Problem: Let $\{ v_1, \ldots, v_n \}$ be a linearly independent subset of $V$, a vector space. let $$ v= t_1 v_1 + \cdots + t_n v_n $$ where $t_1, \ldots t_n \in \mathbb{R}$. For which $v$ is the set $\{v_1 + v ,\ldots , v_n + v \}$ linearly independent?

My ideas were:

Firstly, any linear combination of the new set is, $$ \sum \lambda_i (v_i + v) = \sum \lambda_i v_i + sv = \sum (\lambda_i + st_i) v_i = 0 $$ where $s = \sum \lambda_i$. By linear independence of $v_i$, we must have $\lambda _i + st_i = 0$ for $i=1, \ldots , n$. Summing up yields $ s + s \sum t_i = 0 \Rightarrow s( 1 + \sum t_i) = 0 $.

My claim is: $\sum t_i \not = -1 \Leftrightarrow$ set is linearly independent.

Proof: If $\sum t_i \not = -1$, then $s = 0$, hence $\lambda _i = 0$ for $i=1, \ldots, n$. So the new set is linearly independent.

On the other hand if $\sum t_i = -1$, then regarding the original system of linear equations, where $$\lambda_i + st_i = \sum_{i \not=j} \lambda_j t_i + \lambda_i (1+t_i) = 0$$ for $i-1, \ldots, n$. We wish to find a non trivial set of solutions for $i=1, \ldots, n$. Written out in matrix form, we have a matrix with all column sums equal to $0$ (column $i$ has sum $\sum t_i +1 = 0$), hence the transpose of this matrix, has a non trivial solution, eigenvector $(1, \ldots, 1 )^T$.

Therefore, the transpose matrix is not invertible - hence, the matrix itself is not invertible. So there exists a non trivial solution.

I think this solution is a bit too long winded - I am hoping if anyone could give a more elmentary solution that does not use matrices. Thank you!


It is easy to explicitly guess a nontrivial solution for the system of equations $$\lambda_i + st_i = 0$$ By setting $\lambda_i = t_i$ for all $i$, we get $\lambda_i + st_i = t_i-t_i = 0$ for all $i$ since $s=-1$ by assumption. Also since $\sum t_i = -1$, atleast one $t_i$ is nonzero, so the solution is nontrivial.


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