# Linear independence of the sets

My problem: Prove that if $\{u_1, u_2, ... u_n\}$ is the linearly independent set in the vector space $V$ over $\Bbb F$ and $r_{ij} \in \Bbb F$, $1\le i \le k$, $\;$$1\le j \le n, then the vectors s_i = (r_{i1}, r_{i2 }, ..., r_{in}) create the linearly independent set \{s_1,s_2,...,s_k\}\subset \Bbb F^n if and only if is linearly idendependent the set \{v_1, v_2, ..., v_k\} \subset V, where v_i =$$\sum_{j=0}^nr_{ij}u_j$.

My attempt: Because $\{v_1, v_2, ..., v_k\}$ is linearly independent set, for

$$\sum_{i=1}^ks _{i}v_i = 0$$ we need $s_1=s_2=...=s_k=0$

and we can rewrite it as:

$$\sum_{i=1}^k\sum_{j=1}^n(s _{i}r_{ij})u_j = 0$$

and $\{u_1, u_2, ... u_n\}$ is also a linearly independent set.

I don't know what next or if I am not totally wrong, because I can't orient myself in those indexes, scalars, vectors, so please help me.

• Firstly do the right implication, writing properly your hypothesis, and trying to see if you can reach the result. Then try with the other implication. I can't write an answer soon, I hope someone helps, otherwise tonight I'll give it a shot. Also note that the start index of the sums is $1$ and not $0$ – Giulio Feb 9 '17 at 17:46
• Ok, I will try. My problem is I actually don't understand it well. – Leif Feb 9 '17 at 17:55

## 1 Answer

For a better idea what happens, you can imagine the set $\{r_{i,j}\}_{i,j}$ as a $k\times n$ matrix, which is from the right hand side multiplied by the "vector" $\left(u_1,\ldots, u_n\right)^T$ and rows of the result are the respective vectors $v_i$ (even though it is not formally correct).

From right to left: suppose $\{v_1, \ldots v_k\}$ are linearly independent and let $\sum_i \alpha_i s_i = \left(0,\ldots 0\right)$ for some coefficients $\alpha_i$; looking at a fixed coordinate $j_0$ of the sum, you get $\sum_i \alpha_i r_{i, j_0} = 0$; hence $$0 = \sum_j\left(\sum_i \alpha_i r_{i,j}u_j\right).$$ Since those sums are finite, you can switch the order of summation and get $$0=\sum_i\sum_j \alpha_i r_{i,j}u_j = \sum_i\alpha_i \left(\sum_j r_{i,j}u_j\right) = \sum_i\alpha_i v_i$$

and from the assumption on the set $\{v_1, \ldots v_k\}$ you get $\alpha_i=0$, i.e. the set $\{s_1, \ldots s_k\}$ is independent.

The other half is just running through this process backwards.

• Ok, and děkuji. – Leif Feb 10 '17 at 18:39