How to proof the linear dependency / independency ONLY using vectors (not through matrixes), as I am not familiar with this concept for now.
The example is the following: Are the following vectors linearly independent?
$a=[1,2,-1], b=[3,-4,5]$ and $c=[1,-8,7]$.
Thanks in advance!
On
Put them into a matrix.
$\begin{bmatrix}1 & 2 & -1\\3 & -4 & 5\\1 & -8 &7\end{bmatrix}$
They are linearly dependent as the determinant comes out to be $0$.
On
Recall that a set of vectors $\{v_1,v_2,\ldots,v_n\}$ is said to be linearly depending if there exists constants $c_1,\ldots,c_n$ where at least one of the constants is nonzero such that $c_1v_1+c_2v_2+\dots+c_nv_n=0$; otherwise, the vectors are linearly independent. So in your case, you have: $$c_1(1,2,-1)+c_2(3,-4,5)+c_3(1,-8,7)=0$$ If you can now solve for $c_1,c_2,c_3$ and at least one of them is nonzero, then your vectors are linearly dependent. Otherwise, they are linearly independent.
On
Suppose that $V$ is a vector space and that $x_1, x_2, . . . , x_k$ are vectors in $V$ . Then the set of vectors $\{x_1, x_2, . . . , x_k\}$ is linearly dependent if $ r_1 x_1 + r_2 x_2 + · · · + r_k x_k = 0$ for some $r_1, r_2, . . . , r_k ∈ \mathbb{R}$ where at least one of $r_1, r_2, . . . , r_k$ is non–zero.
Now $ r_1 (1,2,-1) + r_2 (3,-4,5) + r_3 (1,-8,7) = 0$ for some $r_1, r_2, r_3 ∈ \mathbb{R}$ gives
$r_1+3r_2+r_3=0$ . . . $(1)$
$2r_1-4r_2-8r_3=0$ . . . $(2)$
$-r_1+5r_2+7r_3=0$ . . . $(3)$
By cross multiplication from $(1)$ and $(2)$ we have
$\frac{r_1}{-20}=\frac{r_2}{10}=\frac{r_3}{-10}=k$(say)
$\implies r_1 = -2k, r_2 = k, r_3 = -k$
Putting these values in the LHS of $(3)$ we have, $2k+5k-7k=0\implies 0 =0$
Hence $ r_1 , r_2 , r_3 $ may or may not be zero as $k$ takes any value.
Therefore the given $3$ vectors are linearly dependent.
Note that $$2 (1,2,-1) + (1,-8,7) = (3,-4,5) $$so $$2 (1,2,-1) -(3,-4,5) + (1,-8,7) =0$$ Here $r_1 = 2, r_2 = -1, r_3 = 1$
From this you can also conclude that the given $3$ vectors are linearly dependent.
HINT: Suppose $a = \lambda b + \mu c$ and this should give you three equations in the two unknowns. If you can solve for $\lambda$ and $\mu$, there's your dependency. If not, conclude that the three vectors are linearly independent.