Sum of two subspaces is equal to the span of their union

Question

Let $V_{1}$ and $V_{2}$ be two subspaces of $V$. Define the sum of $V_{1}$ and $V_{2}$ to be the subset of $V$

$V_{1}+V_{2}=${$ \overrightarrow v_{1} + \overrightarrow v_{2}: \overrightarrow v_{1} \in V_{1}, \overrightarrow v_{2} \in V_{2}$}

Prove that $V_{1}+V_{2}=Span(V_{1} \cup V_{2})$

Approach: I want to prove $V_{1}+V_{2} \subset Span(V_{1} \cup V_{2})$ and $Span(V_{1} \cup V_{2}) \subset V_{1}+V_{2}$. Any tips on how to proceed?

Answer 1

If $x\in V_1+V_2$ then $x=v_1+v_2$ for $v_i\in V_i$. However this immediately implies $x\in Span(V_1\cup V_2)$ because $x$ is a sum of elements from $V_1\cup V_2$. This gives the first inclusion.

If $x\in Span(V_1\cup V_2)$, then I can write $x$ as a sum of elements from $V_1\cup V_2$, say $u_1+w_1+u_2+w_2+\dots+u_k+w_k$, where $u_i\in V_1$ and $w_i\in V_2$. Then I can group these so

$$x=(u_1+\dots+u_k)+(w_1+\dots+w_k)\in V_1+V_2$$

because $u_1+\dots+u_k\in V_1$, and $w_1+\dots+w_k\in V_2$. This gives the other inclusion and you are done.

Answer 2

I think you're on the right track:

To demonstrate this $V_{1}+V_{2} \subset Span(V_{1} \cup V_{2})$ you need to pick a vector $w$ in $V_{1}+V_{2}$ and prove that is in the $Span(V_{1} \cup V_{2})$ writing it as a linear combination of vector of $V_{1}$ and $V_{2}$ (indeed is trivially from the definition).

The converse $Span(V_{1} \cup V_{2}) \subset V_{1}+V_{2}$ is similar, i.e. pick a vector in the span write it in a base of $Span(V_{1} \cup V_{2})$ then rearrange the term to show that is in $V_{1}+V_{2}$.

Answer 3

Let $x\in V_1+V_2$. By definition of $V_1+V_2$, you have $x_1\in V_1$ and $x_2\in V_2$ so that $x = x_1 + x_2$. So...

$x=x_1+x_2=1\cdot x_1+1\cdot x_2\in \operatorname{Span}(V_1\cup V_2)$ since $x_1\in V_1 \subseteq V_1\cup V_2$ and $x_2\in V_2 \subseteq V_1\cup V_2$.

So we have $x\in \operatorname{Span}(V_1\cup V_2)$.

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Let $x\in \operatorname{Span}(V_1\cup V_2)$. By definition, there is a finite number (so $1\le i \le n$) of $x_i\in V_1\cup V_2$ and of $\lambda_i\in \mathbb R$ so that $x=\lambda_1\cdot x_1+\dots+\lambda_nx_n$. But each $x_i$ is either in $V_1$ or $V_2$. Let $\alpha_1,\dots,\alpha_p$ be the indices so that $x_{\alpha_i}\in V_1$ and $\beta_1,\dots,\beta_q$ the remaining indices. We have $x=(\lambda_{\alpha_1}\cdot x_{\alpha_1}+\dots+\lambda_{\alpha_p}\cdot x_{\alpha_p})+(\lambda_{\beta_1}\cdot x_{\beta_1}+\dots+\lambda_{\beta_q}\cdot x_{\beta_q})$. And so...

We know that $x_{\beta_i}$ will be in $V_1\cup V_2$ and not in $V_1$ so in $V_2$.

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Since $V_1$ is a subspace and all the $x_{\alpha_i}$s are in $V_1$, $(\lambda_{\alpha_1}\cdot x_{\alpha_1}+\dots+\lambda_{\alpha_p}\cdot x_{\alpha_p})\in V_1$. For the same reasons, $(\lambda_{\beta_1}\cdot x_{\beta_1}+\dots+\lambda_{\beta_q}\cdot x_{\beta_q})\in V_2$.

And so $x\in V_1+V_2$.

Answer 4

This is the kind of question that is hard because of how trivial it is. Just rewrite the RHS and the LHS in terms of their definitions and you basically have a proof.

Answer 5

Since the union is contained in the sum and the sum is a vector space (i.e. stable under linear combinations), the span of the union cannot exceed the sum.

The other direction is just the observation that a sum of vectors is a linear combination.