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. 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?
 A: 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.
A: 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.
A: 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 $u\in V$
