1. ## Set proof

Q: Prove that if $S_{1}$ is a nonempty subset of the finite set $S_{2}$, and $S_{1}$ is linearly dependent, then so is $S_{2}$

A: By definition $S_{1}\subset{S_{2}}$ $\iff$ $\forall{\vec{v}}\in{U}$ $(\vec{v}\in{S_{1}}$ $\Rightarrow$ $\vec{v}\in{S_{2}})$. Since we know every vector $\vec{v_{i}}\in{S_{1}}$ can be written as a linear combination of at least one vector $\vec{u}\in{S_{1}}$ and $S_{1}\subset{S_{2}}$ we are guaranteed $\vec{u}\in{S_2}$. So, we have shown exists a linear dependent vector in $S_{2}$.

$\therefore$ $S_{2}$ must also be linearly dependent.

Have I shown enough work to prove my claim?

2. ## Linear dependence

Hello Danneedshelp
Originally Posted by Danneedshelp
Q: Prove that if $S_{1}$ is a nonempty subset of the finite set $S_{2}$, and $S_{1}$ is linearly dependent, then so is $S_{2}$

A: By definition $S_{1}\subset{S_{2}}$ $\iff$ $\forall{\vec{v}}\in{U}$ $(\vec{v}\in{S_{1}}$ $\Rightarrow$ $\vec{v}\in{S_{2}})$. Since we know every vector $\vec{v_{i}}\in{S_{1}}$ can be written as a linear combination of at least one vector $\vec{u}\in{S_{1}}$ and $S_{1}\subset{S_{2}}$ we are guaranteed $\vec{u}\in{S_2}$. So, we have shown exists a linear dependent vector in $S_{2}$.

$\therefore$ $S_{2}$ must also be linearly dependent.

Have I shown enough work to prove my claim?
This is OK, except that it's not every vector $\vec{v_{i}}\in{S_{1}}$, but $\exists\,\vec{v_{i}}\in{S_{1}}$.