if $v$ is a member of $H$ and $v$ is not a member of $M$ then $u$ is member of $K$. How is this possible?

Question

Let $(V,K)$ and $u,v$ is a member of $V$. Suppose that $M$ is a subset of $V$ is a subspace of $V$ with basis $B_m=\{m_1,...,m_r\}$ with $r$ less than and equal to $n$. Let $H$ be a subspace spanned by $B_m$ and $u$. Moreover, let $K$ be the subspace spanned by $B_m$ and $v$. Prove that if $v$ is a member of $H$ and $v$ is not a member of $M$ then $u$ is member of $K$.

Answer 1

$v$ is a member of $H$, means $v$ can be expressed as a linear combination of $m_1,m_2,...,m_r$ and $u$. Let $v=c_1m_1+...+c_rm_r+cu$. Here, $c\neq0$ as $v\notin M$. Thus $u=c^{-1}(v-c_1m_1-...-c_rm_r)$. Hence, $u\in K$.

Answer 2

If $v$ is not an element of $M$, then $K$ is one dimension larger than $M$ and contains $v$. We know that $H$ is at most one dimension larger than $M$ and also contains $v$. Thus $K=H$.