Reduction of basis and dimensions If $S$ is a $5-$dimensional subspace of $\mathbb{R}^6$ .Is it true that every basis of $\mathbb{R}^6$ can be reduced to a basis of $S$ by removing one vector ?

Removing a vector from a basis of $\mathbb{R}^n$ you always have a basis of some subspace $S$ of dimension $n-1$. This is true because you have $n-1$ linearly independent vectors that spans a subspace.  But If you want a particular subspace $S$ then the statement is not true in general and you have to find  $n-1$ linearly independent vectors that span this space, and these vectors can be a linear combinations of the starting basis vectors.
As  an example, if $(1,0)^T$ and $(0,1)^T$ are the canonical basis of $\mathbb{R}^2$ than the subspace $S$ spanned by $(1,3)$ ( the line of points $(x,3x)^T$ that has dimension $1$) has a basis vector that is not $(1,0)^T$ nor $(0,1)^T$
