Suppose $𝑉$ is a vector space over $𝐹$, and let $𝐻\subseteq 𝑉$. Also suppose that $𝐻$ is closed under linear combinations. We will show that $𝐻$ satisfies the three Subspace Criteria.

1. Existence of additive identity. ${\mathrm{𝟎}}_{𝑉}$ is the trivial linear combination, so ${\mathrm{𝟎}}_{𝑉}\in 𝐻$.
2. Closure under vector addition. Let $\mathrm{𝐮},\mathrm{𝐯}\in 𝐻$. Then $\mathrm{𝐮}+\mathrm{𝐯}$ is a linear combination of $\mathrm{𝐮}$ and $\mathrm{𝐯}$ with weights $𝛼=𝛽=1_{𝐹}$, so $\mathrm{𝐮}+\mathrm{𝐯}\in 𝐻$.
3. Closure under scalar multiplication. Let $\mathrm{𝐯}\in 𝐻$ and $𝑘\in 𝐹$. Then $𝑘\mathrm{𝐯}$ is a linear combination of $\mathrm{𝐯}$ with weight $𝑘$, so $𝑘\mathrm{𝐯}\in 𝐻$. ∎

Let $𝑉$ be a vector space over $𝐹$, and let $𝑆=\left\{{\mathrm{𝐯}}_1,{\mathrm{𝐯}}_2,\dots ,{\mathrm{𝐯}}_{𝑛}\right\}\subseteq 𝑉$. Checking the three Subspace Criteria:

1. Existence of additive identity. Fix all weights ${𝑎}_1={𝑎}_2=\dots ={𝑎}_{𝑛}=0_{𝐹}$. Then for any linear combination of $𝑆$, we have:

   $$\sum _{𝑖=1}^{𝑛}{𝑎}_{𝑖}{\mathrm{𝐯}}_{𝑖}=\sum _{𝑖=1}^{𝑛}{0}_{𝐹}{\mathrm{𝐯}}_{𝑖}=\sum _{𝑖=1}^{𝑛}{\mathrm{𝟎}}_{𝑉}={\mathrm{𝟎}}_{𝑉}.$$

2. Closure under vector addition. Let $\mathrm{𝐮},\mathrm{𝐯}\in \mathrm{span}\left(𝑆\right)$, so there exist weights ${𝑎}_1,{𝑎}_2,\dots ,{𝑎}_{𝑛}$ and ${𝑏}_1,{𝑏}_2,\dots ,{𝑏}_{𝑛}$ such that:

   $$\mathrm{𝐮}=\sum _{𝑖=1}^{𝑛}{𝑎}_{𝑖}{\mathrm{𝐯}}_{𝑖},\mathrm{𝐯}=\sum _{𝑖=1}^{𝑛}{𝑏}_{𝑖}{\mathrm{𝐯}}_{𝑖}.$$

   Then:

   $$\mathrm{𝐮}+\mathrm{𝐯}=\sum _{𝑖=1}^{𝑛}{𝑎}_{𝑖}{\mathrm{𝐯}}_{𝑖}+\sum _{𝑖=1}^{𝑛}{𝑏}_{𝑖}{\mathrm{𝐯}}_{𝑖}=\sum _{𝑖=1}^{𝑛}\left({𝑎}_{𝑖}+{𝑏}_{𝑖}\right){\mathrm{𝐯}}_{𝑖}.$$

   $𝐹$ is closed under field addition, so ${𝑎}_{𝑖}+{𝑏}_{𝑖}\in 𝐹$ for all $𝑖$. Thus, $\mathrm{𝐮}+\mathrm{𝐯}$ is a linear combination of $𝑆$, so $\mathrm{𝐮}+\mathrm{𝐯}\in \mathrm{span}\left(𝑆\right)$.

3. Closure under scalar multiplication. Let $\mathrm{𝐯}\in \mathrm{span}\left(𝑆\right)$, so there exist weights ${𝑎}_1,{𝑎}_2,\dots ,{𝑎}_{𝑛}$ such that:

   $$\mathrm{𝐯}=\sum _{𝑖=1}^{𝑛}{𝑎}_{𝑖}{\mathrm{𝐯}}_{𝑖}.$$

   Let $𝑘\in 𝐹$. Then:

   $$𝑘\mathrm{𝐯}=𝑘\sum _{𝑖=1}^{𝑛}{𝑎}_{𝑖}{\mathrm{𝐯}}_{𝑖}=\sum _{𝑖=1}^{𝑛}\left(𝑘{𝑎}_{𝑖}\right){\mathrm{𝐯}}_{𝑖}.$$

   $𝐹$ is closed under field multiplication, so $𝑘{𝑎}_{𝑖}\in 𝐹$ for all $𝑖$. Thus, $𝑘\mathrm{𝐯}$ is a linear combination of $𝑆$, so $𝑘\mathrm{𝐯}\in \mathrm{span}\left(𝑆\right)$. ∎

Let $𝑉$ be a vector space, and let $𝐻$ be a subspace of $𝑉$. If $𝐻=\left\{{\mathrm{𝟎}}_{𝑉}\right\}$, then we can take $𝑆=\varnothing$, so assume that $𝐻\ne \left\{{\mathrm{𝟎}}_{𝑉}\right\}$.

TODO ∎

Let ${𝑆}_1$ and ${𝑆}_2$ be sets of vectors, and let ${𝐻}_1=\mathrm{span}\left({𝑆}_1\right)$ and ${𝐻}_2=\mathrm{span}\left({𝑆}_2\right)$ be the subspaces generated by ${𝑆}_1$ and ${𝑆}_2$, respectively.

Suppose $\mathrm{𝐯}\in \mathrm{span}\left({𝑆}_1\cup {𝑆}_2\right)$. Then there exist weights ${𝑎}_1,{𝑎}_2,\dots ,{𝑎}_{𝑛}\in 𝐹$ and vectors ${\mathrm{𝐮}}_1,{\mathrm{𝐮}}_2,\dots ,{\mathrm{𝐮}}_{𝑛}\in {𝑆}_1\cup {𝑆}_2$ such that:

We can partition the vectors ${\mathrm{𝐮}}_1,{\mathrm{𝐮}}_2,\dots ,{\mathrm{𝐮}}_{𝑛}$ into two groups: those that are in ${𝑆}_1$ and those that are in ${𝑆}_2$. Let the vectors in the first group be denoted by ${\mathrm{𝐬}}_1,{\mathrm{𝐬}}_2,\dots ,{\mathrm{𝐬}}_{𝑘}$, and let the vectors in the second group be denoted by ${\mathrm{𝐭}}_1,{\mathrm{𝐭}}_2,\dots ,{\mathrm{𝐭}}_{𝑛-𝑘}$. Then we can rewrite the equation above as follows:

The first sum is a linear combination of vectors in ${𝑆}_1$, so the first sum is a vector in $\mathrm{span}\left({𝑆}_1\right)={𝐻}_1$; call it ${\mathrm{𝐡}}_1$. Similarly, the second sum is a linear combination of vectors in ${𝑆}_2$, so the second sum is a vector in $\mathrm{span}\left({𝑆}_2\right)={𝐻}_2$; call it ${\mathrm{𝐡}}_2$. Thus, we can write:

Therefore, $\mathrm{𝐯}\in {𝐻}_1+{𝐻}_2$, so $\mathrm{span}\left({𝑆}_1\cup {𝑆}_2\right)\subseteq {𝐻}_1+{𝐻}_2$.

Conversely, suppose that $\mathrm{𝐯}\in {𝐻}_1+{𝐻}_2$. Then there exist vectors ${\mathrm{𝐡}}_1\in {𝐻}_1$ and ${\mathrm{𝐡}}_2\in {𝐻}_2$ such that:

Since ${\mathrm{𝐡}}_1\in {𝐻}_1=\mathrm{span}\left({𝑆}_1\right)$, there exist weights ${𝑎}_1,{𝑎}_2,\dots ,{𝑎}_{𝑘}\in 𝐹$ and vectors ${\mathrm{𝐬}}_1,{\mathrm{𝐬}}_2,\dots ,{\mathrm{𝐬}}_{𝑘}\in {𝑆}_1$ such that:

Similarly, since ${\mathrm{𝐡}}_2\in {𝐻}_2=\mathrm{span}\left({𝑆}_2\right)$, there exist weights ${𝑏}_1,{𝑏}_2,\dots ,{𝑏}_{𝑛-𝑘}\in 𝐹$ and vectors ${\mathrm{𝐭}}_1,{\mathrm{𝐭}}_2,\dots ,{\mathrm{𝐭}}_{𝑛-𝑘}\in {𝑆}_2$ such that:

Thus, we can write:

The vectors ${\mathrm{𝐬}}_1,{\mathrm{𝐬}}_2,\dots ,{\mathrm{𝐬}}_{𝑘}$ and ${\mathrm{𝐭}}_1,{\mathrm{𝐭}}_2,\dots ,{\mathrm{𝐭}}_{𝑛-𝑘}$ are all in ${𝑆}_1\cup {𝑆}_2$, so $\mathrm{𝐯}$ is a linear combination of vectors in ${𝑆}_1\cup {𝑆}_2$. Therefore, $\mathrm{𝐯}\in \mathrm{span}\left({𝑆}_1\cup {𝑆}_2\right)$, so ${𝐻}_1+{𝐻}_2\subseteq \mathrm{span}\left({𝑆}_1\cup {𝑆}_2\right)$.

By double containment, $\mathrm{span}\left({𝑆}_1\cup {𝑆}_2\right)={𝐻}_1+{𝐻}_2$. ∎

Let $𝑉$ be a vector space over $𝐹$, and let $𝑆=\left\{{\mathrm{𝐯}}_1,{\mathrm{𝐯}}_2,\dots ,{\mathrm{𝐯}}_{𝑛}\right\}\subseteq 𝑉$. Let $𝑝$ be the statement that $𝑆$ is linearly independent, and let $𝑞$ be the statement that the only solution to the equation ${𝑎}_1{\mathrm{𝐯}}_1+{𝑎}_2{\mathrm{𝐯}}_2+\cdots +{𝑎}_{𝑛}{\mathrm{𝐯}}_{𝑛}={\mathrm{𝟎}}_{𝑉}$ is the trivial solution ${𝑎}_1={𝑎}_2=\cdots ={𝑎}_{𝑛}=0_{𝐹}$.

$𝒑\Rightarrow 𝒒$. Assume $𝑆$ is linearly independent. Let ${𝑎}_1,{𝑎}_2,\dots ,{𝑎}_{𝑛}$ be scalars such that:

Then, by Theorem 3, realize that:

Since $𝑆$ is linearly independent, the weights of these two linear combinations must be equal, so ${𝑎}_1={𝑎}_2=\dots ={𝑎}_{𝑛}=0_{𝐹}$. Therefore, the only solution to the equation is the trivial solution.

$𝒒\Rightarrow 𝒑.$ Assume that the only solution to the equation is the trivial solution. Let ${𝑎}_1,{𝑎}_2,\dots ,{𝑎}_{𝑛}$ and ${𝑏}_1,{𝑏}_2,\dots ,{𝑏}_{𝑛}$ be scalars such that:

Then we can rearrange this equation as follows:

By our assumption, the only solution to this equation is the trivial solution, so ${𝑎}_{𝑖}-{𝑏}_{𝑖}=0_{𝐹}$ for all $𝑖$, so ${𝑎}_1={𝑏}_1$, ${𝑎}_2={𝑏}_2$, …, ${𝑎}_{𝑛}={𝑏}_{𝑛}$. Therefore, if two linear combinations of vectors in $𝑆$ are equal, then their weights must be equal, so $𝑆$ is linearly independent.

Since $𝑝\Rightarrow 𝑞$ and $𝑞\Rightarrow 𝑝$, we have $𝑝\iff 𝑞$. ∎

Let $𝑆=\left\{{\mathrm{𝐯}}_1,{\mathrm{𝐯}}_2,\dots ,{\mathrm{𝐯}}_{𝑛}\right\}$ be an indexed set of vectors with ${\mathrm{𝐯}}_1\ne \mathrm{𝟎}$. Let $𝑝$ be the statement that $𝑆$ is linearly dependent, and let $𝑞$ be the statement that there exists some $𝑘>1$ such that ${\mathrm{𝐯}}_{𝑘}\in \mathrm{span}\left\{{\mathrm{𝐯}}_1,\dots ,{\mathrm{𝐯}}_{𝑘-1}\right\}$.

$𝒑\Rightarrow 𝒒.$ Assume $𝑆$ is linearly dependent. Then there exist nontrivial weights ${𝑎}_1,{𝑎}_2,\dots ,{𝑎}_{𝑛}$ such that:

Since ${\mathrm{𝐯}}_1\ne {\mathrm{𝟎}}_{𝑉}$, we can choose weights such that ${𝑎}_{𝑘}\ne 0_{𝐹}$ and ${𝑎}_{𝑖}=0_{𝐹}$ for all $𝑖>𝑘$ for some $1<𝑘\le 𝑛$. Then we can rearrange the equation as follows:

Since ${𝑎}_{𝑘}\ne 0_{𝐹}$, we can multiply both sides of the equation by ${𝑎}_{𝑘}^{-1}$ to get:

The right-hand side of this equation is a linear combination of the vectors ${\mathrm{𝐯}}_1,\dots ,{\mathrm{𝐯}}_{𝑘-1}$, so ${\mathrm{𝐯}}_{𝑘}\in \mathrm{span}\left\{{\mathrm{𝐯}}_1,\dots ,{\mathrm{𝐯}}_{𝑘-1}\right\}$.

$𝒒\Rightarrow 𝒑.$ Assume that there exists some $𝑘>1$ such that ${\mathrm{𝐯}}_{𝑘}\in \mathrm{span}\left\{{\mathrm{𝐯}}_1,\dots ,{\mathrm{𝐯}}_{𝑘-1}\right\}$. By the definition of span, there exist weights ${𝑎}_1,{𝑎}_2,\dots ,{𝑎}_{𝑘-1}$ such that:

We can rearrange this equation as follows:

No matter what any of the ${𝑎}_{𝑖}$ are, a nontrivial weight $-1_{𝐹}$ is used for ${\mathrm{𝐯}}_{𝑘}$ to get a linear combination that equals ${\mathrm{𝟎}}_{𝑉}$, so by Theorem 16, $𝑆$ is linearly dependent.

Since $𝑝\Rightarrow 𝑞$ and $𝑞\Rightarrow 𝑝$, we have $𝑝\iff 𝑞$. ∎