Subspace spanned by two standard basis vectors and its intersection with other subspaces Let $\mathcal{V}$ be a subspace of $\mathbb{R}^n$ with $dim\mathcal{V}=n-1$ where $n&gt;3$.

For any integer pairs $i$ and $j$ satisfying $1\leq i &lt; j \leq n$, let 
$\mathcal{W}_{ij}$ be a subspace of $\mathbb{R}^n$ spanned by two vectors $\mathbf{e}_i$ and $\mathbf{e}_j$ that are the $i^\mathrm{th}$ and $j^\mathrm{th}$ column in the $n\times n$ identity matrix.

Clearly, $\mathcal{V}\cap\mathcal{W}_{ij}$ is not empty for any $i$ and $j$. But I'm wondering whether I can claim something stronger.


  For any subspace $\mathcal{V}$ of $\mathbb{R}^n$ with $dim\mathcal{V}=n-1$ where $n&gt;3$, there always exist integers $i$ and $j$ satisfying $\mathcal{V} \supset \mathcal{W}_{ij}=span(\mathbf{e}_i,\mathbf{e}_j)$.


Is this statement true?
Since $n&gt;3$ this implies that $dimV &gt; 2$. Now the dimension of the subspace is one less than the whole vector space so there must exist two vectors from the standard basis in the subspace. 

For suppose not then there exists two $e_i,e_j$ such that  $span(e_i,e_j)\cap V=\{0\}$. To see this note that $V$ must contain one memeber of the standard basis or else it is the zero space, since it must contain one there is at least two other members not in the space by the hypothesis that $n&gt;3$ and the assumption being made. This means they are a direct sum implying that there dimension is $dim(V\oplus span(e_i,e_j))=dim(V)+dim(span(e_i,e_j))=n-1+2=n+1$.

Certainly, the direct sum of these subspaces is itself a subspace of $R^n$. But this a contradiction since the dimension of a subspace is always less than or equal to the vector space. 

Therefore the assertion is indeed true. 

In fact, I suspect there may be even a stronger condition which you may assert but will I leave that to you to explore.
Since $n&gt;3$ this implies that $dimV &gt; 2$. Now the dimension of the subspace is one less than the whole vector space so there must exist two vectors from the standard basis in the subspace. 

For suppose not then there exists two $e_i,e_j$ such that  $span(e_i,e_j)\cap V=\{0\}$. To see this note that $V$ must contain one memeber of the standard basis or else it is the zero space, since it must contain one there is at least two other members not in the space by the hypothesis that $n&gt;3$ and the assumption being made. This means they are a direct sum implying that there dimension is $dim(V\oplus span(e_i,e_j))=dim(V)+dim(span(e_i,e_j))=n-1+2=n+1$.

Certainly, the direct sum of these subspaces is itself a subspace of $R^n$. But this a contradiction since the dimension of a subspace is always less than or equal to the vector space. 

Therefore the assertion is indeed true. 

In fact, I suspect there may be even a stronger condition which you may assert but will I leave that to you to explore.