Given two vectors $a$, $b$ with only strictly positive coordinates, can those two vectors be orthogonal?
$begingroup$
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 $langle a,brangle=0$ 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 $0$, $langle ca,dbrangle$ 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!!
linear-algebra vectors
edited Dec 11 '18 at 13:30
Asaf Karagila♦
303k32429761
asked Dec 11 '18 at 9:18
LillysLillys
778
$endgroup$
add a comment |
$begingroup$
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 $langle a,brangle=0$ 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 $0$, $langle ca,dbrangle$ 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!!
linear-algebra vectors
edited Dec 11 '18 at 13:30
Asaf Karagila♦
303k32429761
asked Dec 11 '18 at 9:18
LillysLillys
778
$endgroup$
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
add a comment |
$begingroup$
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 $langle a,brangle=0$ 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 $0$, $langle ca,dbrangle$ 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!!
linear-algebra vectors
edited Dec 11 '18 at 13:30
Asaf Karagila♦
303k32429761
asked Dec 11 '18 at 9:18
LillysLillys
778
$endgroup$
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 $langle a,brangle=0$ 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 $0$, $langle ca,dbrangle$ 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!!
linear-algebra vectors
linear-algebra vectors
edited Dec 11 '18 at 13:30
Asaf Karagila♦
303k32429761
asked Dec 11 '18 at 9:18
LillysLillys
778
edited Dec 11 '18 at 13:30
Asaf Karagila♦
303k32429761
asked Dec 11 '18 at 9:18
LillysLillys
778
edited Dec 11 '18 at 13:30
Asaf Karagila♦
303k32429761
edited Dec 11 '18 at 13:30
Asaf Karagila♦
303k32429761
edited Dec 11 '18 at 13:30
Asaf Karagila♦
303k32429761
303k32429761
asked Dec 11 '18 at 9:18
LillysLillys
778
asked Dec 11 '18 at 9:18
LillysLillys
778
asked Dec 11 '18 at 9:18
LillysLillys
778
778
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
add a comment |
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
1
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
$begingroup$
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
$begingroup$
@GregMartin They’re orthogonal if $c=0$ or $d=0$.
$endgroup$
– amd
Dec 11 '18 at 18:53
add a comment |
2 Answers
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$begingroup$
$cvec a$ 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.
answered Dec 11 '18 at 9:28
Shubham JohriShubham Johri
5,122717
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add a comment |
$begingroup$
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 ($(ctextbf{a}) cdot (dtextbf{b}) = 0$). 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.
answered Dec 11 '18 at 9:37
Daniel Tartaglione minioniceDaniel Tartaglione minionice
111
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add a comment |
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2 Answers
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active
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2 Answers
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$begingroup$
$cvec a$ 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.
answered Dec 11 '18 at 9:28
Shubham JohriShubham Johri
5,122717
$endgroup$
add a comment |
$begingroup$
$cvec a$ 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.
answered Dec 11 '18 at 9:28
Shubham JohriShubham Johri
5,122717
$endgroup$
add a comment |
$begingroup$
$cvec a$ 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.
answered Dec 11 '18 at 9:28
Shubham JohriShubham Johri
5,122717
$endgroup$
$cvec a$ 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.
answered Dec 11 '18 at 9:28
Shubham JohriShubham Johri
5,122717
edited Dec 11 '18 at 9:34
answered Dec 11 '18 at 9:28
Shubham JohriShubham Johri
5,122717
answered Dec 11 '18 at 9:28
Shubham JohriShubham Johri
5,122717
answered Dec 11 '18 at 9:28
Shubham JohriShubham Johri
5,122717
5,122717
add a comment |
add a comment |
$begingroup$
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 ($(ctextbf{a}) cdot (dtextbf{b}) = 0$). 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.
answered Dec 11 '18 at 9:37
Daniel Tartaglione minioniceDaniel Tartaglione minionice
111
$endgroup$
add a comment |
$begingroup$
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 ($(ctextbf{a}) cdot (dtextbf{b}) = 0$). 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.
answered Dec 11 '18 at 9:37
Daniel Tartaglione minioniceDaniel Tartaglione minionice
111
$endgroup$
add a comment |
$begingroup$
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 ($(ctextbf{a}) cdot (dtextbf{b}) = 0$). 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.
answered Dec 11 '18 at 9:37
Daniel Tartaglione minioniceDaniel Tartaglione minionice
111
$endgroup$
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.
answered Dec 11 '18 at 9:37
Daniel Tartaglione minioniceDaniel Tartaglione minionice
111
answered Dec 11 '18 at 9:37
Daniel Tartaglione minioniceDaniel Tartaglione minionice
111
answered Dec 11 '18 at 9:37
Daniel Tartaglione minioniceDaniel Tartaglione minionice
111
answered Dec 11 '18 at 9:37
Daniel Tartaglione minioniceDaniel Tartaglione minionice
111
111
add a comment |
add a comment |
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$begingroup$
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