5
$\begingroup$

Let $U,V$ be subspaces of the vector space $W$. Show that if $U\nsubseteq V$ and $V\nsubseteq U$ then $U \cup V$ is not a subspace.

I know that in order to be considered a subspace, the matrix addition and scalar multiplication operations must hold. However, I can define:

$U = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ and $V = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$

They are not a subset of each other, but their union is a subspace. Adding a matrix that spans $U\cup V$ or performing scalar multiplication seems valid in my constructed $U \cup V$. Clearly I'm missing something important.

Can anyone shed some light on this?

$\endgroup$
  • 1
    $\begingroup$ Yep, I think what you say doesn't make sense: $U,V$ are subspaces in your question and individual matrices in your alleged answer. You cannot talk about the "union" $U\cup V$ of matrices: there is no such a thing. $\endgroup$ – Agustí Roig Jan 24 '14 at 9:43
  • $\begingroup$ Presumably he means the subspaces generated by the corresponding 4-vectors. But if not this could be problematic... $\endgroup$ – Sempliner Jan 24 '14 at 9:45
  • $\begingroup$ @a.r. I thought that might be where I was going wrong. Can you tell me how I should think about subspaces conceptually? My current understanding is only based on the idea that they must have the properties of scalar multiplication and matrix addition. $\endgroup$ – Alan Jan 24 '14 at 9:48
  • $\begingroup$ Well, first of all, subspaces are sets of vectors and, except for the trivial subspace $\overrightarrow{0}$, or vector spaces over finite fields -about which you don't have to bother right now- it's unusual for them to have just ONE vector. $\endgroup$ – Agustí Roig Jan 24 '14 at 11:11
  • 1
    $\begingroup$ Think geometrically. Roughly, a subspace is a plane. If you take two planes, neither of which is entirely within the other, is their union a plane? $\endgroup$ – Jack M Jan 24 '14 at 11:26
2
$\begingroup$

Consider the following: can the sum of two vectors, one from each subspace, both nonzero, be in either of the two subspaces?

In your particular example what you did was take the formal sum of the two subspaces, not the union. The union is not sums of vectors from each set, it is just the set of vectors that are either in one or the other.

$\endgroup$
1
$\begingroup$

On your example, the identity matrix would have to belong to the union of your two subspaces, since it is the sum of both matrices.

In general, for $U \cup V$ to be considered a subspace it has to contain all the linear combinations of the elements in $U$ and $V$. If they are disjoint, take an element $u+v$ which is the sum of vectors from each subspace. Since $U \cup V$ is a subspace, $u+v$ has to belong to either $U$ or $V$. Say it belongs to $U$. Then $(u+v)-u=v$ must also belong to $U$ since it is a subspace, which contradicts the fact they are disjoint.

$\endgroup$
1
$\begingroup$

Let $u\in U-V$ and $v\in V-U$.

If $U\cup V$ is a vectorspace then $u+v\in U\cup V$.

However $u+v\in U$ combined with $u\in U$ implies that $v\in U$ and $u+v\in V$ combined with $v\in V$ implies that $u\in V$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.