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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!

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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.

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  • $\begingroup$ Thank you. How to solve this equation without resorting to matrix? $\endgroup$ – Maria Lavrovskaya May 16 at 7:07
  • $\begingroup$ You will get a system of three equations: $c_1+3c_2+c_3=0$, $2c_1-4c_2-8c_3=0$, $-c_1+5c_2+7c_3=0$. You can solve this system by solving for one variable at a time in one equation, then plugging into the next equation - no matrices needed. You should eventually get a numerical value for one of your constants which will then give you the value for the others. $\endgroup$ – csch2 May 16 at 7:10
  • $\begingroup$ Thank you, got it! $\endgroup$ – Maria Lavrovskaya May 16 at 7:14
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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.

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  • $\begingroup$ Would you mind writing it more precisely in terms of the example? I really feel so confused. $\endgroup$ – Maria Lavrovskaya May 16 at 6:42
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    $\begingroup$ This method assumes that it is not the case that $b$ and $c$ are multiples of each other, but $a$ is not (geometrically, $b$, $c$, and the origin are collinear, but $a$ does not line the line). In such a case, there will be no solution to the above equation, despite the fact that $a, b, c$ are linearly dependent. $\endgroup$ – Theo Bendit May 16 at 6:52
  • $\begingroup$ Good point, Theo. Maria, I believe the other answers have now spelled it out more clearly, but I didn't want to spoil the fun entirely! $\endgroup$ – bounceback May 16 at 16:55
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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 $0$.

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  • $\begingroup$ Thank you a lot for your answer. However, I am not familiar with the dependent in the context of matrices - it's going to come next in my university course, thereby I have to solve it the other way :( $\endgroup$ – Maria Lavrovskaya May 16 at 6:43
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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.

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