linear dependency in a system

Question

Given a vector system $S=\{v_1,v_2,...,v_k\}$, how can I prove that $S$ is dependent iff $S'=\{v_1,...,v_i,...,v_j + mv_i,...,v_k\}$ is also dependent (with arbitrary $i,j \in\{1,...,k\}$ and $m$ a coefficient)?

Same question for $S''=\{v_1,...,v_{i-1},v_j,v_{i+1},...,v_{j-1},v_i,v_{j+1},...,v_k\}$ and $S'''=\{v_1,...,v_{i-1},mv_i,v_{i+1},...,v_k\}$.

All I know is the linear relationship between vectors but I'm clueless on how to apply it in such a situation. I would highly appreciate a detailed answer.

PS: Do all of these systems have the same rank?

Answer 1

I use $S\sim S'$ to denote $S$ is linearly independent iff $S'$ is linear independent.

1. $S \sim S'$
   • $(\Rightarrow)$: Suppose $S$ is linearly independent. $$a_1 v_1 + a_2 v_2 + \cdots + a_{j-1} v_{j-1} + a_j (v_j + mv_i) + a_{j+1} v_{j+1} + \cdots a_k v_k = 0.$$ Then add and subtract $ma_jv_i$ on the LHS. $$a_1 v_1 + a_2 v_2 + \cdots + a_{i-1} v_{i-1} + (a_i+ma_j) v_i + a_{i+1} v_{i+1} + \cdots + a_{j-1} v_{j-1} + a_j v_j + a_{j+1} v_{j+1} + \cdots a_k v_k = 0.$$ Since $S$ is linearly independent, we have $a_n = 0$ for $k = 1,\dots,i-1,i+1,k$ and $a_i+ma_j = a_i+m(0)=0$, so $a_n=0$ for all $n = 1,\dots,k$.
   • $(\Leftarrow)$: Suppose $S'$ is linearly independent. $\sum_n a_n v_n = 0$. Then add and subtract $ma_jv_i$ on the LHS. $$a_1 v_1 + a_2 v_2 + \cdots + a_{i-1} v_{i-1} + (a_i - ma_j) v_i + a_{i+1} v_{i+1} + \cdots + a_{j-1} v_{j-1} + a_j (v_j + m v_i) + a_{j+1} v_{j+1} + \cdots a_k v_k = 0.$$ Since $S'$ is linearly independent, we have $a_n = 0$ for $k = 1,\dots,i-1,i+1,k$ and $a_i - ma_j = a_i - m(0)=0$, so $a_n=0$ for all $n = 1,\dots,k$.
2. $S \sim S''$
   • $(\Rightarrow)$: Suppose $S$ is linearly independent. To prove linear independence of $S''$, write $$a_1 v_1 + a_2 v_2 + \cdots + a_{i-1} v_{i-1} + a_i v_j + a_{i+1} v_{i+1} + \cdots + a_{j-1} v_{j-1} + a_j v_i + a_{j+1} v_{j+1} + \cdots a_k v_k = 0.$$ Swap the terms $a_i v_j$ and $a_j v_i$ to get $$a_1 v_1 + a_2 v_2 + \cdots + a_{i-1} v_{i-1} + a_j v_i + a_{i+1} v_{i+1} + \cdots + a_{j-1} v_{j-1} + a_i v_j + a_{j+1} v_{j+1} + \cdots a_k v_k = 0.$$ Use the linear independence of $S$ to finish this part.
   • $(\Leftarrow)$: Suppose $S''$ is linearly independent. $\sum_n a_n v_n = 0$. To prove linear i'ndependence of $S$, swap the terms $a_i v_j$ and $a_j v_i$ to get $$a_1 v_1 + a_2 v_2 + \cdots + a_{i-1} v_{i-1} + a_i v_j + a_{i+1} v_{i+1} + \cdots + a_{j-1} v_{j-1} + a_j v_i + a_{j+1} v_{j+1} + \cdots a_k v_k = 0.$$ Use the linear independence of $S''$ to finish this part.
3. $S \sim S'''$
   • $(\Rightarrow)$: Suppose $S$ is linearly independent. To prove linear independence of $S'''$, write $$a_1 v_1 + a_2 v_2 + \cdots + a_{i-1} v_{i-1} + a_i (mv_i) + a_{i+1} v_{i+1} + \cdots a_k v_k = 0.$$ To use linear independence of $S$, we need $m\ne0$ so that the term $a_imv_i$ won't vanish. We have $$a_1 v_1 + a_2 v_2 + \cdots + a_{i-1} v_{i-1} + (a_im) v_i + a_{i+1} v_{i+1} + \cdots a_k v_k = 0,$$ so $S'''$ is linearly independent.
   • $(\Leftarrow)$: Reverse the above process. Suppose $S'''$ is linearly independent. To prove linear independence of $S$, write $$ \sum_n a_n v_n = 0.$$ To use linear independence of $S'''$, we need $m\ne0$ so that the coefficient $a_i/m$ is defined. We have $$a_1 v_1 + a_2 v_2 + \cdots + a_{i-1} v_{i-1} + \frac{a_i}{m} mv_i + a_{i+1} v_{i+1} + \cdots a_k v_k = 0,$$ so $S$ is linearly independent.