0
$\begingroup$

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

The answer says:

$S$ 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?

$\endgroup$
  • $\begingroup$ $\vec e_2 = (0, 1, 0)$. 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. $\endgroup$ – Trevor Norton Jun 15 '17 at 20:41
  • $\begingroup$ math.meta.stackexchange.com/questions/5020/… mathjax references. $\endgroup$ – Siong Thye Goh Jun 15 '17 at 20:48
  • $\begingroup$ It’s equivalent to saying “looking at the second component of the vectors...” $\endgroup$ – amd Jun 15 '17 at 22:26
1
$\begingroup$

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

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 $2$,

$$\alpha_1+ \alpha_2 = 0$$

$\endgroup$
  • $\begingroup$ Thanks - so is this $e_2$ the typical notation for scalar vector with a value of 1 in the second row? $\endgroup$ – Haim Jun 16 '17 at 12:22
  • $\begingroup$ yup, it is a common notation. $\endgroup$ – Siong Thye Goh Jun 16 '17 at 12:24
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.