Linear independent vector sets

Question

Let $V$ be a vector $\mathbb{K}$-space and be $v_{1}, v_{2}, v_{3}, v_{4} \in V$. Demonstrate that the following conditions are equivalent:

1. $\left \{ v_{1}, v_{2}, v_{3}, v_{4} \right \}$ is linear independent

2. $\left \{ v_{1} + v_{4} , v_{2} + v_{4}, v_{3} + v_{4}, v_{4} \right \}$ is linear independent

I have done the following to demonstrate equivalence (double implication)

The set $\left \{ v_{1}, v_{2}, v_{3}, v_{4} \right \}$ is linearly independent, so the range of the matrix whose columns are the vectors of that set is equal number of columns. Therefore, the determinant of this matrix, which we will call $A$, cannot be null. In a mathematical way,

$$det(v_{1} | v_{2} | v_{3} | v_{4}) \neq 0$$

Because of the properties of the determinants,

$det( v_{1} + v_{4} | v_{2} + v_{4} | v_{3} + v_{4} | v_{4}) = det(v_{1} | v_{2} | v_{3} | v_{4}) + det(v_{4} | v_{4} | v_{4} | v_{4}) = det(v_{1} | v_{2} | v_{3} | v_{4}) \neq 0$

Then, the set $\left \{ v_{1} + v_{4} , v_{2} + v_{4}, v_{3} + v_{4}, v_{4} \right \}$ is linear independent.

The other implication would be analogous.

$\blacksquare$

Would the demonstration I put forward be correct?

Answer 1

Almost. However, the computation of $\det(v_1+v_4|v_2+v_4|v_3+v_4|v_4)$ is not correct (although your conclusion is correct: it is not $0$). Since $\det$ is multilinear, you should get $8(=2^3)$ determinants, of which the only one which is not $0$ is, in fat, $\det(v_1|v_2|v_3|v_4)$.

On the other hand, I think that it is more natural to solve this exercise using the definition of linear independence.

Answer 2

We can’t use determinant in general, as an alternative we have that

$$c_1(v_1+v_4)+c_2(v_2+v_4)+c_3(v_3+v_4)+c_4v_4=0$$

$$c_1v_1+c_2v_2+c_3v_3+(c_1+c_2+c_3+c_4)v_4=0$$

and then obtain $c_i=0$.