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 $W_2 \subset W_1$.

 
LoadingAdd to solve later

Sponsored Links


 

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.


LoadingAdd to solve later

Sponsored Links

More from my site

  • The Union of Two Subspaces is Not a Subspace in a Vector SpaceThe 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 GroupUnion 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 SpaceThe 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 SubspacesTrue 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.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 SubspaceDetermine 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 SubspaceThe 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 zeroThe 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 } […]

You may also like...

1 Response

  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“. […]

Please Login to Comment.

This site uses Akismet to reduce spam. Learn how your comment data is processed.

More in Linear Algebra
Idempotent Matrix Problems and Solutions in Linear Algebra
If $A$ is an Idempotent Matrix, then When $I-kA$ is an Idempotent Matrix?

A square matrix $A$ is called idempotent if $A^2=A$. (a) Suppose $A$ is an $n \times n$ idempotent matrix and...

Close