePressentiment You're right that $w\in W$ is not in $U$. But $u+w$ is in $U$ (it has to be, because $u\notin W$, hence (1) $u+w\notin W$ and (2) $u+w\neq 0$). $\endgroup$ – Chris Taylor Nov 26 '13 at 16:02

Hint: What vector is in every subspace? Is that vector in the complement of the column space?


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