Union of Subspaces is a Subspace if and only if One is Included in Another

Problems and Solutions of Eigenvalue, Eigenvector in Linear Algebra

Problem 427

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 $W_2 \subset W_1$.

 
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Proof.

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.


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  1. 06/14/2017

    […] For a proof, see the post “Union of Subspaces is a Subspace if and only if One is Included in Another“. […]

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