Vectors as basis for a subspace Here's a statement that I ran into,
The set $\{\mathbf{u}_{1},\mathbf{u}_{2}\cdots,\mathbf{u}_{6}\}$
is a basis for a subspace $\mathcal{S}$ of $\mathbb{R}^{m}$ if and
only if $\{\mathbf{u}_{1}+\mathbf{u}_{2},\mathbf{u}_{2}+\mathbf{u}_{3}\cdots,\mathbf{u}_{6}+\mathbf{u}_{1}\}$
is also a basis for $\mathcal{S}$.
I intuitively think the statement is true, but I don't know how to explain it. How should I start proving the statement? or is there a counter example for the statement?
 A: You can take a linear combination of the second basis and factor it to a linear combination of the first basis and show the linear combination is zero only if all the coefficients are zero. If so the vectors in the second basis are linearly independent, if there is any other values than zero that makes  linear combination of the first basis zero then second basis is not linearly independent. Edit: can do Gaussian elimination on coefficient matrix.
Then show that any linear combination of the first basis can be written as a linear combination of the second basis. Which means the second basis spans the space of the first basis.
Is this clear? Please let me know.
A: If the $\mathbf{u_i}$ are the standard basis for $\mathbb{R}^6$,  $\mathbf{e_1},\ldots,\mathbf{e_6}$, then all the vectors in the second collection are orthogonal to $\left(1,-1,1,-1,1,-1\right)$ so they are not independent.
A: As usual, the approach is to turn it into one of the standard algebra problems for matrices, and then use your tools to find the solution.
A fairly literal translation of the problem is as follows: let
$$ A = \left[
\begin{matrix} \mathbf{u_1} \\ \mathbf{u_2} \\ \cdots \\ \mathbf{u_6} \end{matrix} \right] $$
and 
$$ B = \left[
\begin{matrix} \mathbf{u_1} + \mathbf{u_2}  \\ \mathbf{u_2} + \mathbf{u_3} \\ \cdots \\ \mathbf{u_6} + \mathbf{u_1} \end{matrix} \right] $$
and you want to know if these two matrices have the same rowspace. Clearly the rowspace of $B$ is contained in the rowspace of $A$, so it's the other direction we need to do work to discover.
The usual method for finding a standard form for the rowspace of a matrix is to do row operations to reduce the matrix. While we can't do it in the usual way, it can still be done with $B$.
Another approach is possible: part of the difficulty is that while the form of $B$ is clear, it is complicated by the appearances of containing unknown vectors. So we can factor it:
$$ B = \left[ \begin{matrix} 1 & 1 & 0 & \cdots & 0
\\ 0 & 1 & 1 & \cdots & 0
\\ \vdots & \vdots & \vdots & & \vdots
\\ 1 & 0 & 0 & \cdots & 1 \end{matrix}
\right] A $$
Call this new matrix $C$. $C$ is all numbers, so we can row reduce it without having to devise a new algorithm to do it "symbolically".
The linear transformation induced by $C$ is thus a linear transformation from the rowspace of $B$ to the rowspace of $A$.
If $C$ is an invertible matrix, then we also have
$$ C^{-1} B = A$$
and so the rowspace of $A$ is thus contained in the rowspace of $B$.
If $C$ is not invertible, then it proves $B$ cannot be either. This gives a lead for finding a counterexample....

But maybe it's not clear the full extent of what this is telling us. The following will be a somewhat more sophisticated argument, but gives (IMO) a nice demonstration of the more complex arguments that can be made with the linear algebra tools you're learning about.
Our subgoal was to see if the rowspace of $A$ is contained in the rowspace of $B$. That is, for every vector $\mathbf{v}$, we are wondering if there is a solution for $\mathbf{x}$ in the equation
$$ \mathbf{v} A = \mathbf{x} B $$
Well, we can rewrite this as
$$ \mathbf{v} A = \mathbf{x} C A $$
$$ (\mathbf{v} - \mathbf{x} C) A = 0 $$
or equivalently, we seek a solution in $\mathbf{x}$ and $\mathbf{w}$ to
$$(\mathbf{v} - \mathbf{x} C) = \mathbf{w}$$
where $\mathbf{w}$ is required to be a left nullvector of $A$. Rearranging, we get:
$$\mathbf{v} = \mathbf{w} + \mathbf{x} C$$
Thus, this has solutions for every $\mathbf{v}$ if and only if:
$$ \text{NS}(A) + \text{RS}(C) = \mathbb{R}^6 $$
where $\text{NS}(A)$ is the left nullspace of $A$, and $\text{RS}(C)$ is the rowspace of $C$. Why? Because this is the vector space of all possible values the right hand side can have: $\mathbf{w}$ can be any element of $\text{NS}(A)$, and similarly $\mathbf{x} C$ can be any element of $\text{RS}(C)$.
It turns out that $\mathbf{C}$ has rank $5$. So this condition thus holds if and only if $A$ has any left nullvector that is not in the rowspace of $C$ (because this sum would thus have dimension strictly greater than $5$, and thus must be $6$).
We can even do better than this: we can compute the right nullspace to $C$, and find that it is spanned by the vector
$$ \mathbf{u} = \left[ \begin{matrix} 1 \\ -1 \\ 1 \\ -1 \\ 1 \\ -1 \end{matrix} \right] $$
Conversely, this means the left nullsapce of $\mathbf{u}$ is precisely the rowspace of $C$. Thus, a vector $\mathbf{y}$ is in the rowspace of $C$ if and only if $\mathbf{y} \mathbf{u} = 0$.
So, finally, we can state the following theorem:

$A$ and $B$ have the same rowspace if and only if there exists a vector $\mathbf{x}$ such that $\mathbf{x}A = 0$ and $\mathbf{xu} \neq 0$.

or even more simply

... if and only if $N \mathbf{u} \neq \mathbf{0}$

where $N$ is any matrix whose rows span the left nullspace of $A$.
A: A more "turn crank" approach to this problem borrows from exterior algebra concepts.
A subspace is represented by an element of the exterior algebra.  This element is merely the wedge product of vectors that span the subspace.  So this would be written as $S = u_1 \wedge u_2 \wedge \ldots \wedge u_6$.
The wedge product is distributive over addition, so the basis given would evaluate to
$$S' = (u_1 + u_2) \wedge (u_2 + u_3) \wedge \ldots \wedge (u_6 + u_1)$$
The wedge product of a vector with itself is zero.  The wedge product is also associative and anticommutative.
Now, there are $2^6 = 64$ different terms here that have to be multiplied out and summed.  Lame!  But doing this in a clever way can reduce some of the complexity of the problem.
Consider just $(u_1 + u_2) \wedge (u_2 + u_3) = u_1 \wedge u_2 + u_1 \wedge u_3 + u_2 \wedge u_3$.  We can do this for 3 pairs of vectors to get
$$\begin{align*}S' &= (u_1 \wedge u_2 + u_1 \wedge u_3 + u_2 \wedge u_3) \\ &\wedge (u_3 \wedge u_4 + u_3 \wedge u_5 + u_4 \wedge u_5) \\ &\wedge (u_5 \wedge u_6  + u_5 \wedge u_1 + u_6 \wedge u_1)\end{align*}$$
The key to evaluating this expression is to remember that any term that is nonzero must contain all of the six basis vectors.  Pay special attention to $u_2, u_4, u_6$.  Each of these only appears in one row, and as a result, we can eliminate products that do not obviously contain all of them.  Indeed, only the first and last columns of bivectors contribute to the final result.
$$S' = u_1 \wedge u_2 \wedge u_3 \wedge u_4 \wedge u_5 \wedge u_6 + u_2 \wedge u_3 \wedge u_4 \wedge u_5 \wedge u_6 \wedge u_1$$
So we take the second term, move $u_1$ to the front at the cost of a minus sign for each switch.  There are five such swaps, so the second term has a minus sign compared to the first term
And we get zero?!
Yes, as a matter of fact, $S' = 0$, and the second set of vectors cannot form a basis set for $S$.  This is because
$$u_6 + u_1 = (u_5 + u_6) - (u_4 + u_5) + (u_3 + u_4) - (u_2 + u_3) + (u_1 + u_2)$$
And so one vector in the set is expressable as a linear combination of the others.
This only happens when the set has even dimensionality.  If there were only 5 vectors in the set, the pattern of alternating additions and subtractions would make this linear dependence impossible.
The exterior algebra approach is helpful because one need not try to identify that one vector is a linear combination of the others.  The wedge product gives a "turn crank" test that merely tests one's ability to compute several wedge products to see if the resulting subspace is nonzero.
