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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.

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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?

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    $\begingroup$ 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. $\endgroup$ Jul 24 '14 at 8:55
  • $\begingroup$ 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. $\endgroup$ Jul 24 '14 at 8:58
  • $\begingroup$ ok, give me v1 and v2 for check my algorithm $\endgroup$ Jul 24 '14 at 9:00
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    $\begingroup$ 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}. $\endgroup$ Jul 24 '14 at 9:17
  • $\begingroup$ 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}. $\endgroup$ Jul 24 '14 at 9:22

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