Linear independence without given coordinates

Question

I have a problem where I'm given 3 vectors with the same length and 60 degree angles between any 2 of them. And one of the aims of the problem is for me to prove that those vectors are linearly independent. But how is that done without knowing the coordinates?

Most Google and SO searches yield how to calculate normal linear dependence, but don't touch up much on how to prove that without coordinates.

Answer 1

Suppose they are linearly dependant, then they are all in the same plane. If you have two vectors $u$ and $v$ in the plane, and they form the same angle with vector $x$ then $u=\alpha v$ for some positive $\alpha$, but then the angle between $u$ and $v$ is zero, a contradiction.

Answer 2

Lets call your vectors $\mathbf a, \mathbf b, \mathbf c$

It will make things easier to make them all unit vectors. Unitizing $\mathbf a, \mathbf b, \mathbf c$ will not change their independence (dependency)

If they are linearly dependent then $n \mathbf a + m\mathbf b = \mathbf c$ or $\mathbf a = \pm \mathbf b$

if $\mathbf a = \pm \mathbf b$ then $\mathbf a \cdot \mathbf b = \pm 1$ but $\mathbf a \cdot \mathbf b = \frac 12$

If $n \mathbf a + m\mathbf b = \mathbf c$, then $(n \mathbf a + m\mathbf b)\cdot\mathbf c = \pm \|(n \mathbf a + m\mathbf b)\|$

$(n \mathbf a + m\mathbf b)\cdot\mathbf c = (n \mathbf a\cdot\mathbf c + m\mathbf b\cdot \mathbf c) = \frac 12 (n + m)$

$\|(n \mathbf a + m\mathbf b)\| = \sqrt {n^2\ + m^2 + nm}\\ \sqrt {n^2\ + m^2 + nm} > \frac 12 (n + m)$

that is, squaring both sides,

$n^2\ + m^2 + nm = \frac 34 (n+m)^2 + \frac 14 (n-m)^2 > \frac 14 (n+m)^2$

Answer 3

Suppose $v_1, v_2, v_3$ are such vectors (e.g., every pair is separated by an angle of 60 degrees) in 3D, all of unit magnitude. Their "heads" are three points on the unit sphere. We just need to show that these points are not coplanar. Well, every pair of vectors spans a 2-D subspace (a plane) that does not contain the 3rd (this can be shown geometrically). Hence the three vectors span the 3-D space, and are therefore linearly independent.

No coordinates.:)

Answer 4

Let $\mathbf a$, $\mathbf b$, $\mathbf c$ three unit vectors such that $$\mathbf a \cdot \mathbf b =\mathbf b \cdot \mathbf c =\mathbf c \cdot \mathbf a = t\in(-1,1)\setminus \{-1/2\}$$ which means that every pair is separated by the same angle in $(0^{\circ},180^{\circ})$ and different from $120^{\circ}$ (for $120^{\circ}$ they can be planar and linear dipendent).

Assume that $A\mathbf a+B\mathbf b+C\mathbf c=\mathbf 0$ then by taking the scalar product by $\mathbf a$, $\mathbf b$ and $\mathbf c$, we obtain three linear equations $$A+Bt+Ct=0,\ tA+B+Ct=0,\ tA+tB+C=0.$$ This linear system has a unique solution $A=B=C=0$, because the determinant of the associated matrix is $1+2t^3-3t^2=(t-1)^2(2t+1)\not=0$ for $t\in(-1,1)\setminus \{-1/2\}$.

Hence the three vectors $\mathbf a$, $\mathbf b$, $\mathbf c$ are linearly independent. This approach is independent from the dimension of the space where the vectors are.

Answer 5

Recall that the inner product of $u,v\in\Bbb R^n$ satisfies $$ \langle u,v\rangle =\left\lVert u\right\rVert \left\lVert v\right\rVert\cos\theta_{uv} $$ where $\theta_{uv}$ is the angle between $u$ and $v$.

Now, we are supplied with three vectors $u,v,w\in\Bbb R^n$ such that \begin{align*} \lVert u\rVert=\lVert v\rVert&=\lVert w\rVert & \cos\theta_{uv}=\cos\theta_{uw}=\cos\theta_{vw} =\cos60^{\circ}&=\frac{1}{2} \end{align*} So, to prove that $\{u,v,w\}$ are linearly independent, suppose that $$ a\cdot u+b\cdot v+c\cdot w=0\tag{1} $$ Applying $\langle u-\rangle$ to (1) gives $$ a\cdot\langle u,u\rangle+b\cdot\langle u,v\rangle+c\cdot\langle u,w\rangle=\langle u,0\rangle $$ which is equivalent to $$ a\cdot\lVert u\rVert^2+b\cdot\lVert u\rVert\lVert v\rVert\cos\theta_{uv}+c\cdot\lVert u\rVert\lVert w\rVert\cos\theta_{uw}=0 $$ which is equivalent to $$ a\cdot \lVert u\rVert^2+\frac{1}{2}\,b\cdot\lVert u\rVert^2+\frac{1}{2}\,c\cdot\lVert u\rVert^2=0 $$ which, after dividing through by $\lVert u\rVert^2$ and multiplying through by $2$, is equivalent to $$ 2\,a+b+c=0\tag{2} $$ Similarly, applying $\langle v,-\rangle$ to (1) gives the equation $$ a+2\,b+c=0\tag{3} $$ and applying $\langle w,-\rangle$ to (1) gives $$ a+b+2\,c=0\tag{4} $$ Putting (2), (3), and (4) together gives the system $$ \begin{array}{rcrcrcrc} 2\,a &+& b &+& c &=& 0 \\ a &+& 2\,b &+& c &=& 0 \\ a &+& b &+& 2\,c &=& 0 \end{array} $$ which is equivalent to $A\vec x=0$ where \begin{align*} A &= \left[\begin{array}{rrr} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{array}\right] & \vec x = \left[\begin{array}{r} a \\ b \\ c \end{array}\right] \end{align*} Since $\det(A)=4$, the only solution to $A\vec x=0$ is $\vec x=0$. This proves that the only solution to (1) is $a=b=c=0$. Hence $\{u,v,w\}$ is linearly independent.