# Finding linearly independent combination of vectors with a certain symmetry property

I have two linearly independent vectors $v_1$ and $v_2$. Is there a way for me using Mathematica to find two linear combinations $w_1$ and $w_2$ such that:

1. $w_1$ and $w_2$ are linearly independent.
2. If I cut $w_2$ in half and re-order the halves to get $w_2'$, i.e. $w_2=\{1,2,3,4,5,6\}\rightarrow w_2'=\{4,5,6,1,2,3\}$, then $w_2'=w_1$?

$v_1$ and $v_2$ are chosen such that I am guaranteed $w_1$, $w_2$ exist.

Funny problem. I like it. But there is must be something else in the condition of the problem? Or I misunderstand you. You say that

I am guaranteed that w1, w2 exist

Ok. Let's set

v1 = {0, 0, 0, 1};
v2 = {1, 0, 0, 0};


it's independent vectors.

Then find w1 and w2 in general case for this example:

w1 = v1*a1 + v2*b1
w2 = v1*a2 + v2*b2
(*{b1,0,0,a1}
{b2,0,0,a2}*)


Then find w2trans ($w_2'$)

w2trans = RotateLeft[w2, 2]
(*{0,a2,b2,0}*)


And w2trans!=w1 in any cases. Is w2trans may be {a1,0,b2,0} in this example?

• Thank-you very much for your answer, I will clarify my question. What I meant was that $v_1$, $v_2$ are chosen such that $w_1$, $w_2$ exist, they are not arbitrary. This problem arises in the context of the Bogoliubov Transformation, where I require a change of basis matrix $T$ to have the property that the upper left block equals the (complex conjugate of the) bottom right, and the upper right block equals the bottom left. The rows of $T$ are eigenvectors of some matrix, and I can take linear combinations of degenerate eigenvectors. Jul 24 '14 at 8:55
• A proof that this is possible is provided at the very end of section II in these notes. However it does not give any algorithm for being able to do this. Jul 24 '14 at 8:58
• ok, give me v1 and v2 for check my algorithm Jul 24 '14 at 9:00
• Thanks very much for your assistance. One set of examples is: v1 = {0.65566, 0.262264, 0, 0.0364256, -0.65566, -0.262264, 0, -0.0364256}; v2 = {0, -0.0802896, 0, -0.702534, 0, -0.0802896, 0, -0.702534}. In this case I can construct w1, w2 manually via: {w1,w2}={v1 + v2, v2 - v1}. Another case is: v1 = {0, 0, 0, 0, 0, 0, 0, 0.707107, 0, 0, 0, 0, 0, 0, 0, 0.707107}; v2 = {0, 0, 0, 0.707107, 0, 0, 0, 0, 0, 0, 0, 0.707107, 0, 0, 0, 0}. Jul 24 '14 at 9:17
• Another case: v1={0.707107, 0, 0, 0, -0.707107, 0, 0, 0}; v2={0, 0, 0, -0.707107, 0, 0, 0, -0.707107}, and as before {w1,w2}={v1 + V2, v2 - v1}. Jul 24 '14 at 9:22