# How did they find linear independence through the dot product?

A question from MIT problem set Are the following collections of vectors in $R^3$ linearly independent? Why or why not?

The vectors are:

$$\left\{\begin{bmatrix} 5 \\ 2 \\ 3 \end{bmatrix}, \begin{bmatrix} 3 \\ 2 \\ 5 \end{bmatrix} \right\}$$

http://web.mit.edu/18.06/www/Spring16/pset1_soln.pdf

$S$ is linearly independent. Indeed, suppose

$$\alpha_1\begin{bmatrix} 5 \\ 2 \\ 3 \end{bmatrix} +\alpha_2 \begin{bmatrix} 3 \\ 2 \\ 5 \end{bmatrix} = \vec{0}$$

By taking the dot product of this equation with $\vec{e_2}$ we see that

## $2\alpha_1 + 2\alpha_2 = 0 \implies \alpha_1 = −\alpha_2$.

Can someone explain how they get this last sentence about the dot product?

• $\vec e_2 = (0, 1, 0)$. So taking the dot product of each side gives $(5, 2, 3) \cdot (0, 1, 0) = 5\cdot 0 + 2\cdot 1 + 3\cdot 0 = 2$ for the first vector, which is where $2\alpha_1$ comes from. The other terms follow in this way. – Trevor Norton Jun 15 '17 at 20:41
• math.meta.stackexchange.com/questions/5020/… mathjax references. – Siong Thye Goh Jun 15 '17 at 20:48
• It’s equivalent to saying “looking at the second component of the vectors...” – amd Jun 15 '17 at 22:26

$$\alpha_1 \begin{bmatrix} 5 \\ 2 \\ 3\end{bmatrix} + \alpha_2 \begin{bmatrix} 3 \\ 2 \\ 5\end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0\end{bmatrix}$$

Multiply by $\begin{bmatrix} 0 & 1 & 0\end{bmatrix}$, which is equivalent to looking at the second row we have

$$2 \alpha_1 + 2 \alpha_2 = 0$$

Divided by $2$,

$$\alpha_1+ \alpha_2 = 0$$

• Thanks - so is this $e_2$ the typical notation for scalar vector with a value of 1 in the second row? – Haim Jun 16 '17 at 12:22
• yup, it is a common notation. – Siong Thye Goh Jun 16 '17 at 12:24

Rewriting your vector equation, we have that $$\begin{pmatrix} 5\alpha_1+3\alpha_2\\2\alpha_1+2\alpha_2\\3\alpha_1+5\alpha_2\end{pmatrix} = 0$$ Taking the scalar product with $e_2 = \begin{pmatrix}0\\1\\0\end{pmatrix}$, we are left with the equation $2\alpha_1+2\alpha_2 = 0$.

You need to solve the system of linear equations

$$\alpha_1\cdot (5,2,3)+\alpha_2\cdot (3,2,5)=(0,0,0)$$

what can be written as

$$5\alpha_1=3\alpha_2,\quad 2\alpha_1=2\alpha_2,\quad 3\alpha_1=5\alpha_2$$

But you can check that the system doesnt have solutions.