Vector, orthogonality
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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!!
linear-algebra vectors
asked 3 hours ago
Lillys
667
add a comment |
up vote
3
down vote
favorite
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!!
linear-algebra vectors
asked 3 hours ago
Lillys
667
1
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?
– Greg Martin
3 hours ago
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
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!!
linear-algebra vectors
asked 3 hours ago
Lillys
667
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!!
linear-algebra vectors
linear-algebra vectors
asked 3 hours ago
Lillys
667
asked 3 hours ago
Lillys
667
asked 3 hours ago
Lillys
667
asked 3 hours ago
Lillys
667
asked 3 hours ago
Lillys
667
667
1
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?
– Greg Martin
3 hours ago
add a comment |
1
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?
– Greg Martin
3 hours ago
1
1
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?
– Greg Martin
3 hours ago
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?
– Greg Martin
3 hours ago
add a comment |
2 Answers
2
active
oldest
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up vote
3
down vote
accepted
$cvec a$ 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 3 hours ago
Shubham Johri
1,856412
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up vote
1
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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 3 hours ago
Daniel Tartaglione minionice
111
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Daniel Tartaglione minionice is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
$cvec a$ 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 3 hours ago
Shubham Johri
1,856412
add a comment |
up vote
3
down vote
accepted
$cvec a$ 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 3 hours ago
Shubham Johri
1,856412
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
$cvec a$ 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 3 hours ago
Shubham Johri
1,856412
$cvec a$ 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 3 hours ago
Shubham Johri
1,856412
edited 3 hours ago
answered 3 hours ago
Shubham Johri
1,856412
answered 3 hours ago
Shubham Johri
1,856412
answered 3 hours ago
Shubham Johri
1,856412
1,856412
add a comment |
add a comment |
up vote
1
down vote
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 3 hours ago
Daniel Tartaglione minionice
111
New contributor
Daniel Tartaglione minionice is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
1
down vote
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 3 hours ago
Daniel Tartaglione minionice
111
New contributor
Daniel Tartaglione minionice is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
1
down vote
up vote
1
down vote
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 3 hours ago
Daniel Tartaglione minionice
111
New contributor
Daniel Tartaglione minionice is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 3 hours ago
Daniel Tartaglione minionice
111
New contributor
Daniel Tartaglione minionice is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 3 hours ago
Daniel Tartaglione minionice
111
New contributor
Daniel Tartaglione minionice is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 3 hours ago
Daniel Tartaglione minionice
111
answered 3 hours ago
Daniel Tartaglione minionice
111
111
New contributor
Daniel Tartaglione minionice is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Daniel Tartaglione minionice is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Daniel Tartaglione minionice is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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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?
– Greg Martin
3 hours ago