Q: provided that $\displaystyle S_{0}=\{\vec{u_{1}},...,\vec{u_{n}}\}$ is a linearly independent set of vectors and that the set $\displaystyle S_{1}=\{\vec{u_{1}},...,\vec{u_{n}},\vec{v}\}$ is linearly dependent, prove that $\displaystyle \vec{v}$ is a linear combination of the $\displaystyle \vec{u_{i}}$'s.

A: Since $\displaystyle S_{0}$ is linearly independent the vector equation,

$\displaystyle c_{1}\vec{u_1}+c_{2}\vec{u_2}+. . .+c_{n}\vec{u_n}=\vec{0}$

has only the trival solution, $\displaystyle c_{1}=0, c_{2}=0,. . .,c_{k}=0$. Conversely, the set $\displaystyle S_{1}$ is linearly dependent; thus, there exists a vector in $\displaystyle S_{1}$ that can be written as a linear combination. Since $\displaystyle S_{1}=\{\{S_{0}\},\vec{v}\}$ and $\displaystyle S_{0}$ is known to be linearly independent we can narrow our search to just one vector, the vector $\displaystyle \vec{v}$. So, we have a new vector equation,

$\displaystyle c_{1}\vec{u_1}+c_{2}\vec{u_2}+. . .+c_{n}\vec{u_n}+k\vec{v}=\vec{0}$

where $\displaystyle k\in{\mathbb{R}}$

Solving for $\displaystyle \vec{v}$

$\displaystyle \vec{v}=-\frac{c_{1}}{k}\vec{u_1}-. . .-\frac{c_{n}}{k}\vec{u_n}$

$\displaystyle \therefore$ $\displaystyle \vec{v}$ can be written as linear comination of $\displaystyle \vec{u_{i}}$'s.

I dunno, I feel a chunck is missing. Is that a sufficient proof?

Thanks