Let $x_1,x_2,\dots,x_6$ be variables.

Let $c_1,c_2,c_3,c_4>0$ be positive real numbers.

Consider the four expressions:

  • $w_1:=c_1x_4-c_2x_2+c_3x_1$
  • $w_2:= c_2x_6-c_3x_5+c_4x_4$
  • $w_3:=c_1x_5-c_2x_3+c_4x_1$
  • $w_4:=c_1x_6-c_3x_3+c_4x_2$

Something nice is that:


(we can tediously verify it)

I.e. there exists a nontrivial linear combination of $w_1,w_2,w_3,w_4$ that results in zero.

My question is what is the underlying principle behind this? (the existence of a nontrivial linear combination that results in zero.) Is there any reason from linear algebra or others?

Or is it due to some pattern of the $w_1,w_2,w_3,w_4$?

Hope this question makes sense. Thanks.

Another way to phrase my question is how do we get this relation $$c_4w_1-c_1w_2-c_3w_3+c_2w_4=0$$ (or any other linear combination that results in zero) without doing guessing or complicated solving of system of equations? And in the first place, how do we know if such a relation exists?

One way is to try to match one variable by one variable, say consider $x_4$. Then the hint is to multiply $w_1$ by $c_4$ and $w_2$ by $-c_1$ to "eliminate" $x_4$. But why does this method work out so nicely?

Thanks once again.


The existence of such a solution is easy: it just happens to be that the vectors $$w_1=\begin{bmatrix}c_3\\-c_2\\0\\c_1\\0\\0\\\end{bmatrix},w_2=\begin{bmatrix}0\\0\\0\\c_4\\-c_3\\c_2\end{bmatrix},w_3=\begin{bmatrix}c_4\\0\\-c_2\\0\\c_1\\0\end{bmatrix},w_4=\begin{bmatrix}0\\c_4\\-c_3\\0\\0\\c_1\end{bmatrix}$$ are linearly dependent. This can be shown in a number of ways (look at the invertible matrix theorem to see equivalent ways to say vectors are linearly independent). One way would be to see that the matrix formed by $w_i$ as the $i$-th column has rank less than $4$, so the nullspace is non-trivial, and thus there is a non-trivial solution to $k_1w_1+k_2w_2+k_3w_3+k_4w_4=0$.

  • $\begingroup$ Makes sense.. Thanks! $\endgroup$ – yoyostein Sep 30 '17 at 16:07

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