Intersection of $n-1$ dimensional subspaces... Let $V$ be an $n$-dimensional vector space over $\Bbb R$. Show that every one dimensional subspace is the intersection of all $n-1$ dimensional subspaces containing it.
I honestly have no clue on this one.
 A: Hint: Pick a basis $\{x_1\}$ for the line. Extend this to a basis $\{x_1,\ldots,x_n\}$. Clearly the line is in the intersection of all $n-1$ dimensional subspaces containing it, and to show the converse what if you considered $H_2,\ldots,H_n$ where $H_i=\text{span}\{x_j:j\ne i\}$. 
A: Let the one-dimensional subspace be $V=\mathrm{span}(\mathbf{v})$.  If $\mathbf{w} \not\in V$, then $\mathbf{w}$ is a linear combination of $\mathbf{v}$ and a non-zero vector $\mathbf{u}$ orthogonal to $\mathbf{v}$ ($\mathbf{u}$ is the vector rejection, according to Wikipedia).  If $\mathbf{u}=(u_1,u_2,\ldots,u_n)$, then the $1 \times n$ matrix $$\left(
\begin{matrix}
u_1 & u_2 & \cdots & u_n \\
\end{matrix}
\right)$$ has an $(n-1)$-dimensional null space, by the Rank-Nullity Theorem.  Moreover:


*

*$\mathbf{v}$ is in this null space, since $\mathbf{u}$ and $\mathbf{v}$ are orthogonal.

*$\mathbf{u}$ is not in this null space, otherwise $\mathbf{u} \cdot \mathbf{u}=0$, implying $\mathbf{u}$ is the zero vector.

*Hence $\mathbf{w}$ is not in this null space, otherwise $\mathbf{u}$ could be formed from a linear combination of $\mathbf{w}$ and $\mathbf{v}$.
Thus, for any $\mathbf{w} \not\in V$, we have found an $(n-1)$-dimensional subspace containing $V$ but not $\mathbf{w}$.
The converse is true by definition (if $\mathbf{w} \in V$, then its in all $(n-1)$-subspaces contain $V$, and hence their intersection also contains $V$).
