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 ?
 A: 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 $(0,1)^T$
A: Let's take a simple example and build on it.
Consider the basis of $\mathbb R^6$ : $span[(1,0,0,0,0,0),(0,1,0,0,0,0),(0,0,1,0,0,0),(0,0,0,1,0,0),(0,0,0,0,1,0),(0,0,0,0,0,1)] $
It's clear that if you remove the vector $(0,0,0,0,0,1)$, then none of the other vectors generate a $6th$ dimension, so yes, indeed, removing that vector, you get the standard basis of $\mathbb R^5$. Now, for a more general example, for a particular subspace $S$, in order to take a basis for the $S^{n-1}$ space, from a set of vectors, for example, $span[v_1,v_2,..,v_n$], you will need to find $n-1$ in number linear independent vectors that create the $n-1$ dimension space, from the $n$ dimension space you already have.
