1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.