Will the span of vectors (1,0,0,0,0),(0,1,0,0,0) form a subspace/basis in the set of fields of dimension 5?What is the dimension of the subspace of $P_2 $ given by span $2 + x^2, 4-2x+3x^2, 1+x?$Span and Dimension: A subspaceIs the set a basis for the SubspaceIs $span(v_1, . . . ,v_m)$ a linearly dependent or linearly independent set of vectors? Also, what will happen if we take span of span?Dimension of Basis of SubspaceProve that centering vectors reduces the span by 1 dimensionQuestion on whether certain vectors span a subspaceBasis and vectors spanHow is dimension of a span of a set of vectors related to the number of vectors in the set?Exercise to determine the subspace of $mathbb R^4$ that these 4 vectors span
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Will the span of vectors (1,0,0,0,0),(0,1,0,0,0) form a subspace/basis in the set of fields of dimension 5?
What is the dimension of the subspace of $P_2 $ given by span $2 + x^2, 4-2x+3x^2, 1+x?$Span and Dimension: A subspaceIs the set a basis for the SubspaceIs $span(v_1, . . . ,v_m)$ a linearly dependent or linearly independent set of vectors? Also, what will happen if we take span of span?Dimension of Basis of SubspaceProve that centering vectors reduces the span by 1 dimensionQuestion on whether certain vectors span a subspaceBasis and vectors spanHow is dimension of a span of a set of vectors related to the number of vectors in the set?Exercise to determine the subspace of $mathbb R^4$ that these 4 vectors span
$begingroup$
I have checked a calculator website which checks if a set of vectors is a basis and the one that I put in the title is not. (http://www.mathforyou.net/en/online/vectors/basis/)
By the subspace definition, the vector 0 is contained into the span, 0*(1,0,0,0,0)+0*(0,1,0,0,0), and we can get any other vector from the span with just the independent vectors (1,0,0,0,0) and (0,1,0,0,0) and this span must be a subspace and basis too.
Please tell me if I m doing anything wrong or I misunderstand any concept.
linear-algebra hamel-basis
edited Mar 28 at 16:19
J. W. Tanner
4,2761320
asked Mar 28 at 15:59
ValVal
557
$endgroup$
add a comment |
$begingroup$
I have checked a calculator website which checks if a set of vectors is a basis and the one that I put in the title is not. (http://www.mathforyou.net/en/online/vectors/basis/)
By the subspace definition, the vector 0 is contained into the span, 0*(1,0,0,0,0)+0*(0,1,0,0,0), and we can get any other vector from the span with just the independent vectors (1,0,0,0,0) and (0,1,0,0,0) and this span must be a subspace and basis too.
Please tell me if I m doing anything wrong or I misunderstand any concept.
linear-algebra hamel-basis
edited Mar 28 at 16:19
J. W. Tanner
4,2761320
asked Mar 28 at 15:59
ValVal
557
$endgroup$
add a comment |
$begingroup$
I have checked a calculator website which checks if a set of vectors is a basis and the one that I put in the title is not. (http://www.mathforyou.net/en/online/vectors/basis/)
By the subspace definition, the vector 0 is contained into the span, 0*(1,0,0,0,0)+0*(0,1,0,0,0), and we can get any other vector from the span with just the independent vectors (1,0,0,0,0) and (0,1,0,0,0) and this span must be a subspace and basis too.
Please tell me if I m doing anything wrong or I misunderstand any concept.
linear-algebra hamel-basis
edited Mar 28 at 16:19
J. W. Tanner
4,2761320
asked Mar 28 at 15:59
ValVal
557
$endgroup$
I have checked a calculator website which checks if a set of vectors is a basis and the one that I put in the title is not. (http://www.mathforyou.net/en/online/vectors/basis/)
By the subspace definition, the vector 0 is contained into the span, 0*(1,0,0,0,0)+0*(0,1,0,0,0), and we can get any other vector from the span with just the independent vectors (1,0,0,0,0) and (0,1,0,0,0) and this span must be a subspace and basis too.
Please tell me if I m doing anything wrong or I misunderstand any concept.
linear-algebra hamel-basis
linear-algebra hamel-basis
edited Mar 28 at 16:19
J. W. Tanner
4,2761320
asked Mar 28 at 15:59
ValVal
557
edited Mar 28 at 16:19
J. W. Tanner
4,2761320
asked Mar 28 at 15:59
ValVal
557
edited Mar 28 at 16:19
J. W. Tanner
4,2761320
edited Mar 28 at 16:19
J. W. Tanner
4,2761320
edited Mar 28 at 16:19
J. W. Tanner
4,2761320
4,2761320
asked Mar 28 at 15:59
ValVal
557
asked Mar 28 at 15:59
ValVal
557
asked Mar 28 at 15:59
ValVal
557
557
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The span of those vectors is indeed a subspace of dimension $2$ of $mathbb R^5$, whose dimension is $5$. So, those two vectors do not form a basis of $mathbb R^5$. That's all.
answered Mar 28 at 16:02
José Carlos SantosJosé Carlos Santos
172k23132240
$endgroup$
$begingroup$
@Santos Yes they do not form a basis of $mathbb R^5$, but the span is itself a basis. Is that statement right?
$endgroup$
– Val
Mar 28 at 16:14
$begingroup$
No. It is wrong. The vectors form a basis, but their span is not a basis.
$endgroup$
– José Carlos Santos
Mar 28 at 16:15
$begingroup$
@Santos But the vectors are independent and they span the subspace, why wouldn't they be a basis. Could you please give me the basis of this span then?
$endgroup$
– Val
Mar 28 at 16:19
2
$begingroup$
The correct definition of a basis is a set of vectors that span the whole space and it is a maximal set of linearly independent vectors. Your example is a set of linearly independent vectors (not maximal), and it does not span the whole space.
$endgroup$
– Oscar
Mar 28 at 16:19
$begingroup$
Oh ok thank you I understand it now!
$endgroup$
– Val
Mar 28 at 16:20
|
show 1 more comment
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1 Answer
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1 Answer
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$begingroup$
The span of those vectors is indeed a subspace of dimension $2$ of $mathbb R^5$, whose dimension is $5$. So, those two vectors do not form a basis of $mathbb R^5$. That's all.
answered Mar 28 at 16:02
José Carlos SantosJosé Carlos Santos
172k23132240
$endgroup$
$begingroup$
@Santos Yes they do not form a basis of $mathbb R^5$, but the span is itself a basis. Is that statement right?
$endgroup$
– Val
Mar 28 at 16:14
$begingroup$
No. It is wrong. The vectors form a basis, but their span is not a basis.
$endgroup$
– José Carlos Santos
Mar 28 at 16:15
$begingroup$
@Santos But the vectors are independent and they span the subspace, why wouldn't they be a basis. Could you please give me the basis of this span then?
$endgroup$
– Val
Mar 28 at 16:19
2
$begingroup$
The correct definition of a basis is a set of vectors that span the whole space and it is a maximal set of linearly independent vectors. Your example is a set of linearly independent vectors (not maximal), and it does not span the whole space.
$endgroup$
– Oscar
Mar 28 at 16:19
$begingroup$
Oh ok thank you I understand it now!
$endgroup$
– Val
Mar 28 at 16:20
|
show 1 more comment
$begingroup$
The span of those vectors is indeed a subspace of dimension $2$ of $mathbb R^5$, whose dimension is $5$. So, those two vectors do not form a basis of $mathbb R^5$. That's all.
answered Mar 28 at 16:02
José Carlos SantosJosé Carlos Santos
172k23132240
$endgroup$
$begingroup$
@Santos Yes they do not form a basis of $mathbb R^5$, but the span is itself a basis. Is that statement right?
$endgroup$
– Val
Mar 28 at 16:14
$begingroup$
No. It is wrong. The vectors form a basis, but their span is not a basis.
$endgroup$
– José Carlos Santos
Mar 28 at 16:15
$begingroup$
@Santos But the vectors are independent and they span the subspace, why wouldn't they be a basis. Could you please give me the basis of this span then?
$endgroup$
– Val
Mar 28 at 16:19
2
$begingroup$
The correct definition of a basis is a set of vectors that span the whole space and it is a maximal set of linearly independent vectors. Your example is a set of linearly independent vectors (not maximal), and it does not span the whole space.
$endgroup$
– Oscar
Mar 28 at 16:19
$begingroup$
Oh ok thank you I understand it now!
$endgroup$
– Val
Mar 28 at 16:20
|
show 1 more comment
$begingroup$
The span of those vectors is indeed a subspace of dimension $2$ of $mathbb R^5$, whose dimension is $5$. So, those two vectors do not form a basis of $mathbb R^5$. That's all.
answered Mar 28 at 16:02
José Carlos SantosJosé Carlos Santos
172k23132240
$endgroup$
The span of those vectors is indeed a subspace of dimension $2$ of $mathbb R^5$, whose dimension is $5$. So, those two vectors do not form a basis of $mathbb R^5$. That's all.
answered Mar 28 at 16:02
José Carlos SantosJosé Carlos Santos
172k23132240
answered Mar 28 at 16:02
José Carlos SantosJosé Carlos Santos
172k23132240
answered Mar 28 at 16:02
José Carlos SantosJosé Carlos Santos
172k23132240
answered Mar 28 at 16:02
José Carlos SantosJosé Carlos Santos
172k23132240
172k23132240
$begingroup$
@Santos Yes they do not form a basis of $mathbb R^5$, but the span is itself a basis. Is that statement right?
$endgroup$
– Val
Mar 28 at 16:14
$begingroup$
No. It is wrong. The vectors form a basis, but their span is not a basis.
$endgroup$
– José Carlos Santos
Mar 28 at 16:15
$begingroup$
@Santos But the vectors are independent and they span the subspace, why wouldn't they be a basis. Could you please give me the basis of this span then?
$endgroup$
– Val
Mar 28 at 16:19
2
$begingroup$
The correct definition of a basis is a set of vectors that span the whole space and it is a maximal set of linearly independent vectors. Your example is a set of linearly independent vectors (not maximal), and it does not span the whole space.
$endgroup$
– Oscar
Mar 28 at 16:19
$begingroup$
Oh ok thank you I understand it now!
$endgroup$
– Val
Mar 28 at 16:20
|
show 1 more comment
$begingroup$
@Santos Yes they do not form a basis of $mathbb R^5$, but the span is itself a basis. Is that statement right?
$endgroup$
– Val
Mar 28 at 16:14
$begingroup$
No. It is wrong. The vectors form a basis, but their span is not a basis.
$endgroup$
– José Carlos Santos
Mar 28 at 16:15
$begingroup$
@Santos But the vectors are independent and they span the subspace, why wouldn't they be a basis. Could you please give me the basis of this span then?
$endgroup$
– Val
Mar 28 at 16:19
2
$begingroup$
The correct definition of a basis is a set of vectors that span the whole space and it is a maximal set of linearly independent vectors. Your example is a set of linearly independent vectors (not maximal), and it does not span the whole space.
$endgroup$
– Oscar
Mar 28 at 16:19
$begingroup$
Oh ok thank you I understand it now!
$endgroup$
– Val
Mar 28 at 16:20
$begingroup$
@Santos Yes they do not form a basis of $mathbb R^5$, but the span is itself a basis. Is that statement right?
$endgroup$
– Val
Mar 28 at 16:14
$begingroup$
@Santos Yes they do not form a basis of $mathbb R^5$, but the span is itself a basis. Is that statement right?
$endgroup$
– Val
Mar 28 at 16:14
$begingroup$
No. It is wrong. The vectors form a basis, but their span is not a basis.
$endgroup$
– José Carlos Santos
Mar 28 at 16:15
$begingroup$
No. It is wrong. The vectors form a basis, but their span is not a basis.
$endgroup$
– José Carlos Santos
Mar 28 at 16:15
$begingroup$
@Santos But the vectors are independent and they span the subspace, why wouldn't they be a basis. Could you please give me the basis of this span then?
$endgroup$
– Val
Mar 28 at 16:19
$begingroup$
@Santos But the vectors are independent and they span the subspace, why wouldn't they be a basis. Could you please give me the basis of this span then?
$endgroup$
– Val
Mar 28 at 16:19
2
2
$begingroup$
The correct definition of a basis is a set of vectors that span the whole space and it is a maximal set of linearly independent vectors. Your example is a set of linearly independent vectors (not maximal), and it does not span the whole space.
$endgroup$
– Oscar
Mar 28 at 16:19
$begingroup$
The correct definition of a basis is a set of vectors that span the whole space and it is a maximal set of linearly independent vectors. Your example is a set of linearly independent vectors (not maximal), and it does not span the whole space.
$endgroup$
– Oscar
Mar 28 at 16:19
$begingroup$
Oh ok thank you I understand it now!
$endgroup$
– Val
Mar 28 at 16:20
$begingroup$
Oh ok thank you I understand it now!
$endgroup$
– Val
Mar 28 at 16:20
|
show 1 more comment
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