Qfyilyi

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 §=0§ this can only be the case if all ab-coordinates are 0, which is not the case bc 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 R.

No if a 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 would also be 0 would this be than count as orthogonal??

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

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

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

Many thanks for your help!!

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

This is because:

$1| cvec acdot dvec b=0implies cd=0 (vec acdotvec bne0)$

$2| cd=0implies cdcdot(vec acdotvec b)=0implies cvec acdot dvec b=0$

Your answers and reasoning are fine.

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 ($(ctextbf{a}) cdot (dtextbf{b}) = 0$). Using the properties of the dot product, because $ctextbf{a} cdot dtextbf{b} = cd(textbf{a} cdot textbf{b})$, 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.