Proof linear dependency for vectors in $3$-D space

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 $$c=[1,-8,7]$$.

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.

• Thank you. How to solve this equation without resorting to matrix? – Maria Lavrovskaya May 16 at 7:07
• 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. – csch2 May 16 at 7:10
• Thank you, got it! – Maria Lavrovskaya May 16 at 7:14

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.

• Would you mind writing it more precisely in terms of the example? I really feel so confused. – Maria Lavrovskaya May 16 at 6:42
• 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. – Theo Bendit May 16 at 6:52
• 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! – bounceback May 16 at 16:55

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 $$0$$.

• 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 :( – Maria Lavrovskaya May 16 at 6:43

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.