The question I have is: Suppose $x_1, x_2, ... , x_k$ are linearly independent n-dimensional vectors. Show that $x_1+x_2, x_2+x_3, ... ,x_{k-1}+x_k, x_k+x_1$ are also linearly independent. Here k is an odd number.

I thought I had a proof done, but was told that I showed linear dependence instead. What I did was took a set of constants $a_1, a_2, ..., a_k$ and set $a_1(x_1+x_2)+a_2(x_2+x_3)+...+a_k(x_k+x_1)=0$ and proved that each constant was 0. Is that right? If not, how can I prove this?


Yes, what you disclose to us are correct moves.

That is if $$a_1(x_1+x_2) +a_2(x_2+x_3) + \ldots + a_k(x_k+x_1)=0$$ implies $a_1=\ldots = a_k=0$, then you have shown that the set of vectors are linearly independent.

$$(a_1+a_k)x_1+(a_1+a_2)x_2+ \ldots +(a_{k-1}+a_k)x_k=0$$

We get $$a_1+a_k=a_1+a_2=\ldots =a_{k-1}+a_k = 0$$

From this $a_i=a_{i+2}$ for $i=1, \ldots, k-2$. $a_{k-1}=a_1$, and $a_k=a_2$.

The first part $a_i=a_j$ is $i$ and $j$ share the same parity. However, $a_{k-1}=a_1$ enable us to conclude that all numbers are equal and since they sum to $0$, all numbers must be zero.

Alternatively, it is possible to form linear system involving $a_i$ and show that its determinant is non-zero using Laplacian expansion and determinant of triangular matrices.

  • $\begingroup$ So does this work: (ak+a1)x1+(a1+a2)x2+...+(a(k-1)ak)xk=0 $\endgroup$ – Alvin Jan 28 '18 at 3:06
  • $\begingroup$ yes, your approach is correct. $\endgroup$ – Siong Thye Goh Jan 28 '18 at 3:06
  • $\begingroup$ then this is a linear combination of the xi's which are linearly independent, so each pair sum of ai must be 0? $\endgroup$ – Alvin Jan 28 '18 at 3:07
  • $\begingroup$ yes, the deduction is correct. $\endgroup$ – Siong Thye Goh Jan 28 '18 at 3:09

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