The Sum of Subspaces is a Subspace of a Vector Space

Linear Algebra Problems and Solutions

Problem 430

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 $V$.

 
LoadingAdd to solve later

Sponsored Links

Proof.

We prove the following subspace criteria:

  1. The zero vector $\mathbf{0}$ of $V$ is in $W_1+W_2$.
  2. For any $\mathbf{u}, \mathbf{v}\in W_1+W_2$, we have $\mathbf{u}+\mathbf{v}\in W_1+W_2$.
  3. For any $\mathbf{v}\in W_1+W_2$ and $r\in K$, we have $r\mathbf{v}\in W_1+W_2$.

Since $W_1$ and $W_2$ are subspaces of $V$, the zero vector $\mathbf{0}$ of $V$ is in both $W_1$ and $W_2$.
Thus we have
\[\mathbf{0}=\mathbf{0}+\mathbf{0}\in W_1+W_2.\] So condition 1 is met.


Next, let $\mathbf{u}, \mathbf{v}\in W_1+W_2$.
Since $\mathbf{u}\in W_1+W_2$, we can write
\[\mathbf{u}=\mathbf{x}+\mathbf{y}\] for some $\mathbf{x}\in W_1$ and $\mathbf{y}\in W_2$.
Similarly, we write
\[\mathbf{v}=\mathbf{x}’+\mathbf{y}’\] for some $\mathbf{x}’\in W_1$ and $\mathbf{y}’\in W_2$.

Then we have
\begin{align*}
\mathbf{u}+\mathbf{v}&=(\mathbf{x}+\mathbf{y})+(\mathbf{x}’+\mathbf{y}’)\\
&=(\mathbf{x}+\mathbf{x}’)+(\mathbf{y}+\mathbf{y}’).
\end{align*}
Since $\mathbf{x}$ and $\mathbf{x}’$ are both in the vector space $W_1$, their sum $\mathbf{x}+\mathbf{x}’$ is also in $W_1$.
Similarly we have $\mathbf{y}+\mathbf{y}’\in W_2$ since $\mathbf{y}, \mathbf{y}’\in W_2$.

Thus from the expression above, we see that
\[\mathbf{u}+\mathbf{v}\in W_1+W_2,\] hence condition 2 is met.


Finally, let $\mathbf{v}\in W_1+W_2$ and $r\in K$.
Then there exist $\mathbf{x}\in W_1$ and $\mathbf{y}\in W_2$ such that
\[\mathbf{v}=\mathbf{x}+\mathbf{y}.\] Since $W_1$ is a subspace, it is closed under scalar multiplication. Hence we have $r\mathbf{x}\in W_1$.
Similarly, we have $r\mathbf{y}\in W_2$.

It follows from this observation that
\begin{align*}
r\mathbf{v}&=r(\mathbf{x}+\mathbf{y})\\
&=r\mathbf{x}+r\mathbf{y}\in W_1+W_2,
\end{align*}
and thus condition 3 is met.


Therefore, by the subspace criteria $W_1+W_2$ is a subspace of $V$.

Related Question.

Let $U$ and $V$ be finite dimensional subspaces in a vector space over a scalar field $K$.
Then prove that
\[\dim(U+V) \leq \dim(U)+\dim(V).\]

For a proof, see the post “Dimension of the sum of two subspaces“.


LoadingAdd to solve later

Sponsored Links

More from my site

  • Dimension of the Sum of Two SubspacesDimension of the Sum of Two Subspaces Let $U$ and $V$ be finite dimensional subspaces in a vector space over a scalar field $K$. Then prove that \[\dim(U+V) \leq \dim(U)+\dim(V).\]   Definition (The sum of subspaces). Recall that the sum of subspaces $U$ and $V$ is \[U+V=\{\mathbf{x}+\mathbf{y} \mid […]
  • 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 } […]
  • Vector Space of 2 by 2 Traceless MatricesVector Space of 2 by 2 Traceless Matrices Let $V$ be the vector space of all $2\times 2$ matrices whose entries are real numbers. Let \[W=\left\{\, A\in V \quad \middle | \quad A=\begin{bmatrix} a & b\\ c& -a \end{bmatrix} \text{ for any } a, b, c\in \R \,\right\}.\] (a) Show that $W$ is a subspace of […]
  • Determine Whether a Set of Functions $f(x)$ such that $f(x)=f(1-x)$ is a SubspaceDetermine Whether a Set of Functions $f(x)$ such that $f(x)=f(1-x)$ is a Subspace Let $V$ be the vector space over $\R$ of all real valued function on the interval $[0, 1]$ and let \[W=\{ f(x)\in V \mid f(x)=f(1-x) \text{ for } x\in [0,1]\}\] be a subset of $V$. Determine whether the subset $W$ is a subspace of the vector space $V$.   Proof. […]
  • The Subset Consisting of the Zero Vector is a Subspace and its Dimension is ZeroThe Subset Consisting of the Zero Vector is a Subspace and its Dimension is Zero Let $V$ be a subset of the vector space $\R^n$ consisting only of the zero vector of $\R^n$. Namely $V=\{\mathbf{0}\}$. Then prove that $V$ is a subspace of $\R^n$.   Proof. To prove that $V=\{\mathbf{0}\}$ is a subspace of $\R^n$, we check the following subspace […]
  • Quiz 8. Determine Subsets are Subspaces: Functions Taking Integer Values / Set of Skew-Symmetric MatricesQuiz 8. Determine Subsets are Subspaces: Functions Taking Integer Values / Set of Skew-Symmetric Matrices (a) Let $C[-1,1]$ be the vector space over $\R$ of all real-valued continuous functions defined on the interval $[-1, 1]$. Consider the subset $F$ of $C[-1, 1]$ defined by \[F=\{ f(x)\in C[-1, 1] \mid f(0) \text{ is an integer}\}.\] Prove or disprove that $F$ is a subspace of […]

You may also like...

1 Response

  1. 06/05/2017

    […] mid mathbf{x}in U, mathbf{y}in V}.] The sum $U+V$ is a subspace. (See the post “The sum of subspaces is a subspace of a vector space” for a […]

Leave a Reply

Your email address will not be published. Required fields are marked *

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

More in Linear Algebra
Problems and Solutions of Eigenvalue, Eigenvector in Linear Algebra
Idempotent Matrices are Diagonalizable

Let $A$ be an $n\times n$ idempotent matrix, that is, $A^2=A$. Then prove that $A$ is diagonalizable.  

Close