Linearly Indept set of n vectors implies size of vector space to be at least n.
Theorem:
If a set of n vectors is linearly independent in a vector space, then the dimension of the space is at least n.
Proof:
Let $S = {vec{v}_{i}}_{i=1}^{n}$ be a L.I set of n vectors in a vector space $V$.
Certainly, the S is a basis for the vector space $V$. The order of the set $S$ is $n$, and by definition, dim$left(Vright) = |S| = n$.
It might suffice to show that the union of the set $S$ with another set $bar{S}, bar{S} neq 0$
does not affect the span of $S$.
I.e, $S = S cup bar{S}$
Any help to take me further is greatly appreciated.
Thanks in advance.
linear-algebra proof-verification
asked Nov 25 at 4:32
Mathematicing
2,44121851
add a comment |
Theorem:
If a set of n vectors is linearly independent in a vector space, then the dimension of the space is at least n.
Proof:
Let $S = {vec{v}_{i}}_{i=1}^{n}$ be a L.I set of n vectors in a vector space $V$.
Certainly, the S is a basis for the vector space $V$. The order of the set $S$ is $n$, and by definition, dim$left(Vright) = |S| = n$.
It might suffice to show that the union of the set $S$ with another set $bar{S}, bar{S} neq 0$
does not affect the span of $S$.
I.e, $S = S cup bar{S}$
Any help to take me further is greatly appreciated.
Thanks in advance.
linear-algebra proof-verification
asked Nov 25 at 4:32
Mathematicing
2,44121851
2
Why is $S$ a basis for $V$? The dimension of $V$ was not given.
– David Peterson
Nov 25 at 4:39
add a comment |
Theorem:
If a set of n vectors is linearly independent in a vector space, then the dimension of the space is at least n.
Proof:
Let $S = {vec{v}_{i}}_{i=1}^{n}$ be a L.I set of n vectors in a vector space $V$.
Certainly, the S is a basis for the vector space $V$. The order of the set $S$ is $n$, and by definition, dim$left(Vright) = |S| = n$.
It might suffice to show that the union of the set $S$ with another set $bar{S}, bar{S} neq 0$
does not affect the span of $S$.
I.e, $S = S cup bar{S}$
Any help to take me further is greatly appreciated.
Thanks in advance.
linear-algebra proof-verification
asked Nov 25 at 4:32
Mathematicing
2,44121851
Theorem:
If a set of n vectors is linearly independent in a vector space, then the dimension of the space is at least n.
Proof:
Let $S = {vec{v}_{i}}_{i=1}^{n}$ be a L.I set of n vectors in a vector space $V$.
Certainly, the S is a basis for the vector space $V$. The order of the set $S$ is $n$, and by definition, dim$left(Vright) = |S| = n$.
It might suffice to show that the union of the set $S$ with another set $bar{S}, bar{S} neq 0$
does not affect the span of $S$.
I.e, $S = S cup bar{S}$
Any help to take me further is greatly appreciated.
Thanks in advance.
linear-algebra proof-verification
linear-algebra proof-verification
asked Nov 25 at 4:32
Mathematicing
2,44121851
asked Nov 25 at 4:32
Mathematicing
2,44121851
asked Nov 25 at 4:32
Mathematicing
2,44121851
asked Nov 25 at 4:32
Mathematicing
2,44121851
asked Nov 25 at 4:32
Mathematicing
2,44121851
2,44121851
2
Why is $S$ a basis for $V$? The dimension of $V$ was not given.
– David Peterson
Nov 25 at 4:39
add a comment |
2
Why is $S$ a basis for $V$? The dimension of $V$ was not given.
– David Peterson
Nov 25 at 4:39
2
2
Why is $S$ a basis for $V$? The dimension of $V$ was not given.
– David Peterson
Nov 25 at 4:39
Why is $S$ a basis for $V$? The dimension of $V$ was not given.
– David Peterson
Nov 25 at 4:39
add a comment |
2 Answers
2
active
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votes
Pick your set $S$ of linearly independent vectors in $V$, then if they don't span $V$, find a vector that is not in their span and add it to the set. You are still going to get a set of linearly independent vectors. Do this until you get a spanning set and then $S$ becomes a basis for $V$. In particular, you don't remove vectors from your initial set $S$ to get a spanning set. So the dimension of $V$ is at least $|S|$. Notice that it is not true that a linearly independent set of vectors is a basis for $V$. Take for example $V=mathbb{R^3}$ and $S={(1,0,0),(0,1,0) }$, $S$ is not a basis!
answered Nov 25 at 4:49
mathnoob
1,794422
add a comment |
$S$ need not be a basis set. For example, in $Bbb{R}^3$, we can have $S=left{begin{bmatrix}1\0\0end{bmatrix}, begin{bmatrix}1\1\0end{bmatrix}right}$ a set of $2$ linearly independent vectors but definitely it doesn't span $Bbb{R}^3$. So it is not a basis of $Bbb{R}^3$. However, one can have a basis of $Bbb{R}^3$ which can contain (at least) these two vectors.
In general, the best we can say is that one can have a basis $mathcal{B}$ that can contain at least these $n$ linearly independent vectors (from $S$). Thus $text{dim} geq n$.
answered Nov 25 at 4:46
Anurag A
25.6k12249
add a comment |
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2 Answers
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2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
Pick your set $S$ of linearly independent vectors in $V$, then if they don't span $V$, find a vector that is not in their span and add it to the set. You are still going to get a set of linearly independent vectors. Do this until you get a spanning set and then $S$ becomes a basis for $V$. In particular, you don't remove vectors from your initial set $S$ to get a spanning set. So the dimension of $V$ is at least $|S|$. Notice that it is not true that a linearly independent set of vectors is a basis for $V$. Take for example $V=mathbb{R^3}$ and $S={(1,0,0),(0,1,0) }$, $S$ is not a basis!
answered Nov 25 at 4:49
mathnoob
1,794422
add a comment |
Pick your set $S$ of linearly independent vectors in $V$, then if they don't span $V$, find a vector that is not in their span and add it to the set. You are still going to get a set of linearly independent vectors. Do this until you get a spanning set and then $S$ becomes a basis for $V$. In particular, you don't remove vectors from your initial set $S$ to get a spanning set. So the dimension of $V$ is at least $|S|$. Notice that it is not true that a linearly independent set of vectors is a basis for $V$. Take for example $V=mathbb{R^3}$ and $S={(1,0,0),(0,1,0) }$, $S$ is not a basis!
answered Nov 25 at 4:49
mathnoob
1,794422
add a comment |
Pick your set $S$ of linearly independent vectors in $V$, then if they don't span $V$, find a vector that is not in their span and add it to the set. You are still going to get a set of linearly independent vectors. Do this until you get a spanning set and then $S$ becomes a basis for $V$. In particular, you don't remove vectors from your initial set $S$ to get a spanning set. So the dimension of $V$ is at least $|S|$. Notice that it is not true that a linearly independent set of vectors is a basis for $V$. Take for example $V=mathbb{R^3}$ and $S={(1,0,0),(0,1,0) }$, $S$ is not a basis!
answered Nov 25 at 4:49
mathnoob
1,794422
Pick your set $S$ of linearly independent vectors in $V$, then if they don't span $V$, find a vector that is not in their span and add it to the set. You are still going to get a set of linearly independent vectors. Do this until you get a spanning set and then $S$ becomes a basis for $V$. In particular, you don't remove vectors from your initial set $S$ to get a spanning set. So the dimension of $V$ is at least $|S|$. Notice that it is not true that a linearly independent set of vectors is a basis for $V$. Take for example $V=mathbb{R^3}$ and $S={(1,0,0),(0,1,0) }$, $S$ is not a basis!
answered Nov 25 at 4:49
mathnoob
1,794422
answered Nov 25 at 4:49
mathnoob
1,794422
answered Nov 25 at 4:49
mathnoob
1,794422
answered Nov 25 at 4:49
mathnoob
1,794422
1,794422
add a comment |
add a comment |
$S$ need not be a basis set. For example, in $Bbb{R}^3$, we can have $S=left{begin{bmatrix}1\0\0end{bmatrix}, begin{bmatrix}1\1\0end{bmatrix}right}$ a set of $2$ linearly independent vectors but definitely it doesn't span $Bbb{R}^3$. So it is not a basis of $Bbb{R}^3$. However, one can have a basis of $Bbb{R}^3$ which can contain (at least) these two vectors.
In general, the best we can say is that one can have a basis $mathcal{B}$ that can contain at least these $n$ linearly independent vectors (from $S$). Thus $text{dim} geq n$.
answered Nov 25 at 4:46
Anurag A
25.6k12249
add a comment |
$S$ need not be a basis set. For example, in $Bbb{R}^3$, we can have $S=left{begin{bmatrix}1\0\0end{bmatrix}, begin{bmatrix}1\1\0end{bmatrix}right}$ a set of $2$ linearly independent vectors but definitely it doesn't span $Bbb{R}^3$. So it is not a basis of $Bbb{R}^3$. However, one can have a basis of $Bbb{R}^3$ which can contain (at least) these two vectors.
In general, the best we can say is that one can have a basis $mathcal{B}$ that can contain at least these $n$ linearly independent vectors (from $S$). Thus $text{dim} geq n$.
answered Nov 25 at 4:46
Anurag A
25.6k12249
add a comment |
$S$ need not be a basis set. For example, in $Bbb{R}^3$, we can have $S=left{begin{bmatrix}1\0\0end{bmatrix}, begin{bmatrix}1\1\0end{bmatrix}right}$ a set of $2$ linearly independent vectors but definitely it doesn't span $Bbb{R}^3$. So it is not a basis of $Bbb{R}^3$. However, one can have a basis of $Bbb{R}^3$ which can contain (at least) these two vectors.
In general, the best we can say is that one can have a basis $mathcal{B}$ that can contain at least these $n$ linearly independent vectors (from $S$). Thus $text{dim} geq n$.
answered Nov 25 at 4:46
Anurag A
25.6k12249
$S$ need not be a basis set. For example, in $Bbb{R}^3$, we can have $S=left{begin{bmatrix}1\0\0end{bmatrix}, begin{bmatrix}1\1\0end{bmatrix}right}$ a set of $2$ linearly independent vectors but definitely it doesn't span $Bbb{R}^3$. So it is not a basis of $Bbb{R}^3$. However, one can have a basis of $Bbb{R}^3$ which can contain (at least) these two vectors.
In general, the best we can say is that one can have a basis $mathcal{B}$ that can contain at least these $n$ linearly independent vectors (from $S$). Thus $text{dim} geq n$.
answered Nov 25 at 4:46
Anurag A
25.6k12249
edited Nov 25 at 7:51
answered Nov 25 at 4:46
Anurag A
25.6k12249
answered Nov 25 at 4:46
Anurag A
25.6k12249
answered Nov 25 at 4:46
Anurag A
25.6k12249
25.6k12249
add a comment |
add a comment |
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Why is $S$ a basis for $V$? The dimension of $V$ was not given.
– David Peterson
Nov 25 at 4:39