3
$\begingroup$

I have a couple of questions to answer, and I am unsure if i argue correctly:

Given two vectors $a$, $b$ with only strictly positive coordinates, can those two vectors be orthogonal? My answer would be no. As $\langle a,b\rangle=0$ this can only be the case if all $ab$-coordinates are $0$, which is not the case because the coordinates have to be strictly positive, so in ordered to get to $0$ some ab products have to be negative.

Can $ca$ and $db$ be orthogonal, for $c$,$d$ elements of $\mathbb R$?

No if $a$ and $b$ are not orthogonal? I'm not sure if this question refers to the question above....my second question would be if $c$ and $d$ are $0$, $\langle ca,db\rangle$ would also be $0$ would this be than count as orthogonal??

How many vectors build a orthonormal basis in $\mathbb R^n$ - n

How many of these vectors (of the basis above) can have strictly positive Coordinates, how many strictly negative?

I would guess only $1$, because if all vectors are orthogonal than there can be only one with strictly positive coordinates..

Many thanks for your help!!

$\endgroup$
  • 1
    $\begingroup$ In your second question, yes, if $c=d=0$ then $ca$ and $bd$ are orthogonal; indeed, the zero vector is always orthogonal to everything. There are other possibilities though; what if $c=-d$ for example? $\endgroup$ – Greg Martin Dec 11 '18 at 9:22
  • $\begingroup$ @GregMartin They’re orthogonal if $c=0$ or $d=0$. $\endgroup$ – amd Dec 11 '18 at 18:53
3
$\begingroup$

$c\vec a$ and $d\vec b$ are orthogonal for non-orthogonal vectors $\vec a,\vec b$ iff $c=0$ or $d=0$.

This is because:

$1|\ \ \ \ c\vec a\cdot d\vec b=0\implies cd=0\ (\vec a\cdot\vec b\ne0)$

$2|\ \ \ \ cd=0\implies cd\cdot(\vec a\cdot\vec b)=0\implies c\vec a\cdot d\vec b=0$

Your answers and reasoning are fine.

$\endgroup$
1
$\begingroup$

If the coordinates are strictly positive, they cannot be $0$! Therefore, the dot product between the two vectors is always a sum of positive products, so it is never $0$.

The second question is asking whether, if you multiply each vector by a real number, it is possible that the vectors are orthogonal ($(c\textbf{a}) \cdot (d\textbf{b}) = 0$). Using the properties of the dot product, because $c\textbf{a} \cdot d\textbf{b} = cd(\textbf{a} \cdot \textbf{b})$, and because $\textbf{a} \cdot \textbf{b} \neq 0$, the answer is no unless one of $c$ and $d$ is $0$.

The rest of your analysis makes sense and can be verified using similar properties.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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