0
$\begingroup$

Can two perpendicular vectors to each other be linearly dependent and can two parallel vectors to each other be linearly independent ?

$\endgroup$

closed as off-topic by Namaste, user99914, egreg, Mohammad Riazi-Kermani, Brian Borchers Mar 24 '18 at 23:20

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Community, egreg, Mohammad Riazi-Kermani, Brian Borchers
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 3
    $\begingroup$ 1) Yes, if one of them is zero. 2) No. $\endgroup$ – Qiaochu Yuan Jan 16 '18 at 20:29
  • $\begingroup$ So if non of two perpendicular vectors is 0 then they cant be linearly dependent right ? $\endgroup$ – Endrit Shabani Jan 16 '18 at 20:32
  • $\begingroup$ ^ Yes that's right $\endgroup$ – John Doe Jan 16 '18 at 20:32
  • 1
    $\begingroup$ @QiaochuYuan Slightly strengthening 1): if and only if. $\endgroup$ – Thomas Jan 16 '18 at 20:32
  • $\begingroup$ but isn't the 0 vector perpendicular to any vector ? $\endgroup$ – Endrit Shabani Jan 16 '18 at 20:33
2
$\begingroup$

Every set which contains mutually perpendicular vectors is a independent set. All the vectors in this set are independent. You can search for Gram-Schmidt process. In that process it makes an orthonormal basis for $\mathbb R^n$. There you can easily see why a set of mutually perpendicular vectors are independent.

If two vectors are parallel then each of them is a non-zero scalar multiple of other one unless one of them is zero. In both situation they form a dependent set

$\endgroup$
0
$\begingroup$

Note that for $v_1\neq 0$ and $v_2\neq 0$ and $v_1\cdot v_2=0$

$$av_1+bv_2=0 \iff av_1\cdot v_1+bv_1\cdot v_2=0\iff a|v_1|^2=0\iff a=0$$

$$av_1+bv_2=0 \iff av_1\cdot v_2+bv_2\cdot v_2=0\iff b|v_2|^2=0\iff b=0$$

thus $v_1$ and $v_2$ are linearly independent.

$\endgroup$

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