A set of vectors in $R^2$ for which $x + y$ stays in the set but $frac12x$ may be outside.
$begingroup$
These questions refer to subspace.
One requirement can be met while the other fails. Show this by finding
(a) A set of vectors in $R^2$ for which $x + y$ stays in the set but $frac12x$ may be outside.
(b) A set of vectors in $R^2$ (other than two quarter-planes) for which every $cx$ stays in the set but $x + y$ may be outside.
these are the answers my textbook gives,
(a) The vectors with integer components allow addition, but not multiplication by $frac12$
This is saying that you can't multiply a vector by a non-integer? why not?
(b) Remove the x axis from the xy plane (but leave the origin). Multiplication by any c is allowed but not all vector additions.
If you remove the x-axis then you are only left with vectors on the y-axis, so shouldn't it still be closed under addition?
linear-algebra
$endgroup$
add a comment |
$begingroup$
These questions refer to subspace.
One requirement can be met while the other fails. Show this by finding
(a) A set of vectors in $R^2$ for which $x + y$ stays in the set but $frac12x$ may be outside.
(b) A set of vectors in $R^2$ (other than two quarter-planes) for which every $cx$ stays in the set but $x + y$ may be outside.
these are the answers my textbook gives,
(a) The vectors with integer components allow addition, but not multiplication by $frac12$
This is saying that you can't multiply a vector by a non-integer? why not?
(b) Remove the x axis from the xy plane (but leave the origin). Multiplication by any c is allowed but not all vector additions.
If you remove the x-axis then you are only left with vectors on the y-axis, so shouldn't it still be closed under addition?
linear-algebra
$endgroup$
add a comment |
$begingroup$
These questions refer to subspace.
One requirement can be met while the other fails. Show this by finding
(a) A set of vectors in $R^2$ for which $x + y$ stays in the set but $frac12x$ may be outside.
(b) A set of vectors in $R^2$ (other than two quarter-planes) for which every $cx$ stays in the set but $x + y$ may be outside.
these are the answers my textbook gives,
(a) The vectors with integer components allow addition, but not multiplication by $frac12$
This is saying that you can't multiply a vector by a non-integer? why not?
(b) Remove the x axis from the xy plane (but leave the origin). Multiplication by any c is allowed but not all vector additions.
If you remove the x-axis then you are only left with vectors on the y-axis, so shouldn't it still be closed under addition?
linear-algebra
$endgroup$
These questions refer to subspace.
One requirement can be met while the other fails. Show this by finding
(a) A set of vectors in $R^2$ for which $x + y$ stays in the set but $frac12x$ may be outside.
(b) A set of vectors in $R^2$ (other than two quarter-planes) for which every $cx$ stays in the set but $x + y$ may be outside.
these are the answers my textbook gives,
(a) The vectors with integer components allow addition, but not multiplication by $frac12$
This is saying that you can't multiply a vector by a non-integer? why not?
(b) Remove the x axis from the xy plane (but leave the origin). Multiplication by any c is allowed but not all vector additions.
If you remove the x-axis then you are only left with vectors on the y-axis, so shouldn't it still be closed under addition?
linear-algebra
linear-algebra
asked Feb 26 '16 at 22:21
idknuttinidknuttin
1,2071631
1,2071631
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
- $frac12(1, 1)$ is not an integer vector.
- The plane with the $x$-axis removed contains all points with the first coordinates non-zero.
$endgroup$
$begingroup$
a set of integer vectors can be considered a subset?
$endgroup$
– idknuttin
Feb 26 '16 at 22:26
$begingroup$
@idknuttin ??? If it is a set, why would it not be a subset?
$endgroup$
– Igor Rivin
Feb 26 '16 at 22:27
$begingroup$
I meant a set of integer vectors can be considered a subspace?
$endgroup$
– idknuttin
Feb 26 '16 at 22:30
$begingroup$
@idknuttin No, but it does not say it should be a subspace.
$endgroup$
– Igor Rivin
Feb 26 '16 at 22:30
add a comment |
$begingroup$
(a) If you multiply a nonzero vector with integer components by a noninteger, the result is not in "the set of vectors with integer components" anymore.
(b) if you remove the $x$-axis, you still have points like $(1,1)$ and $(1,-1)$ even though you don't have their sum.
$endgroup$
$begingroup$
Technically scaling by 0 makes b) not work.
$endgroup$
– Improve
Feb 26 '16 at 22:44
$begingroup$
@improve since the book said "but leave the origin" and the question was about closure under addition, I didn't think that was worth mentioning here. But you're certainly correct.
$endgroup$
– Mark S.
Feb 26 '16 at 22:46
$begingroup$
I did not read the entire question.
$endgroup$
– Improve
Feb 26 '16 at 22:47
add a comment |
$begingroup$
They are asking for sets that satisfy some property of vector spaces, but that are not vector spaces.
(a) the only operation they are allowing is addition, thus clearly vectors with entire componends only add to vectors with entire components, it's not a vector space since $mathbb{Z}$ is not a field (you don't have $n^{-1}$.
(b) You are assuming it's a vector space, it is not, thats why the set doesn't collapses to only the y-axis.
$endgroup$
add a comment |
$begingroup$
b)
imagine a line that is exactly on the y-axis.
you have a vector v = (0,1) , so cv always on the y-axis but for example
(0,1) + (1,1) = (1,2) wich is not on the line (y-axix)
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
- $frac12(1, 1)$ is not an integer vector.
- The plane with the $x$-axis removed contains all points with the first coordinates non-zero.
$endgroup$
$begingroup$
a set of integer vectors can be considered a subset?
$endgroup$
– idknuttin
Feb 26 '16 at 22:26
$begingroup$
@idknuttin ??? If it is a set, why would it not be a subset?
$endgroup$
– Igor Rivin
Feb 26 '16 at 22:27
$begingroup$
I meant a set of integer vectors can be considered a subspace?
$endgroup$
– idknuttin
Feb 26 '16 at 22:30
$begingroup$
@idknuttin No, but it does not say it should be a subspace.
$endgroup$
– Igor Rivin
Feb 26 '16 at 22:30
add a comment |
$begingroup$
- $frac12(1, 1)$ is not an integer vector.
- The plane with the $x$-axis removed contains all points with the first coordinates non-zero.
$endgroup$
$begingroup$
a set of integer vectors can be considered a subset?
$endgroup$
– idknuttin
Feb 26 '16 at 22:26
$begingroup$
@idknuttin ??? If it is a set, why would it not be a subset?
$endgroup$
– Igor Rivin
Feb 26 '16 at 22:27
$begingroup$
I meant a set of integer vectors can be considered a subspace?
$endgroup$
– idknuttin
Feb 26 '16 at 22:30
$begingroup$
@idknuttin No, but it does not say it should be a subspace.
$endgroup$
– Igor Rivin
Feb 26 '16 at 22:30
add a comment |
$begingroup$
- $frac12(1, 1)$ is not an integer vector.
- The plane with the $x$-axis removed contains all points with the first coordinates non-zero.
$endgroup$
- $frac12(1, 1)$ is not an integer vector.
- The plane with the $x$-axis removed contains all points with the first coordinates non-zero.
answered Feb 26 '16 at 22:24
Igor RivinIgor Rivin
16.1k11234
16.1k11234
$begingroup$
a set of integer vectors can be considered a subset?
$endgroup$
– idknuttin
Feb 26 '16 at 22:26
$begingroup$
@idknuttin ??? If it is a set, why would it not be a subset?
$endgroup$
– Igor Rivin
Feb 26 '16 at 22:27
$begingroup$
I meant a set of integer vectors can be considered a subspace?
$endgroup$
– idknuttin
Feb 26 '16 at 22:30
$begingroup$
@idknuttin No, but it does not say it should be a subspace.
$endgroup$
– Igor Rivin
Feb 26 '16 at 22:30
add a comment |
$begingroup$
a set of integer vectors can be considered a subset?
$endgroup$
– idknuttin
Feb 26 '16 at 22:26
$begingroup$
@idknuttin ??? If it is a set, why would it not be a subset?
$endgroup$
– Igor Rivin
Feb 26 '16 at 22:27
$begingroup$
I meant a set of integer vectors can be considered a subspace?
$endgroup$
– idknuttin
Feb 26 '16 at 22:30
$begingroup$
@idknuttin No, but it does not say it should be a subspace.
$endgroup$
– Igor Rivin
Feb 26 '16 at 22:30
$begingroup$
a set of integer vectors can be considered a subset?
$endgroup$
– idknuttin
Feb 26 '16 at 22:26
$begingroup$
a set of integer vectors can be considered a subset?
$endgroup$
– idknuttin
Feb 26 '16 at 22:26
$begingroup$
@idknuttin ??? If it is a set, why would it not be a subset?
$endgroup$
– Igor Rivin
Feb 26 '16 at 22:27
$begingroup$
@idknuttin ??? If it is a set, why would it not be a subset?
$endgroup$
– Igor Rivin
Feb 26 '16 at 22:27
$begingroup$
I meant a set of integer vectors can be considered a subspace?
$endgroup$
– idknuttin
Feb 26 '16 at 22:30
$begingroup$
I meant a set of integer vectors can be considered a subspace?
$endgroup$
– idknuttin
Feb 26 '16 at 22:30
$begingroup$
@idknuttin No, but it does not say it should be a subspace.
$endgroup$
– Igor Rivin
Feb 26 '16 at 22:30
$begingroup$
@idknuttin No, but it does not say it should be a subspace.
$endgroup$
– Igor Rivin
Feb 26 '16 at 22:30
add a comment |
$begingroup$
(a) If you multiply a nonzero vector with integer components by a noninteger, the result is not in "the set of vectors with integer components" anymore.
(b) if you remove the $x$-axis, you still have points like $(1,1)$ and $(1,-1)$ even though you don't have their sum.
$endgroup$
$begingroup$
Technically scaling by 0 makes b) not work.
$endgroup$
– Improve
Feb 26 '16 at 22:44
$begingroup$
@improve since the book said "but leave the origin" and the question was about closure under addition, I didn't think that was worth mentioning here. But you're certainly correct.
$endgroup$
– Mark S.
Feb 26 '16 at 22:46
$begingroup$
I did not read the entire question.
$endgroup$
– Improve
Feb 26 '16 at 22:47
add a comment |
$begingroup$
(a) If you multiply a nonzero vector with integer components by a noninteger, the result is not in "the set of vectors with integer components" anymore.
(b) if you remove the $x$-axis, you still have points like $(1,1)$ and $(1,-1)$ even though you don't have their sum.
$endgroup$
$begingroup$
Technically scaling by 0 makes b) not work.
$endgroup$
– Improve
Feb 26 '16 at 22:44
$begingroup$
@improve since the book said "but leave the origin" and the question was about closure under addition, I didn't think that was worth mentioning here. But you're certainly correct.
$endgroup$
– Mark S.
Feb 26 '16 at 22:46
$begingroup$
I did not read the entire question.
$endgroup$
– Improve
Feb 26 '16 at 22:47
add a comment |
$begingroup$
(a) If you multiply a nonzero vector with integer components by a noninteger, the result is not in "the set of vectors with integer components" anymore.
(b) if you remove the $x$-axis, you still have points like $(1,1)$ and $(1,-1)$ even though you don't have their sum.
$endgroup$
(a) If you multiply a nonzero vector with integer components by a noninteger, the result is not in "the set of vectors with integer components" anymore.
(b) if you remove the $x$-axis, you still have points like $(1,1)$ and $(1,-1)$ even though you don't have their sum.
answered Feb 26 '16 at 22:26
Mark S.Mark S.
12.5k22772
12.5k22772
$begingroup$
Technically scaling by 0 makes b) not work.
$endgroup$
– Improve
Feb 26 '16 at 22:44
$begingroup$
@improve since the book said "but leave the origin" and the question was about closure under addition, I didn't think that was worth mentioning here. But you're certainly correct.
$endgroup$
– Mark S.
Feb 26 '16 at 22:46
$begingroup$
I did not read the entire question.
$endgroup$
– Improve
Feb 26 '16 at 22:47
add a comment |
$begingroup$
Technically scaling by 0 makes b) not work.
$endgroup$
– Improve
Feb 26 '16 at 22:44
$begingroup$
@improve since the book said "but leave the origin" and the question was about closure under addition, I didn't think that was worth mentioning here. But you're certainly correct.
$endgroup$
– Mark S.
Feb 26 '16 at 22:46
$begingroup$
I did not read the entire question.
$endgroup$
– Improve
Feb 26 '16 at 22:47
$begingroup$
Technically scaling by 0 makes b) not work.
$endgroup$
– Improve
Feb 26 '16 at 22:44
$begingroup$
Technically scaling by 0 makes b) not work.
$endgroup$
– Improve
Feb 26 '16 at 22:44
$begingroup$
@improve since the book said "but leave the origin" and the question was about closure under addition, I didn't think that was worth mentioning here. But you're certainly correct.
$endgroup$
– Mark S.
Feb 26 '16 at 22:46
$begingroup$
@improve since the book said "but leave the origin" and the question was about closure under addition, I didn't think that was worth mentioning here. But you're certainly correct.
$endgroup$
– Mark S.
Feb 26 '16 at 22:46
$begingroup$
I did not read the entire question.
$endgroup$
– Improve
Feb 26 '16 at 22:47
$begingroup$
I did not read the entire question.
$endgroup$
– Improve
Feb 26 '16 at 22:47
add a comment |
$begingroup$
They are asking for sets that satisfy some property of vector spaces, but that are not vector spaces.
(a) the only operation they are allowing is addition, thus clearly vectors with entire componends only add to vectors with entire components, it's not a vector space since $mathbb{Z}$ is not a field (you don't have $n^{-1}$.
(b) You are assuming it's a vector space, it is not, thats why the set doesn't collapses to only the y-axis.
$endgroup$
add a comment |
$begingroup$
They are asking for sets that satisfy some property of vector spaces, but that are not vector spaces.
(a) the only operation they are allowing is addition, thus clearly vectors with entire componends only add to vectors with entire components, it's not a vector space since $mathbb{Z}$ is not a field (you don't have $n^{-1}$.
(b) You are assuming it's a vector space, it is not, thats why the set doesn't collapses to only the y-axis.
$endgroup$
add a comment |
$begingroup$
They are asking for sets that satisfy some property of vector spaces, but that are not vector spaces.
(a) the only operation they are allowing is addition, thus clearly vectors with entire componends only add to vectors with entire components, it's not a vector space since $mathbb{Z}$ is not a field (you don't have $n^{-1}$.
(b) You are assuming it's a vector space, it is not, thats why the set doesn't collapses to only the y-axis.
$endgroup$
They are asking for sets that satisfy some property of vector spaces, but that are not vector spaces.
(a) the only operation they are allowing is addition, thus clearly vectors with entire componends only add to vectors with entire components, it's not a vector space since $mathbb{Z}$ is not a field (you don't have $n^{-1}$.
(b) You are assuming it's a vector space, it is not, thats why the set doesn't collapses to only the y-axis.
answered Feb 26 '16 at 22:29
panchopancho
573312
573312
add a comment |
add a comment |
$begingroup$
b)
imagine a line that is exactly on the y-axis.
you have a vector v = (0,1) , so cv always on the y-axis but for example
(0,1) + (1,1) = (1,2) wich is not on the line (y-axix)
$endgroup$
add a comment |
$begingroup$
b)
imagine a line that is exactly on the y-axis.
you have a vector v = (0,1) , so cv always on the y-axis but for example
(0,1) + (1,1) = (1,2) wich is not on the line (y-axix)
$endgroup$
add a comment |
$begingroup$
b)
imagine a line that is exactly on the y-axis.
you have a vector v = (0,1) , so cv always on the y-axis but for example
(0,1) + (1,1) = (1,2) wich is not on the line (y-axix)
$endgroup$
b)
imagine a line that is exactly on the y-axis.
you have a vector v = (0,1) , so cv always on the y-axis but for example
(0,1) + (1,1) = (1,2) wich is not on the line (y-axix)
answered Dec 24 '18 at 18:59
Bishoy AbdBishoy Abd
1012
1012
add a comment |
add a comment |
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