Why isn't a linearly dependent set in $mathbb R^n$ a spanning set in $mathbb R^n$?
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Aren't there some cases where they do in fact span $mathbb R^n$?
E.g: the set of vectors $(1, 0, 0), (0, 1, 0), (0,0,1), (0,0,2)$ in $mathbb R^3$
Clearly the first three vectors form a basis for $mathbb R^3$ and span all of $mathbb R^3$. The vector $(0,0,2)$ can also be used in different scalar combinations with other vectors in the set to reach any point $mathbb R^3$ so why would adding it to the set suddenly make the set not span $mathbb R^3$?
Might it be due to some formal definition of a spanning set of a subspace being the basis of that subspace or something alike?
linear-algebra
asked Dec 2 at 14:12
Mango164
383
add a comment |
up vote
1
down vote
favorite
Aren't there some cases where they do in fact span $mathbb R^n$?
E.g: the set of vectors $(1, 0, 0), (0, 1, 0), (0,0,1), (0,0,2)$ in $mathbb R^3$
Clearly the first three vectors form a basis for $mathbb R^3$ and span all of $mathbb R^3$. The vector $(0,0,2)$ can also be used in different scalar combinations with other vectors in the set to reach any point $mathbb R^3$ so why would adding it to the set suddenly make the set not span $mathbb R^3$?
Might it be due to some formal definition of a spanning set of a subspace being the basis of that subspace or something alike?
linear-algebra
asked Dec 2 at 14:12
Mango164
383
Spanning, yes, but not minimally spanning
– MPW
Dec 2 at 14:50
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Aren't there some cases where they do in fact span $mathbb R^n$?
E.g: the set of vectors $(1, 0, 0), (0, 1, 0), (0,0,1), (0,0,2)$ in $mathbb R^3$
Clearly the first three vectors form a basis for $mathbb R^3$ and span all of $mathbb R^3$. The vector $(0,0,2)$ can also be used in different scalar combinations with other vectors in the set to reach any point $mathbb R^3$ so why would adding it to the set suddenly make the set not span $mathbb R^3$?
Might it be due to some formal definition of a spanning set of a subspace being the basis of that subspace or something alike?
linear-algebra
asked Dec 2 at 14:12
Mango164
383
Aren't there some cases where they do in fact span $mathbb R^n$?
E.g: the set of vectors $(1, 0, 0), (0, 1, 0), (0,0,1), (0,0,2)$ in $mathbb R^3$
Clearly the first three vectors form a basis for $mathbb R^3$ and span all of $mathbb R^3$. The vector $(0,0,2)$ can also be used in different scalar combinations with other vectors in the set to reach any point $mathbb R^3$ so why would adding it to the set suddenly make the set not span $mathbb R^3$?
Might it be due to some formal definition of a spanning set of a subspace being the basis of that subspace or something alike?
linear-algebra
linear-algebra
asked Dec 2 at 14:12
Mango164
383
asked Dec 2 at 14:12
Mango164
383
asked Dec 2 at 14:12
Mango164
383
asked Dec 2 at 14:12
Mango164
383
asked Dec 2 at 14:12
Mango164
383
383
Spanning, yes, but not minimally spanning
– MPW
Dec 2 at 14:50
add a comment |
Spanning, yes, but not minimally spanning
– MPW
Dec 2 at 14:50
Spanning, yes, but not minimally spanning
– MPW
Dec 2 at 14:50
Spanning, yes, but not minimally spanning
– MPW
Dec 2 at 14:50
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
${0,0,1),(0,0,2)}$ is not linearly independent and they do not span $mathbb{R}^3$.
If you already have a set that span $mathbb{R}^n$, adding more vectors do not change the property that it span $mathbb{R}^n$.
answered Dec 2 at 14:15
Siong Thye Goh
96.4k1462116
Therefore a linearly dependent set can in fact span $mathbb R^3$?
– Mango164
Dec 2 at 14:18
1
yes, it is possible.
– Siong Thye Goh
Dec 2 at 14:23
add a comment |
up vote
1
down vote
That set of vectors does span $mathbb{R}^3$ but it is not linear independent.
So if some set is not linear independant it can still span some space.
If you add any vector to base set of space $V$ it still spans $V$.
answered Dec 2 at 14:15
greedoid
36.1k114591
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
${0,0,1),(0,0,2)}$ is not linearly independent and they do not span $mathbb{R}^3$.
If you already have a set that span $mathbb{R}^n$, adding more vectors do not change the property that it span $mathbb{R}^n$.
answered Dec 2 at 14:15
Siong Thye Goh
96.4k1462116
Therefore a linearly dependent set can in fact span $mathbb R^3$?
– Mango164
Dec 2 at 14:18
1
yes, it is possible.
– Siong Thye Goh
Dec 2 at 14:23
add a comment |
up vote
1
down vote
accepted
${0,0,1),(0,0,2)}$ is not linearly independent and they do not span $mathbb{R}^3$.
If you already have a set that span $mathbb{R}^n$, adding more vectors do not change the property that it span $mathbb{R}^n$.
answered Dec 2 at 14:15
Siong Thye Goh
96.4k1462116
Therefore a linearly dependent set can in fact span $mathbb R^3$?
– Mango164
Dec 2 at 14:18
1
yes, it is possible.
– Siong Thye Goh
Dec 2 at 14:23
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
${0,0,1),(0,0,2)}$ is not linearly independent and they do not span $mathbb{R}^3$.
If you already have a set that span $mathbb{R}^n$, adding more vectors do not change the property that it span $mathbb{R}^n$.
answered Dec 2 at 14:15
Siong Thye Goh
96.4k1462116
${0,0,1),(0,0,2)}$ is not linearly independent and they do not span $mathbb{R}^3$.
If you already have a set that span $mathbb{R}^n$, adding more vectors do not change the property that it span $mathbb{R}^n$.
answered Dec 2 at 14:15
Siong Thye Goh
96.4k1462116
answered Dec 2 at 14:15
Siong Thye Goh
96.4k1462116
answered Dec 2 at 14:15
Siong Thye Goh
96.4k1462116
answered Dec 2 at 14:15
Siong Thye Goh
96.4k1462116
96.4k1462116
Therefore a linearly dependent set can in fact span $mathbb R^3$?
– Mango164
Dec 2 at 14:18
1
yes, it is possible.
– Siong Thye Goh
Dec 2 at 14:23
add a comment |
Therefore a linearly dependent set can in fact span $mathbb R^3$?
– Mango164
Dec 2 at 14:18
1
yes, it is possible.
– Siong Thye Goh
Dec 2 at 14:23
Therefore a linearly dependent set can in fact span $mathbb R^3$?
– Mango164
Dec 2 at 14:18
Therefore a linearly dependent set can in fact span $mathbb R^3$?
– Mango164
Dec 2 at 14:18
1
1
yes, it is possible.
– Siong Thye Goh
Dec 2 at 14:23
yes, it is possible.
– Siong Thye Goh
Dec 2 at 14:23
add a comment |
up vote
1
down vote
That set of vectors does span $mathbb{R}^3$ but it is not linear independent.
So if some set is not linear independant it can still span some space.
If you add any vector to base set of space $V$ it still spans $V$.
answered Dec 2 at 14:15
greedoid
36.1k114591
add a comment |
up vote
1
down vote
That set of vectors does span $mathbb{R}^3$ but it is not linear independent.
So if some set is not linear independant it can still span some space.
If you add any vector to base set of space $V$ it still spans $V$.
answered Dec 2 at 14:15
greedoid
36.1k114591
add a comment |
up vote
1
down vote
up vote
1
down vote
That set of vectors does span $mathbb{R}^3$ but it is not linear independent.
So if some set is not linear independant it can still span some space.
If you add any vector to base set of space $V$ it still spans $V$.
answered Dec 2 at 14:15
greedoid
36.1k114591
That set of vectors does span $mathbb{R}^3$ but it is not linear independent.
So if some set is not linear independant it can still span some space.
If you add any vector to base set of space $V$ it still spans $V$.
answered Dec 2 at 14:15
greedoid
36.1k114591
answered Dec 2 at 14:15
greedoid
36.1k114591
answered Dec 2 at 14:15
greedoid
36.1k114591
answered Dec 2 at 14:15
greedoid
36.1k114591
36.1k114591
add a comment |
add a comment |
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Spanning, yes, but not minimally spanning
– MPW
Dec 2 at 14:50