Let $a_1,.. a_n$ a set of vectors above vector space $V$. It's given that $a_1,...a_n$ are linearly independent. Find all $\lambda$'s such that the following set is linearly independent:

$$ a_1 + a_2, ..., a_{n-1} + a_n, a_n + \lambda a_1 $$

What I did:
Assume that the set is linearly independent.
$$ \beta_1(a_1 + a_2) + ... + \beta_{n-1}(a_{n-1} + a_n) + \beta_n(a_n + \lambda a_1) = 0$$

rearranging the terms, we have:
$$ a_1(\beta_1 + \beta_n\lambda) + a_2(\beta_1 + \beta_2) + ... + a_n(\beta_{n-1}+\beta_n) =0 $$

but we know that $a_1,...,a_n$ are linearly independent, so the expressions inside the brackets are zeros.

I'm not sure how to proceed from that point, but I think I need to get a contradiction and conclude the set is linearly dependent for all $\lambda$'s.

  • $\begingroup$ Thank you all for the detailed answers! $\endgroup$ – Elimination Jun 28 '14 at 12:46

Probably easier to look for $\lambda$ which makes the set linearly dependant. Then you can correctly say that $ \beta_1(a_1 + a_2) + ... + \beta_{n-1}(a_{n-1} + a_n) + \beta_n(a_n + \lambda a_1) = 0$, where at least one of the $\beta_i \ne 0$.

WLG, let $\beta_1 = B$ (this can be arranged by re-labelling the $a_i$.)

Now re-arrange and $a_1(\beta_1 + \beta_n\lambda) + a_2(\beta_1 + \beta_2) + ... + a_n(\beta_{n-1}+\beta_n) =0$

Since the $a_i$ are linearly independent this requires every coefficient of the $a_i$ = 0, so that $\beta_2 = - B$, .... $\beta_n = (-1)^{n-1} B$. The coefficient of $a_1$ must also be zero so that $\beta_1 = B = -\lambda (-1)^{n-1} B$, and hence $\lambda = (-1)^{n}$.

So, the set is linearly independent for all other values of lambda, i.e. $\lambda \ne (-1)^{n}$.


You started well.

Essentially we have that $\beta_i=-\beta_{i+1}$ for all $i=1,\dots,n-1$, and $\beta_1=-\lambda \beta_n$. Combining inductively, $\beta_1=-\beta_2=\cdots =(-1)^{n-1}\beta_n=-\lambda\beta_n$. So $\beta_n((-1)^{n-1}+\lambda)=0$.

Now the "tricky" part is wrapping your head around the logic. The new set is linearly independent iff these equations imply $\beta_n=0$ which is true iff its coefficient is nonzero, i.e. $\lambda \neq (-1)^n$, so this is the required condition.

A quick sanity check: $a_1=(0,1)^T,\ a_2=(1,0)^T$ are linearly independent in $\mathbb R^2$. $a_1+a_2=(1,1)^T,\ a_1+\lambda a_2=(\lambda,1)^T$ will be linearly independent iff $\lambda\neq 1=(-1)^2=(-1)^n$


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