Linearly Indept set of n vectors implies size of vector space to be at least n.












0















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.










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  • 2




    Why is $S$ a basis for $V$? The dimension of $V$ was not given.
    – David Peterson
    Nov 25 at 4:39
















0















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.










share|cite|improve this question


















  • 2




    Why is $S$ a basis for $V$? The dimension of $V$ was not given.
    – David Peterson
    Nov 25 at 4:39














0












0








0








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.










share|cite|improve this question














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






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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














  • 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










2 Answers
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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!






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    0














    $S$ 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 $text{dim} geq n$.






    share|cite|improve this answer























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      2 Answers
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      2 Answers
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      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!






      share|cite|improve this answer


























        0














        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!






        share|cite|improve this answer
























          0












          0








          0






          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!






          share|cite|improve this answer












          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!







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 25 at 4:49









          mathnoob

          1,794422




          1,794422























              0














              $S$ 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 $text{dim} geq n$.






              share|cite|improve this answer




























                0














                $S$ 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 $text{dim} geq n$.






                share|cite|improve this answer


























                  0












                  0








                  0






                  $S$ 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 $text{dim} geq n$.






                  share|cite|improve this answer














                  $S$ 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 $text{dim} geq n$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 25 at 7:51

























                  answered Nov 25 at 4:46









                  Anurag A

                  25.6k12249




                  25.6k12249






























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