Mixing
Is this a correct way to think of a subspace?
$begingroup$
A subspace in $R^n$ is just a set with conditions that create a hyperplane (so a line, a plane, etc.) in $R^n$?
And of course it has to pass the subspace test.
But if what I said is correct, because a subspace has to contain the zero vector, so that the scalar multiplication test passes, does the line or plane HAVE to cross the origin? What's the difference between a line passing through the origin and one that is just shifted one unit up so it doesn't touch to origin? What makes the first a subspace and the second not?
If any of my questions were hard to follow I think it's because I have a poor understanding of what a subspace is so any help on that would be nice. Thanks!
linear-algebra
asked Feb 2 at 23:32
mingming
4606
$endgroup$
add a comment |
$begingroup$
A subspace in $R^n$ is just a set with conditions that create a hyperplane (so a line, a plane, etc.) in $R^n$?
And of course it has to pass the subspace test.
But if what I said is correct, because a subspace has to contain the zero vector, so that the scalar multiplication test passes, does the line or plane HAVE to cross the origin? What's the difference between a line passing through the origin and one that is just shifted one unit up so it doesn't touch to origin? What makes the first a subspace and the second not?
If any of my questions were hard to follow I think it's because I have a poor understanding of what a subspace is so any help on that would be nice. Thanks!
linear-algebra
asked Feb 2 at 23:32
mingming
4606
$endgroup$
$begingroup$
In general, a hyperplane is defined as a maximal proper subspace, and this in a finite dimensional space as $;Bbb R^n;$ means any subspace of dimension $;n-1;$ . There are many, many other subspace that are not hyperplanes if for example the dimension is$;3,4,...etc.;$
$endgroup$
– DonAntonio
Feb 2 at 23:35
$begingroup$
Go read about "affine spaces". They are essentially vector spaces but with the stipulation of having a zero vector removed. Revelant: math.stackexchange.com/questions/884666/…
$endgroup$
– Theo C.
Feb 2 at 23:35
$begingroup$
What makes a line through the origin a subspace, contrary to what happens with a line not passing through the origin, is that in the first case the sum of two vectors on the line in again on the line, and the same happens when you multiply any vector on the line by any scalar...
$endgroup$
– DonAntonio
Feb 2 at 23:37
add a comment |
$begingroup$
A subspace in $R^n$ is just a set with conditions that create a hyperplane (so a line, a plane, etc.) in $R^n$?
And of course it has to pass the subspace test.
But if what I said is correct, because a subspace has to contain the zero vector, so that the scalar multiplication test passes, does the line or plane HAVE to cross the origin? What's the difference between a line passing through the origin and one that is just shifted one unit up so it doesn't touch to origin? What makes the first a subspace and the second not?
If any of my questions were hard to follow I think it's because I have a poor understanding of what a subspace is so any help on that would be nice. Thanks!
linear-algebra
asked Feb 2 at 23:32
mingming
4606
$endgroup$
A subspace in $R^n$ is just a set with conditions that create a hyperplane (so a line, a plane, etc.) in $R^n$?
And of course it has to pass the subspace test.
But if what I said is correct, because a subspace has to contain the zero vector, so that the scalar multiplication test passes, does the line or plane HAVE to cross the origin? What's the difference between a line passing through the origin and one that is just shifted one unit up so it doesn't touch to origin? What makes the first a subspace and the second not?
If any of my questions were hard to follow I think it's because I have a poor understanding of what a subspace is so any help on that would be nice. Thanks!
linear-algebra
linear-algebra
asked Feb 2 at 23:32
mingming
4606
asked Feb 2 at 23:32
mingming
4606
asked Feb 2 at 23:32
mingming
4606
asked Feb 2 at 23:32
mingming
4606
asked Feb 2 at 23:32
mingming
4606
4606
$begingroup$
In general, a hyperplane is defined as a maximal proper subspace, and this in a finite dimensional space as $;Bbb R^n;$ means any subspace of dimension $;n-1;$ . There are many, many other subspace that are not hyperplanes if for example the dimension is$;3,4,...etc.;$
$endgroup$
– DonAntonio
Feb 2 at 23:35
$begingroup$
Go read about "affine spaces". They are essentially vector spaces but with the stipulation of having a zero vector removed. Revelant: math.stackexchange.com/questions/884666/…
$endgroup$
– Theo C.
Feb 2 at 23:35
$begingroup$
What makes a line through the origin a subspace, contrary to what happens with a line not passing through the origin, is that in the first case the sum of two vectors on the line in again on the line, and the same happens when you multiply any vector on the line by any scalar...
$endgroup$
– DonAntonio
Feb 2 at 23:37
add a comment |
$begingroup$
In general, a hyperplane is defined as a maximal proper subspace, and this in a finite dimensional space as $;Bbb R^n;$ means any subspace of dimension $;n-1;$ . There are many, many other subspace that are not hyperplanes if for example the dimension is$;3,4,...etc.;$
$endgroup$
– DonAntonio
Feb 2 at 23:35
$begingroup$
Go read about "affine spaces". They are essentially vector spaces but with the stipulation of having a zero vector removed. Revelant: math.stackexchange.com/questions/884666/…
$endgroup$
– Theo C.
Feb 2 at 23:35
$begingroup$
What makes a line through the origin a subspace, contrary to what happens with a line not passing through the origin, is that in the first case the sum of two vectors on the line in again on the line, and the same happens when you multiply any vector on the line by any scalar...
$endgroup$
– DonAntonio
Feb 2 at 23:37
$begingroup$
In general, a hyperplane is defined as a maximal proper subspace, and this in a finite dimensional space as $;Bbb R^n;$ means any subspace of dimension $;n-1;$ . There are many, many other subspace that are not hyperplanes if for example the dimension is$;3,4,...etc.;$
$endgroup$
– DonAntonio
Feb 2 at 23:35
$begingroup$
In general, a hyperplane is defined as a maximal proper subspace, and this in a finite dimensional space as $;Bbb R^n;$ means any subspace of dimension $;n-1;$ . There are many, many other subspace that are not hyperplanes if for example the dimension is$;3,4,...etc.;$
$endgroup$
– DonAntonio
Feb 2 at 23:35
$begingroup$
Go read about "affine spaces". They are essentially vector spaces but with the stipulation of having a zero vector removed. Revelant: math.stackexchange.com/questions/884666/…
$endgroup$
– Theo C.
Feb 2 at 23:35
$begingroup$
Go read about "affine spaces". They are essentially vector spaces but with the stipulation of having a zero vector removed. Revelant: math.stackexchange.com/questions/884666/…
$endgroup$
– Theo C.
Feb 2 at 23:35
$begingroup$
What makes a line through the origin a subspace, contrary to what happens with a line not passing through the origin, is that in the first case the sum of two vectors on the line in again on the line, and the same happens when you multiply any vector on the line by any scalar...
$endgroup$
– DonAntonio
Feb 2 at 23:37
$begingroup$
What makes a line through the origin a subspace, contrary to what happens with a line not passing through the origin, is that in the first case the sum of two vectors on the line in again on the line, and the same happens when you multiply any vector on the line by any scalar...
$endgroup$
– DonAntonio
Feb 2 at 23:37
add a comment |
1 Answer
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$begingroup$
To define a subspace let us first define a vector space. A vector space (over $mathbb R$) is a set $V$, such that for any elements $v,win V$ and $a,binmathbb R$, we have that $av+bwin V$. I.e. I can multiply a vector by a number and get another vector in my space, and I can add arbitrary vectors together and get a vector in my space. (See https://en.wikipedia.org/wiki/Vector_space for the full list of definitions).
Now, a subspace of a vector space is a subset of $V$, $W$, such that $W$ is a vector space. So, as you said a line through the origin is a subspace, but a line that does not pass through the origin is not a vector space. This is because it does not contain 0. Since we can multiply vectors by arbitrary numbers in $mathbb R$, we can multiply any vector by zero. So every vector space must contain 0.
Now, if we have a finite dimensional vector space $V=mathbb R^n$ and a subspace $W$ of dimension $m$, then one can show (using https://en.wikipedia.org/wiki/Gram–Schmidt_process) that there will be an orthonormal basis $mathbf e_1,dots,mathbf e_n$ of $V$ such that $mathbf e_1,dots,mathbf e_m$ is a basis for $W$. Then the $ntimes n$ matrix
$$
P=(mathbf e_1,dots,mathbf e_m,mathbf 0,dots,mathbf 0),
$$
gives $m$ linear equations that define $W$. Namely
$$
W={xin V|Px=x}.
$$
answered Feb 2 at 23:42
Alec B-GAlec B-G
52019
$endgroup$
add a comment |
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$begingroup$
To define a subspace let us first define a vector space. A vector space (over $mathbb R$) is a set $V$, such that for any elements $v,win V$ and $a,binmathbb R$, we have that $av+bwin V$. I.e. I can multiply a vector by a number and get another vector in my space, and I can add arbitrary vectors together and get a vector in my space. (See https://en.wikipedia.org/wiki/Vector_space for the full list of definitions).
Now, a subspace of a vector space is a subset of $V$, $W$, such that $W$ is a vector space. So, as you said a line through the origin is a subspace, but a line that does not pass through the origin is not a vector space. This is because it does not contain 0. Since we can multiply vectors by arbitrary numbers in $mathbb R$, we can multiply any vector by zero. So every vector space must contain 0.
Now, if we have a finite dimensional vector space $V=mathbb R^n$ and a subspace $W$ of dimension $m$, then one can show (using https://en.wikipedia.org/wiki/Gram–Schmidt_process) that there will be an orthonormal basis $mathbf e_1,dots,mathbf e_n$ of $V$ such that $mathbf e_1,dots,mathbf e_m$ is a basis for $W$. Then the $ntimes n$ matrix
$$
P=(mathbf e_1,dots,mathbf e_m,mathbf 0,dots,mathbf 0),
$$
gives $m$ linear equations that define $W$. Namely
$$
W={xin V|Px=x}.
$$
answered Feb 2 at 23:42
Alec B-GAlec B-G
52019
$endgroup$
add a comment |
$begingroup$
To define a subspace let us first define a vector space. A vector space (over $mathbb R$) is a set $V$, such that for any elements $v,win V$ and $a,binmathbb R$, we have that $av+bwin V$. I.e. I can multiply a vector by a number and get another vector in my space, and I can add arbitrary vectors together and get a vector in my space. (See https://en.wikipedia.org/wiki/Vector_space for the full list of definitions).
Now, a subspace of a vector space is a subset of $V$, $W$, such that $W$ is a vector space. So, as you said a line through the origin is a subspace, but a line that does not pass through the origin is not a vector space. This is because it does not contain 0. Since we can multiply vectors by arbitrary numbers in $mathbb R$, we can multiply any vector by zero. So every vector space must contain 0.
Now, if we have a finite dimensional vector space $V=mathbb R^n$ and a subspace $W$ of dimension $m$, then one can show (using https://en.wikipedia.org/wiki/Gram–Schmidt_process) that there will be an orthonormal basis $mathbf e_1,dots,mathbf e_n$ of $V$ such that $mathbf e_1,dots,mathbf e_m$ is a basis for $W$. Then the $ntimes n$ matrix
$$
P=(mathbf e_1,dots,mathbf e_m,mathbf 0,dots,mathbf 0),
$$
gives $m$ linear equations that define $W$. Namely
$$
W={xin V|Px=x}.
$$
answered Feb 2 at 23:42
Alec B-GAlec B-G
52019
$endgroup$
add a comment |
$begingroup$
To define a subspace let us first define a vector space. A vector space (over $mathbb R$) is a set $V$, such that for any elements $v,win V$ and $a,binmathbb R$, we have that $av+bwin V$. I.e. I can multiply a vector by a number and get another vector in my space, and I can add arbitrary vectors together and get a vector in my space. (See https://en.wikipedia.org/wiki/Vector_space for the full list of definitions).
Now, a subspace of a vector space is a subset of $V$, $W$, such that $W$ is a vector space. So, as you said a line through the origin is a subspace, but a line that does not pass through the origin is not a vector space. This is because it does not contain 0. Since we can multiply vectors by arbitrary numbers in $mathbb R$, we can multiply any vector by zero. So every vector space must contain 0.
Now, if we have a finite dimensional vector space $V=mathbb R^n$ and a subspace $W$ of dimension $m$, then one can show (using https://en.wikipedia.org/wiki/Gram–Schmidt_process) that there will be an orthonormal basis $mathbf e_1,dots,mathbf e_n$ of $V$ such that $mathbf e_1,dots,mathbf e_m$ is a basis for $W$. Then the $ntimes n$ matrix
$$
P=(mathbf e_1,dots,mathbf e_m,mathbf 0,dots,mathbf 0),
$$
gives $m$ linear equations that define $W$. Namely
$$
W={xin V|Px=x}.
$$
answered Feb 2 at 23:42
Alec B-GAlec B-G
52019
$endgroup$
To define a subspace let us first define a vector space. A vector space (over $mathbb R$) is a set $V$, such that for any elements $v,win V$ and $a,binmathbb R$, we have that $av+bwin V$. I.e. I can multiply a vector by a number and get another vector in my space, and I can add arbitrary vectors together and get a vector in my space. (See https://en.wikipedia.org/wiki/Vector_space for the full list of definitions).
Now, a subspace of a vector space is a subset of $V$, $W$, such that $W$ is a vector space. So, as you said a line through the origin is a subspace, but a line that does not pass through the origin is not a vector space. This is because it does not contain 0. Since we can multiply vectors by arbitrary numbers in $mathbb R$, we can multiply any vector by zero. So every vector space must contain 0.
Now, if we have a finite dimensional vector space $V=mathbb R^n$ and a subspace $W$ of dimension $m$, then one can show (using https://en.wikipedia.org/wiki/Gram–Schmidt_process) that there will be an orthonormal basis $mathbf e_1,dots,mathbf e_n$ of $V$ such that $mathbf e_1,dots,mathbf e_m$ is a basis for $W$. Then the $ntimes n$ matrix
$$
P=(mathbf e_1,dots,mathbf e_m,mathbf 0,dots,mathbf 0),
$$
gives $m$ linear equations that define $W$. Namely
$$
W={xin V|Px=x}.
$$
answered Feb 2 at 23:42
Alec B-GAlec B-G
52019
answered Feb 2 at 23:42
Alec B-GAlec B-G
52019
answered Feb 2 at 23:42
Alec B-GAlec B-G
52019
answered Feb 2 at 23:42
Alec B-GAlec B-G
52019
52019
add a comment |
add a comment |
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$begingroup$
In general, a hyperplane is defined as a maximal proper subspace, and this in a finite dimensional space as $;Bbb R^n;$ means any subspace of dimension $;n-1;$ . There are many, many other subspace that are not hyperplanes if for example the dimension is$;3,4,...etc.;$
$endgroup$
– DonAntonio
Feb 2 at 23:35
$begingroup$
Go read about "affine spaces". They are essentially vector spaces but with the stipulation of having a zero vector removed. Revelant: math.stackexchange.com/questions/884666/…
$endgroup$
– Theo C.
Feb 2 at 23:35
$begingroup$
What makes a line through the origin a subspace, contrary to what happens with a line not passing through the origin, is that in the first case the sum of two vectors on the line in again on the line, and the same happens when you multiply any vector on the line by any scalar...
$endgroup$
– DonAntonio
Feb 2 at 23:37