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!!

  • 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

$c\vec a$ 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.


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.


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