Let $W_1, W_2$ be subspaces of a vector space $V$. Then prove that $W_1 \cup W_2$ is a subspace of $V$ if and only if $W_1 \subset W_2$ or $W_2 \subset W_1$.

If $W_1 \cup W_2$ is a subspace, then $W_1 \subset W_2$ or $W_2 \subset W_1$.

$(\implies)$ Suppose that the union $W_1\cup W_2$ is a subspace of $V$.
Seeking a contradiction, assume that $W_1 \not \subset W_2$ and $W_2 \not \subset W_1$.
This means that there are elements
\[x\in W_1\setminus W_2 \text{ and } y \in W_2 \setminus W_1.\]

Since $W_1 \cup W_2$ is a subspace, it is closed under addition. Thus, we have $x+y\in W_1 \cup W_2$.

It follows that we have either
\[x+y\in W_1 \text{ or } x+y\in W_2.\]
Suppose that $x+y\in W_1$. Then we write \begin{align*}
y=(x+y)-x.
\end{align*}
Since both $x+y$ and $x$ are elements of the subspace $W_1$, their difference $y=(x+y)-x$ is also in $W_1$. However, this contradicts the choice of $y \in W_2 \setminus W_1$.

Similarly, when $x+y\in W_2$, then we have
\[x=(x+y)-y\in W_2,\]
and this contradicts the choice of $x \in W_1 \setminus W_2$.

In either case, we have reached a contradiction.
Therefore, we have either $W_1 \subset W_2$ or $W_2 \subset W_1$.

If $W_1 \subset W_2$ or $W_2 \subset W_1$, then $W_1 \cup W_2$ is a subspace.

$(\impliedby)$ If we have $W_1 \subset W_2$, then it yields that $W_1 \cup W_2=W_2$ and it is a subspace of $V$.

Similarly, if $W_2 \subset W_1$, then we have $W_1\cup W_2=W_2$ and it is a subspace of $V$.
In either case, the union $W_1 \cup W_2$ is a subspace.

The Union of Two Subspaces is Not a Subspace in a Vector Space
Let $U$ and $V$ be subspaces of the vector space $\R^n$.
If neither $U$ nor $V$ is a subset of the other, then prove that the union $U \cup V$ is not a subspace of $\R^n$.
Proof.
Since $U$ is not contained in $V$, there exists a vector $\mathbf{u}\in U$ but […]

Union of Two Subgroups is Not a Group
Let $G$ be a group and let $H_1, H_2$ be subgroups of $G$ such that $H_1 \not \subset H_2$ and $H_2 \not \subset H_1$.
(a) Prove that the union $H_1 \cup H_2$ is never a subgroup in $G$.
(b) Prove that a group cannot be written as the union of two proper […]

The Sum of Subspaces is a Subspace of a Vector Space
Let $V$ be a vector space over a field $K$.
If $W_1$ and $W_2$ are subspaces of $V$, then prove that the subset
\[W_1+W_2:=\{\mathbf{x}+\mathbf{y} \mid \mathbf{x}\in W_1, \mathbf{y}\in W_2\}\]
is a subspace of the vector space $V$.
Proof.
We prove the […]

True or False. The Intersection of Bases is a Basis of the Intersection of Subspaces
Determine whether the following is true or false. If it is true, then give a proof. If it is false, then give a counterexample.
Let $W_1$ and $W_2$ be subspaces of the vector space $\R^n$.
If $B_1$ and $B_2$ are bases for $W_1$ and $W_2$, respectively, then $B_1\cap B_2$ is a […]

Two Subspaces Intersecting Trivially, and the Direct Sum of Vector Spaces.
Let $V$ and $W$ be subspaces of $\R^n$ such that $V \cap W =\{\mathbf{0}\}$ and $\dim(V)+\dim(W)=n$.
(a) If $\mathbf{v}+\mathbf{w}=\mathbf{0}$, where $\mathbf{v}\in V$ and $\mathbf{w}\in W$, then show that $\mathbf{v}=\mathbf{0}$ and $\mathbf{w}=\mathbf{0}$.
(b) If $B_1$ is a […]

Determine the Values of $a$ so that $W_a$ is a Subspace
For what real values of $a$ is the set
\[W_a = \{ f \in C(\mathbb{R}) \mid f(0) = a \}\]
a subspace of the vector space $C(\mathbb{R})$ of all real-valued functions?
Solution.
The zero element of $C(\mathbb{R})$ is the function $\mathbf{0}$ defined by […]

The Intersection of Two Subspaces is also a Subspace
Let $U$ and $V$ be subspaces of the $n$-dimensional vector space $\R^n$.
Prove that the intersection $U\cap V$ is also a subspace of $\R^n$.
Definition (Intersection).
Recall that the intersection $U\cap V$ is the set of elements that are both elements of $U$ […]

The Subspace of Linear Combinations whose Sums of Coefficients are zero
Let $V$ be a vector space over a scalar field $K$.
Let $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k$ be vectors in $V$ and consider the subset
\[W=\{a_1\mathbf{v}_1+a_2\mathbf{v}_2+\cdots+ a_k\mathbf{v}_k \mid a_1, a_2, \dots, a_k \in K \text{ and } […]

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