why is $ {(0,x,z)|x,zin R}$ is a two dimensional subspace space of $R^{3}$ over $R$ but ${(0,0,z)|zin R} $...
$begingroup$
why is $ A={(0,x,z)|x,zin R}$ is a two dimensional subspace space of $R^{3}$ over $R$
but $B={(0,0,z)|zin R} $ $cup$ ${(0,x,0)|xin R}$ is not?
i Think both are two dimensional as A has Basis={(0,1,0),(0,0,1)} and
B is direct sum of two subspace of $ R^{3}$ with same basis .
linear-algebra matrices linear-transformations
asked Jan 21 at 12:32
sejysejy
1539
$endgroup$
add a comment |
$begingroup$
why is $ A={(0,x,z)|x,zin R}$ is a two dimensional subspace space of $R^{3}$ over $R$
but $B={(0,0,z)|zin R} $ $cup$ ${(0,x,0)|xin R}$ is not?
i Think both are two dimensional as A has Basis={(0,1,0),(0,0,1)} and
B is direct sum of two subspace of $ R^{3}$ with same basis .
linear-algebra matrices linear-transformations
asked Jan 21 at 12:32
sejysejy
1539
$endgroup$
2
$begingroup$
"B is direct sum of two subspace of $ R^{3}$ with same basis" No, $B$ is a union of subspaces.
$endgroup$
– Arthur
Jan 21 at 12:35
1
$begingroup$
I suspect your confusion lies in the fact that every $v in B$ has a unique decomposition $v = (0,0,a) +(0,b,0)$. The issue is not in the existence or uniqueness of this decomposition, it is that the space $B$ isn't closed under addition.
$endgroup$
– Calvin Khor
Jan 21 at 12:38
add a comment |
$begingroup$
why is $ A={(0,x,z)|x,zin R}$ is a two dimensional subspace space of $R^{3}$ over $R$
but $B={(0,0,z)|zin R} $ $cup$ ${(0,x,0)|xin R}$ is not?
i Think both are two dimensional as A has Basis={(0,1,0),(0,0,1)} and
B is direct sum of two subspace of $ R^{3}$ with same basis .
linear-algebra matrices linear-transformations
asked Jan 21 at 12:32
sejysejy
1539
$endgroup$
why is $ A={(0,x,z)|x,zin R}$ is a two dimensional subspace space of $R^{3}$ over $R$
but $B={(0,0,z)|zin R} $ $cup$ ${(0,x,0)|xin R}$ is not?
i Think both are two dimensional as A has Basis={(0,1,0),(0,0,1)} and
B is direct sum of two subspace of $ R^{3}$ with same basis .
linear-algebra matrices linear-transformations
linear-algebra matrices linear-transformations
asked Jan 21 at 12:32
sejysejy
1539
asked Jan 21 at 12:32
sejysejy
1539
asked Jan 21 at 12:32
sejysejy
1539
asked Jan 21 at 12:32
sejysejy
1539
asked Jan 21 at 12:32
sejysejy
1539
1539
2
$begingroup$
"B is direct sum of two subspace of $ R^{3}$ with same basis" No, $B$ is a union of subspaces.
$endgroup$
– Arthur
Jan 21 at 12:35
1
$begingroup$
I suspect your confusion lies in the fact that every $v in B$ has a unique decomposition $v = (0,0,a) +(0,b,0)$. The issue is not in the existence or uniqueness of this decomposition, it is that the space $B$ isn't closed under addition.
$endgroup$
– Calvin Khor
Jan 21 at 12:38
add a comment |
2
$begingroup$
"B is direct sum of two subspace of $ R^{3}$ with same basis" No, $B$ is a union of subspaces.
$endgroup$
– Arthur
Jan 21 at 12:35
1
$begingroup$
I suspect your confusion lies in the fact that every $v in B$ has a unique decomposition $v = (0,0,a) +(0,b,0)$. The issue is not in the existence or uniqueness of this decomposition, it is that the space $B$ isn't closed under addition.
$endgroup$
– Calvin Khor
Jan 21 at 12:38
2
2
$begingroup$
"B is direct sum of two subspace of $ R^{3}$ with same basis" No, $B$ is a union of subspaces.
$endgroup$
– Arthur
Jan 21 at 12:35
$begingroup$
"B is direct sum of two subspace of $ R^{3}$ with same basis" No, $B$ is a union of subspaces.
$endgroup$
– Arthur
Jan 21 at 12:35
1
1
$begingroup$
I suspect your confusion lies in the fact that every $v in B$ has a unique decomposition $v = (0,0,a) +(0,b,0)$. The issue is not in the existence or uniqueness of this decomposition, it is that the space $B$ isn't closed under addition.
$endgroup$
– Calvin Khor
Jan 21 at 12:38
$begingroup$
I suspect your confusion lies in the fact that every $v in B$ has a unique decomposition $v = (0,0,a) +(0,b,0)$. The issue is not in the existence or uniqueness of this decomposition, it is that the space $B$ isn't closed under addition.
$endgroup$
– Calvin Khor
Jan 21 at 12:38
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$B$ is defined as the union of two subsets of $mathbb R^{3}$, not as a sum of subspaces. $(0,0,1)+(0,1,0)=(0,1,1)$ is not in $B$ so it is not a subspace of $mathbb R^{3}$.
answered Jan 21 at 12:33
Kavi Rama MurthyKavi Rama Murthy
62.3k42262
$endgroup$
add a comment |
$begingroup$
$B$ is a union of two subspaces and is missing all of the vectors that are “in between” the subspaces. More formally, the union isn’t closed under vector addition: $(0,0,z)+(0,x,0)notin B$ unless at least one of those two vectors is zero. On the other hand, $A$ contains all of those sums and then some: if you add any two vectors of the form $(0,x,z)$ to each other, you get another vector of the same form, so $A$ is closed under vector addition. It’s easy to verify that it also contains the zero vector and is closed under scalar multiplication, so it is a subspace of $mathbb R^3$.
To illustrate this geometrically, $B$ is the union of the $y$-and $z$-axes. This just gets you a pair of lines, which you probably know is not a vector space. On the other hand, $A$ describes the entire $y$-$z$ plane, which contains those two lines, but also all of the points “in between” them on that plane. This plane is isomorphic to $mathbb R^2$, which you know to be a vector space.
answered Jan 21 at 21:35
amdamd
30.6k21050
$endgroup$
$begingroup$
U cleared my confusion insightful !!
$endgroup$
– sejy
Jan 23 at 0:38
add a comment |
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2 Answers
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$begingroup$
$B$ is defined as the union of two subsets of $mathbb R^{3}$, not as a sum of subspaces. $(0,0,1)+(0,1,0)=(0,1,1)$ is not in $B$ so it is not a subspace of $mathbb R^{3}$.
answered Jan 21 at 12:33
Kavi Rama MurthyKavi Rama Murthy
62.3k42262
$endgroup$
add a comment |
$begingroup$
$B$ is defined as the union of two subsets of $mathbb R^{3}$, not as a sum of subspaces. $(0,0,1)+(0,1,0)=(0,1,1)$ is not in $B$ so it is not a subspace of $mathbb R^{3}$.
answered Jan 21 at 12:33
Kavi Rama MurthyKavi Rama Murthy
62.3k42262
$endgroup$
add a comment |
$begingroup$
$B$ is defined as the union of two subsets of $mathbb R^{3}$, not as a sum of subspaces. $(0,0,1)+(0,1,0)=(0,1,1)$ is not in $B$ so it is not a subspace of $mathbb R^{3}$.
answered Jan 21 at 12:33
Kavi Rama MurthyKavi Rama Murthy
62.3k42262
$endgroup$
$B$ is defined as the union of two subsets of $mathbb R^{3}$, not as a sum of subspaces. $(0,0,1)+(0,1,0)=(0,1,1)$ is not in $B$ so it is not a subspace of $mathbb R^{3}$.
answered Jan 21 at 12:33
Kavi Rama MurthyKavi Rama Murthy
62.3k42262
answered Jan 21 at 12:33
Kavi Rama MurthyKavi Rama Murthy
62.3k42262
answered Jan 21 at 12:33
Kavi Rama MurthyKavi Rama Murthy
62.3k42262
answered Jan 21 at 12:33
Kavi Rama MurthyKavi Rama Murthy
62.3k42262
62.3k42262
add a comment |
add a comment |
$begingroup$
$B$ is a union of two subspaces and is missing all of the vectors that are “in between” the subspaces. More formally, the union isn’t closed under vector addition: $(0,0,z)+(0,x,0)notin B$ unless at least one of those two vectors is zero. On the other hand, $A$ contains all of those sums and then some: if you add any two vectors of the form $(0,x,z)$ to each other, you get another vector of the same form, so $A$ is closed under vector addition. It’s easy to verify that it also contains the zero vector and is closed under scalar multiplication, so it is a subspace of $mathbb R^3$.
To illustrate this geometrically, $B$ is the union of the $y$-and $z$-axes. This just gets you a pair of lines, which you probably know is not a vector space. On the other hand, $A$ describes the entire $y$-$z$ plane, which contains those two lines, but also all of the points “in between” them on that plane. This plane is isomorphic to $mathbb R^2$, which you know to be a vector space.
answered Jan 21 at 21:35
amdamd
30.6k21050
$endgroup$
$begingroup$
U cleared my confusion insightful !!
$endgroup$
– sejy
Jan 23 at 0:38
add a comment |
$begingroup$
$B$ is a union of two subspaces and is missing all of the vectors that are “in between” the subspaces. More formally, the union isn’t closed under vector addition: $(0,0,z)+(0,x,0)notin B$ unless at least one of those two vectors is zero. On the other hand, $A$ contains all of those sums and then some: if you add any two vectors of the form $(0,x,z)$ to each other, you get another vector of the same form, so $A$ is closed under vector addition. It’s easy to verify that it also contains the zero vector and is closed under scalar multiplication, so it is a subspace of $mathbb R^3$.
To illustrate this geometrically, $B$ is the union of the $y$-and $z$-axes. This just gets you a pair of lines, which you probably know is not a vector space. On the other hand, $A$ describes the entire $y$-$z$ plane, which contains those two lines, but also all of the points “in between” them on that plane. This plane is isomorphic to $mathbb R^2$, which you know to be a vector space.
answered Jan 21 at 21:35
amdamd
30.6k21050
$endgroup$
$begingroup$
U cleared my confusion insightful !!
$endgroup$
– sejy
Jan 23 at 0:38
add a comment |
$begingroup$
$B$ is a union of two subspaces and is missing all of the vectors that are “in between” the subspaces. More formally, the union isn’t closed under vector addition: $(0,0,z)+(0,x,0)notin B$ unless at least one of those two vectors is zero. On the other hand, $A$ contains all of those sums and then some: if you add any two vectors of the form $(0,x,z)$ to each other, you get another vector of the same form, so $A$ is closed under vector addition. It’s easy to verify that it also contains the zero vector and is closed under scalar multiplication, so it is a subspace of $mathbb R^3$.
To illustrate this geometrically, $B$ is the union of the $y$-and $z$-axes. This just gets you a pair of lines, which you probably know is not a vector space. On the other hand, $A$ describes the entire $y$-$z$ plane, which contains those two lines, but also all of the points “in between” them on that plane. This plane is isomorphic to $mathbb R^2$, which you know to be a vector space.
answered Jan 21 at 21:35
amdamd
30.6k21050
$endgroup$
$B$ is a union of two subspaces and is missing all of the vectors that are “in between” the subspaces. More formally, the union isn’t closed under vector addition: $(0,0,z)+(0,x,0)notin B$ unless at least one of those two vectors is zero. On the other hand, $A$ contains all of those sums and then some: if you add any two vectors of the form $(0,x,z)$ to each other, you get another vector of the same form, so $A$ is closed under vector addition. It’s easy to verify that it also contains the zero vector and is closed under scalar multiplication, so it is a subspace of $mathbb R^3$.
To illustrate this geometrically, $B$ is the union of the $y$-and $z$-axes. This just gets you a pair of lines, which you probably know is not a vector space. On the other hand, $A$ describes the entire $y$-$z$ plane, which contains those two lines, but also all of the points “in between” them on that plane. This plane is isomorphic to $mathbb R^2$, which you know to be a vector space.
answered Jan 21 at 21:35
amdamd
30.6k21050
edited Jan 23 at 0:52
answered Jan 21 at 21:35
amdamd
30.6k21050
answered Jan 21 at 21:35
amdamd
30.6k21050
answered Jan 21 at 21:35
amdamd
30.6k21050
30.6k21050
$begingroup$
U cleared my confusion insightful !!
$endgroup$
– sejy
Jan 23 at 0:38
add a comment |
$begingroup$
U cleared my confusion insightful !!
$endgroup$
– sejy
Jan 23 at 0:38
$begingroup$
U cleared my confusion insightful !!
$endgroup$
– sejy
Jan 23 at 0:38
$begingroup$
U cleared my confusion insightful !!
$endgroup$
– sejy
Jan 23 at 0:38
add a comment |
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2
$begingroup$
"B is direct sum of two subspace of $ R^{3}$ with same basis" No, $B$ is a union of subspaces.
$endgroup$
– Arthur
Jan 21 at 12:35
1
$begingroup$
I suspect your confusion lies in the fact that every $v in B$ has a unique decomposition $v = (0,0,a) +(0,b,0)$. The issue is not in the existence or uniqueness of this decomposition, it is that the space $B$ isn't closed under addition.
$endgroup$
– Calvin Khor
Jan 21 at 12:38