Set of vectors spanning an n - dim vector space implies set is a basis
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Theorem:
If a set of n vectors spans an n-dimensional vector space, then the set is a basis for that vector space.
Attempt:
Let S be a set of n vectors spanning an n-dimensional vector space.
This implies that any vector in the vector space $left(V, R^{n}right)$ is a linear combination of vectors in the set S.
It suffice to show that S is linearly independent.
Suppose that S is linearly dependent.
Then for some vectors $vec{v}_{i}, exists i in mathbb{Z}_{1}^{n}$ in S that may be expressed as a linear combination of some vectors in S, the removal of $vec{v}_{i}$ does not affect the span of set S.
Any hints to bring me forward is highly appreciated.
Thanks in advance.
linear-algebra proof-verification
asked Nov 23 at 14:19
Mathematicing
2,43621851
add a comment |
up vote
0
down vote
favorite
Theorem:
If a set of n vectors spans an n-dimensional vector space, then the set is a basis for that vector space.
Attempt:
Let S be a set of n vectors spanning an n-dimensional vector space.
This implies that any vector in the vector space $left(V, R^{n}right)$ is a linear combination of vectors in the set S.
It suffice to show that S is linearly independent.
Suppose that S is linearly dependent.
Then for some vectors $vec{v}_{i}, exists i in mathbb{Z}_{1}^{n}$ in S that may be expressed as a linear combination of some vectors in S, the removal of $vec{v}_{i}$ does not affect the span of set S.
Any hints to bring me forward is highly appreciated.
Thanks in advance.
linear-algebra proof-verification
asked Nov 23 at 14:19
Mathematicing
2,43621851
1
You're almost done - say $span(S) = span(S setminus lbrace v_1 rbrace)$, then $n = dim(span(S)) = dim(span(S setminus lbrace v_1 rbrace))$, but what is $|S setminus lbrace v_1 rbrace|$?
– Stockfish
Nov 23 at 14:34
@Stockfish And so a contradiction exists for the order of the set S is n is n -1.
– Mathematicing
Nov 23 at 14:40
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Theorem:
If a set of n vectors spans an n-dimensional vector space, then the set is a basis for that vector space.
Attempt:
Let S be a set of n vectors spanning an n-dimensional vector space.
This implies that any vector in the vector space $left(V, R^{n}right)$ is a linear combination of vectors in the set S.
It suffice to show that S is linearly independent.
Suppose that S is linearly dependent.
Then for some vectors $vec{v}_{i}, exists i in mathbb{Z}_{1}^{n}$ in S that may be expressed as a linear combination of some vectors in S, the removal of $vec{v}_{i}$ does not affect the span of set S.
Any hints to bring me forward is highly appreciated.
Thanks in advance.
linear-algebra proof-verification
asked Nov 23 at 14:19
Mathematicing
2,43621851
Theorem:
If a set of n vectors spans an n-dimensional vector space, then the set is a basis for that vector space.
Attempt:
Let S be a set of n vectors spanning an n-dimensional vector space.
This implies that any vector in the vector space $left(V, R^{n}right)$ is a linear combination of vectors in the set S.
It suffice to show that S is linearly independent.
Suppose that S is linearly dependent.
Then for some vectors $vec{v}_{i}, exists i in mathbb{Z}_{1}^{n}$ in S that may be expressed as a linear combination of some vectors in S, the removal of $vec{v}_{i}$ does not affect the span of set S.
Any hints to bring me forward is highly appreciated.
Thanks in advance.
linear-algebra proof-verification
linear-algebra proof-verification
asked Nov 23 at 14:19
Mathematicing
2,43621851
asked Nov 23 at 14:19
Mathematicing
2,43621851
asked Nov 23 at 14:19
Mathematicing
2,43621851
asked Nov 23 at 14:19
Mathematicing
2,43621851
asked Nov 23 at 14:19
Mathematicing
2,43621851
2,43621851
1
You're almost done - say $span(S) = span(S setminus lbrace v_1 rbrace)$, then $n = dim(span(S)) = dim(span(S setminus lbrace v_1 rbrace))$, but what is $|S setminus lbrace v_1 rbrace|$?
– Stockfish
Nov 23 at 14:34
@Stockfish And so a contradiction exists for the order of the set S is n is n -1.
– Mathematicing
Nov 23 at 14:40
add a comment |
1
You're almost done - say $span(S) = span(S setminus lbrace v_1 rbrace)$, then $n = dim(span(S)) = dim(span(S setminus lbrace v_1 rbrace))$, but what is $|S setminus lbrace v_1 rbrace|$?
– Stockfish
Nov 23 at 14:34
@Stockfish And so a contradiction exists for the order of the set S is n is n -1.
– Mathematicing
Nov 23 at 14:40
1
1
You're almost done - say $span(S) = span(S setminus lbrace v_1 rbrace)$, then $n = dim(span(S)) = dim(span(S setminus lbrace v_1 rbrace))$, but what is $|S setminus lbrace v_1 rbrace|$?
– Stockfish
Nov 23 at 14:34
You're almost done - say $span(S) = span(S setminus lbrace v_1 rbrace)$, then $n = dim(span(S)) = dim(span(S setminus lbrace v_1 rbrace))$, but what is $|S setminus lbrace v_1 rbrace|$?
– Stockfish
Nov 23 at 14:34
@Stockfish And so a contradiction exists for the order of the set S is n is n -1.
– Mathematicing
Nov 23 at 14:40
@Stockfish And so a contradiction exists for the order of the set S is n is n -1.
– Mathematicing
Nov 23 at 14:40
add a comment |
2 Answers
2
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oldest
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up vote
3
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accepted
Let $S={x_1,x_2,ldots,x_n}$ be a set spanning the vector space $V$ of dimension $n$. Suppose $S$ is not a basis of $V$. Then $S$ is linearly dependent. Thus there exists $x_iin S$ such that $x_i$ is a linear combination of remaining $n-1$ vectors as $Ssetminus {x_i}$ also spans $V$. Thus $n-1$ vectors can span $V$, which is a contradiction. Hence $S$ is a basis.
answered Nov 23 at 14:37
Anupam
2,3701823
1
Which is a contradiction or not, depending on what previous results we're allowed to use...
– David C. Ullrich
Nov 23 at 14:39
Up for the effort
– Mathematicing
Nov 23 at 14:41
add a comment |
up vote
1
down vote
It's impossible to say what's a correct solution without knowing what previously proved results you're allowed to use (for example, the other answer is fine if you're given that $n-1$ vectors cannot span $V$.)
Since the word "dimension" appears it seems reasonable to conjecture that you've already shown that any basis for $V$ must have exactly $n$ elements; this is required before the typical definition of "dimension" makes sense. Assuming that it's an easy exercise:
Show that $S$ contains a minimal spanning set: There exists $S_1subset S$ such that $S_1$ spans $V$ and if $S_2subset S_1$ and $S_2$ spans $V$ then $S_2=S_1$. Show that it follows that $S_1$ is independent.
So $S_1$ is a basis. Hence that previous result show that $S_1$ contains exactly $n$ elements; since $S_1subset S$ this shows that $S_1=S$.
answered Nov 23 at 14:47
David C. Ullrich
57.6k43891
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Let $S={x_1,x_2,ldots,x_n}$ be a set spanning the vector space $V$ of dimension $n$. Suppose $S$ is not a basis of $V$. Then $S$ is linearly dependent. Thus there exists $x_iin S$ such that $x_i$ is a linear combination of remaining $n-1$ vectors as $Ssetminus {x_i}$ also spans $V$. Thus $n-1$ vectors can span $V$, which is a contradiction. Hence $S$ is a basis.
answered Nov 23 at 14:37
Anupam
2,3701823
1
Which is a contradiction or not, depending on what previous results we're allowed to use...
– David C. Ullrich
Nov 23 at 14:39
Up for the effort
– Mathematicing
Nov 23 at 14:41
add a comment |
up vote
3
down vote
accepted
Let $S={x_1,x_2,ldots,x_n}$ be a set spanning the vector space $V$ of dimension $n$. Suppose $S$ is not a basis of $V$. Then $S$ is linearly dependent. Thus there exists $x_iin S$ such that $x_i$ is a linear combination of remaining $n-1$ vectors as $Ssetminus {x_i}$ also spans $V$. Thus $n-1$ vectors can span $V$, which is a contradiction. Hence $S$ is a basis.
answered Nov 23 at 14:37
Anupam
2,3701823
1
Which is a contradiction or not, depending on what previous results we're allowed to use...
– David C. Ullrich
Nov 23 at 14:39
Up for the effort
– Mathematicing
Nov 23 at 14:41
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Let $S={x_1,x_2,ldots,x_n}$ be a set spanning the vector space $V$ of dimension $n$. Suppose $S$ is not a basis of $V$. Then $S$ is linearly dependent. Thus there exists $x_iin S$ such that $x_i$ is a linear combination of remaining $n-1$ vectors as $Ssetminus {x_i}$ also spans $V$. Thus $n-1$ vectors can span $V$, which is a contradiction. Hence $S$ is a basis.
answered Nov 23 at 14:37
Anupam
2,3701823
Let $S={x_1,x_2,ldots,x_n}$ be a set spanning the vector space $V$ of dimension $n$. Suppose $S$ is not a basis of $V$. Then $S$ is linearly dependent. Thus there exists $x_iin S$ such that $x_i$ is a linear combination of remaining $n-1$ vectors as $Ssetminus {x_i}$ also spans $V$. Thus $n-1$ vectors can span $V$, which is a contradiction. Hence $S$ is a basis.
answered Nov 23 at 14:37
Anupam
2,3701823
answered Nov 23 at 14:37
Anupam
2,3701823
answered Nov 23 at 14:37
Anupam
2,3701823
answered Nov 23 at 14:37
Anupam
2,3701823
2,3701823
1
Which is a contradiction or not, depending on what previous results we're allowed to use...
– David C. Ullrich
Nov 23 at 14:39
Up for the effort
– Mathematicing
Nov 23 at 14:41
add a comment |
1
Which is a contradiction or not, depending on what previous results we're allowed to use...
– David C. Ullrich
Nov 23 at 14:39
Up for the effort
– Mathematicing
Nov 23 at 14:41
1
1
Which is a contradiction or not, depending on what previous results we're allowed to use...
– David C. Ullrich
Nov 23 at 14:39
Which is a contradiction or not, depending on what previous results we're allowed to use...
– David C. Ullrich
Nov 23 at 14:39
Up for the effort
– Mathematicing
Nov 23 at 14:41
Up for the effort
– Mathematicing
Nov 23 at 14:41
add a comment |
up vote
1
down vote
It's impossible to say what's a correct solution without knowing what previously proved results you're allowed to use (for example, the other answer is fine if you're given that $n-1$ vectors cannot span $V$.)
Since the word "dimension" appears it seems reasonable to conjecture that you've already shown that any basis for $V$ must have exactly $n$ elements; this is required before the typical definition of "dimension" makes sense. Assuming that it's an easy exercise:
Show that $S$ contains a minimal spanning set: There exists $S_1subset S$ such that $S_1$ spans $V$ and if $S_2subset S_1$ and $S_2$ spans $V$ then $S_2=S_1$. Show that it follows that $S_1$ is independent.
So $S_1$ is a basis. Hence that previous result show that $S_1$ contains exactly $n$ elements; since $S_1subset S$ this shows that $S_1=S$.
answered Nov 23 at 14:47
David C. Ullrich
57.6k43891
add a comment |
up vote
1
down vote
It's impossible to say what's a correct solution without knowing what previously proved results you're allowed to use (for example, the other answer is fine if you're given that $n-1$ vectors cannot span $V$.)
Since the word "dimension" appears it seems reasonable to conjecture that you've already shown that any basis for $V$ must have exactly $n$ elements; this is required before the typical definition of "dimension" makes sense. Assuming that it's an easy exercise:
Show that $S$ contains a minimal spanning set: There exists $S_1subset S$ such that $S_1$ spans $V$ and if $S_2subset S_1$ and $S_2$ spans $V$ then $S_2=S_1$. Show that it follows that $S_1$ is independent.
So $S_1$ is a basis. Hence that previous result show that $S_1$ contains exactly $n$ elements; since $S_1subset S$ this shows that $S_1=S$.
answered Nov 23 at 14:47
David C. Ullrich
57.6k43891
add a comment |
up vote
1
down vote
up vote
1
down vote
It's impossible to say what's a correct solution without knowing what previously proved results you're allowed to use (for example, the other answer is fine if you're given that $n-1$ vectors cannot span $V$.)
Since the word "dimension" appears it seems reasonable to conjecture that you've already shown that any basis for $V$ must have exactly $n$ elements; this is required before the typical definition of "dimension" makes sense. Assuming that it's an easy exercise:
Show that $S$ contains a minimal spanning set: There exists $S_1subset S$ such that $S_1$ spans $V$ and if $S_2subset S_1$ and $S_2$ spans $V$ then $S_2=S_1$. Show that it follows that $S_1$ is independent.
So $S_1$ is a basis. Hence that previous result show that $S_1$ contains exactly $n$ elements; since $S_1subset S$ this shows that $S_1=S$.
answered Nov 23 at 14:47
David C. Ullrich
57.6k43891
It's impossible to say what's a correct solution without knowing what previously proved results you're allowed to use (for example, the other answer is fine if you're given that $n-1$ vectors cannot span $V$.)
Since the word "dimension" appears it seems reasonable to conjecture that you've already shown that any basis for $V$ must have exactly $n$ elements; this is required before the typical definition of "dimension" makes sense. Assuming that it's an easy exercise:
Show that $S$ contains a minimal spanning set: There exists $S_1subset S$ such that $S_1$ spans $V$ and if $S_2subset S_1$ and $S_2$ spans $V$ then $S_2=S_1$. Show that it follows that $S_1$ is independent.
So $S_1$ is a basis. Hence that previous result show that $S_1$ contains exactly $n$ elements; since $S_1subset S$ this shows that $S_1=S$.
answered Nov 23 at 14:47
David C. Ullrich
57.6k43891
answered Nov 23 at 14:47
David C. Ullrich
57.6k43891
answered Nov 23 at 14:47
David C. Ullrich
57.6k43891
answered Nov 23 at 14:47
David C. Ullrich
57.6k43891
57.6k43891
add a comment |
add a comment |
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You're almost done - say $span(S) = span(S setminus lbrace v_1 rbrace)$, then $n = dim(span(S)) = dim(span(S setminus lbrace v_1 rbrace))$, but what is $|S setminus lbrace v_1 rbrace|$?
– Stockfish
Nov 23 at 14:34
@Stockfish And so a contradiction exists for the order of the set S is n is n -1.
– Mathematicing
Nov 23 at 14:40