Is the set of all vectors in the span of $pmatrix{1 \ 2}$ and $pmatrix{2 \ -1}$ a subspace or not?
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I'm trying to figure out whether or not the set of all vectors in the span of $pmatrix{1 \ 2}$ and $pmatrix{2 \ -1}$ is a subspace or not (I know the answer to be yes, it is a subspace but I want to understand how to prove it).
So far I have:
$U=(1(X_1),2(Y_1))$
$V=(2(X_2),-1(Y_2))$
$U+V=(1(X_1) , 2(Y_1)) + (2(X_2) , -1(Y_2))=(3(X_1+X_2) , 1(Y_1+Y_2))$
Not sure if I'm going down the right path here $dots$
Thanks in advance
linear-algebra vector-spaces invariant-subspace
$endgroup$
add a comment |
$begingroup$
I'm trying to figure out whether or not the set of all vectors in the span of $pmatrix{1 \ 2}$ and $pmatrix{2 \ -1}$ is a subspace or not (I know the answer to be yes, it is a subspace but I want to understand how to prove it).
So far I have:
$U=(1(X_1),2(Y_1))$
$V=(2(X_2),-1(Y_2))$
$U+V=(1(X_1) , 2(Y_1)) + (2(X_2) , -1(Y_2))=(3(X_1+X_2) , 1(Y_1+Y_2))$
Not sure if I'm going down the right path here $dots$
Thanks in advance
linear-algebra vector-spaces invariant-subspace
$endgroup$
2
$begingroup$
Explicitly writing down the definition of the span of a set of vectors will help a ton!
$endgroup$
– wilkersmon
Dec 10 '18 at 21:35
$begingroup$
I'm guessing here you mean it is a subspace of $mathbb{R}^2$ where we consider the standard canonical basis ${e_1,e_2}$ such that by $v= pmatrix{1 \ 2}$ you mean $v= e_1 + 2 e_2$
$endgroup$
– Wesley Strik
Dec 10 '18 at 22:47
add a comment |
$begingroup$
I'm trying to figure out whether or not the set of all vectors in the span of $pmatrix{1 \ 2}$ and $pmatrix{2 \ -1}$ is a subspace or not (I know the answer to be yes, it is a subspace but I want to understand how to prove it).
So far I have:
$U=(1(X_1),2(Y_1))$
$V=(2(X_2),-1(Y_2))$
$U+V=(1(X_1) , 2(Y_1)) + (2(X_2) , -1(Y_2))=(3(X_1+X_2) , 1(Y_1+Y_2))$
Not sure if I'm going down the right path here $dots$
Thanks in advance
linear-algebra vector-spaces invariant-subspace
$endgroup$
I'm trying to figure out whether or not the set of all vectors in the span of $pmatrix{1 \ 2}$ and $pmatrix{2 \ -1}$ is a subspace or not (I know the answer to be yes, it is a subspace but I want to understand how to prove it).
So far I have:
$U=(1(X_1),2(Y_1))$
$V=(2(X_2),-1(Y_2))$
$U+V=(1(X_1) , 2(Y_1)) + (2(X_2) , -1(Y_2))=(3(X_1+X_2) , 1(Y_1+Y_2))$
Not sure if I'm going down the right path here $dots$
Thanks in advance
linear-algebra vector-spaces invariant-subspace
linear-algebra vector-spaces invariant-subspace
edited Dec 10 '18 at 23:07
Wesley Strik
1,858423
1,858423
asked Dec 10 '18 at 21:33
Paul RomanoPaul Romano
111
111
2
$begingroup$
Explicitly writing down the definition of the span of a set of vectors will help a ton!
$endgroup$
– wilkersmon
Dec 10 '18 at 21:35
$begingroup$
I'm guessing here you mean it is a subspace of $mathbb{R}^2$ where we consider the standard canonical basis ${e_1,e_2}$ such that by $v= pmatrix{1 \ 2}$ you mean $v= e_1 + 2 e_2$
$endgroup$
– Wesley Strik
Dec 10 '18 at 22:47
add a comment |
2
$begingroup$
Explicitly writing down the definition of the span of a set of vectors will help a ton!
$endgroup$
– wilkersmon
Dec 10 '18 at 21:35
$begingroup$
I'm guessing here you mean it is a subspace of $mathbb{R}^2$ where we consider the standard canonical basis ${e_1,e_2}$ such that by $v= pmatrix{1 \ 2}$ you mean $v= e_1 + 2 e_2$
$endgroup$
– Wesley Strik
Dec 10 '18 at 22:47
2
2
$begingroup$
Explicitly writing down the definition of the span of a set of vectors will help a ton!
$endgroup$
– wilkersmon
Dec 10 '18 at 21:35
$begingroup$
Explicitly writing down the definition of the span of a set of vectors will help a ton!
$endgroup$
– wilkersmon
Dec 10 '18 at 21:35
$begingroup$
I'm guessing here you mean it is a subspace of $mathbb{R}^2$ where we consider the standard canonical basis ${e_1,e_2}$ such that by $v= pmatrix{1 \ 2}$ you mean $v= e_1 + 2 e_2$
$endgroup$
– Wesley Strik
Dec 10 '18 at 22:47
$begingroup$
I'm guessing here you mean it is a subspace of $mathbb{R}^2$ where we consider the standard canonical basis ${e_1,e_2}$ such that by $v= pmatrix{1 \ 2}$ you mean $v= e_1 + 2 e_2$
$endgroup$
– Wesley Strik
Dec 10 '18 at 22:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Actually, any span will form a subspace by definition of the span, but maybe you have not reached this result yet. We know that the span of two vectors is given by all combinations, say:
$$W= left{ a pmatrix{1 \2}+bpmatrix{2 \-1} big| a, b in mathbb{R} right} $$
Subspaces must satisfy the subspace test, that is a subspace
- Is nonempty (Often we show it contains the zero vector, this is a nice test).
Clearly if we take $a=b=0$ the zero vector is contained in this set.
- Is closed under linear combinations.
You need to check that if we combine two vectors of this form, we again get vectors of this form. Take two arbitrary vectors $v= a_1 pmatrix{1 \2}+b_1pmatrix{2 \-1}$ and $w=a_2 pmatrix{1 \2}+b_2pmatrix{2 \-1}$ in $W$, we now consider:
$$ lambda v + mu w= lambda (a_1 pmatrix{1 \2}+b_1pmatrix{2 \-1}) + mu ( a_2 pmatrix{1 \2}+b_2pmatrix{2 \-1})$$
Now we use the fact that we defined our vectors over a scalar field and we can use the distributive and associative properties together with the properties of vectors (check!):
$$ lambda v + mu w= (lambda a_1 + mu b_1) pmatrix{1 \2}+(lambda a_2 + mu b_2)pmatrix{2 \-1})$$
Which is again of the same form so it is contained in $W$, hence $W$ is closed under linear combinations and is a linear subspace.
If you look closely at this proof you might notice that I do not use anything explicit for the vectors, so it will work for any vector. This is why the span is always a subspace.
$endgroup$
1
$begingroup$
Right on, thank you!
$endgroup$
– Paul Romano
Dec 11 '18 at 16:49
$begingroup$
No problem ;) it's what we're here for.
$endgroup$
– Wesley Strik
Dec 11 '18 at 17:01
add a comment |
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$begingroup$
Actually, any span will form a subspace by definition of the span, but maybe you have not reached this result yet. We know that the span of two vectors is given by all combinations, say:
$$W= left{ a pmatrix{1 \2}+bpmatrix{2 \-1} big| a, b in mathbb{R} right} $$
Subspaces must satisfy the subspace test, that is a subspace
- Is nonempty (Often we show it contains the zero vector, this is a nice test).
Clearly if we take $a=b=0$ the zero vector is contained in this set.
- Is closed under linear combinations.
You need to check that if we combine two vectors of this form, we again get vectors of this form. Take two arbitrary vectors $v= a_1 pmatrix{1 \2}+b_1pmatrix{2 \-1}$ and $w=a_2 pmatrix{1 \2}+b_2pmatrix{2 \-1}$ in $W$, we now consider:
$$ lambda v + mu w= lambda (a_1 pmatrix{1 \2}+b_1pmatrix{2 \-1}) + mu ( a_2 pmatrix{1 \2}+b_2pmatrix{2 \-1})$$
Now we use the fact that we defined our vectors over a scalar field and we can use the distributive and associative properties together with the properties of vectors (check!):
$$ lambda v + mu w= (lambda a_1 + mu b_1) pmatrix{1 \2}+(lambda a_2 + mu b_2)pmatrix{2 \-1})$$
Which is again of the same form so it is contained in $W$, hence $W$ is closed under linear combinations and is a linear subspace.
If you look closely at this proof you might notice that I do not use anything explicit for the vectors, so it will work for any vector. This is why the span is always a subspace.
$endgroup$
1
$begingroup$
Right on, thank you!
$endgroup$
– Paul Romano
Dec 11 '18 at 16:49
$begingroup$
No problem ;) it's what we're here for.
$endgroup$
– Wesley Strik
Dec 11 '18 at 17:01
add a comment |
$begingroup$
Actually, any span will form a subspace by definition of the span, but maybe you have not reached this result yet. We know that the span of two vectors is given by all combinations, say:
$$W= left{ a pmatrix{1 \2}+bpmatrix{2 \-1} big| a, b in mathbb{R} right} $$
Subspaces must satisfy the subspace test, that is a subspace
- Is nonempty (Often we show it contains the zero vector, this is a nice test).
Clearly if we take $a=b=0$ the zero vector is contained in this set.
- Is closed under linear combinations.
You need to check that if we combine two vectors of this form, we again get vectors of this form. Take two arbitrary vectors $v= a_1 pmatrix{1 \2}+b_1pmatrix{2 \-1}$ and $w=a_2 pmatrix{1 \2}+b_2pmatrix{2 \-1}$ in $W$, we now consider:
$$ lambda v + mu w= lambda (a_1 pmatrix{1 \2}+b_1pmatrix{2 \-1}) + mu ( a_2 pmatrix{1 \2}+b_2pmatrix{2 \-1})$$
Now we use the fact that we defined our vectors over a scalar field and we can use the distributive and associative properties together with the properties of vectors (check!):
$$ lambda v + mu w= (lambda a_1 + mu b_1) pmatrix{1 \2}+(lambda a_2 + mu b_2)pmatrix{2 \-1})$$
Which is again of the same form so it is contained in $W$, hence $W$ is closed under linear combinations and is a linear subspace.
If you look closely at this proof you might notice that I do not use anything explicit for the vectors, so it will work for any vector. This is why the span is always a subspace.
$endgroup$
1
$begingroup$
Right on, thank you!
$endgroup$
– Paul Romano
Dec 11 '18 at 16:49
$begingroup$
No problem ;) it's what we're here for.
$endgroup$
– Wesley Strik
Dec 11 '18 at 17:01
add a comment |
$begingroup$
Actually, any span will form a subspace by definition of the span, but maybe you have not reached this result yet. We know that the span of two vectors is given by all combinations, say:
$$W= left{ a pmatrix{1 \2}+bpmatrix{2 \-1} big| a, b in mathbb{R} right} $$
Subspaces must satisfy the subspace test, that is a subspace
- Is nonempty (Often we show it contains the zero vector, this is a nice test).
Clearly if we take $a=b=0$ the zero vector is contained in this set.
- Is closed under linear combinations.
You need to check that if we combine two vectors of this form, we again get vectors of this form. Take two arbitrary vectors $v= a_1 pmatrix{1 \2}+b_1pmatrix{2 \-1}$ and $w=a_2 pmatrix{1 \2}+b_2pmatrix{2 \-1}$ in $W$, we now consider:
$$ lambda v + mu w= lambda (a_1 pmatrix{1 \2}+b_1pmatrix{2 \-1}) + mu ( a_2 pmatrix{1 \2}+b_2pmatrix{2 \-1})$$
Now we use the fact that we defined our vectors over a scalar field and we can use the distributive and associative properties together with the properties of vectors (check!):
$$ lambda v + mu w= (lambda a_1 + mu b_1) pmatrix{1 \2}+(lambda a_2 + mu b_2)pmatrix{2 \-1})$$
Which is again of the same form so it is contained in $W$, hence $W$ is closed under linear combinations and is a linear subspace.
If you look closely at this proof you might notice that I do not use anything explicit for the vectors, so it will work for any vector. This is why the span is always a subspace.
$endgroup$
Actually, any span will form a subspace by definition of the span, but maybe you have not reached this result yet. We know that the span of two vectors is given by all combinations, say:
$$W= left{ a pmatrix{1 \2}+bpmatrix{2 \-1} big| a, b in mathbb{R} right} $$
Subspaces must satisfy the subspace test, that is a subspace
- Is nonempty (Often we show it contains the zero vector, this is a nice test).
Clearly if we take $a=b=0$ the zero vector is contained in this set.
- Is closed under linear combinations.
You need to check that if we combine two vectors of this form, we again get vectors of this form. Take two arbitrary vectors $v= a_1 pmatrix{1 \2}+b_1pmatrix{2 \-1}$ and $w=a_2 pmatrix{1 \2}+b_2pmatrix{2 \-1}$ in $W$, we now consider:
$$ lambda v + mu w= lambda (a_1 pmatrix{1 \2}+b_1pmatrix{2 \-1}) + mu ( a_2 pmatrix{1 \2}+b_2pmatrix{2 \-1})$$
Now we use the fact that we defined our vectors over a scalar field and we can use the distributive and associative properties together with the properties of vectors (check!):
$$ lambda v + mu w= (lambda a_1 + mu b_1) pmatrix{1 \2}+(lambda a_2 + mu b_2)pmatrix{2 \-1})$$
Which is again of the same form so it is contained in $W$, hence $W$ is closed under linear combinations and is a linear subspace.
If you look closely at this proof you might notice that I do not use anything explicit for the vectors, so it will work for any vector. This is why the span is always a subspace.
edited Dec 10 '18 at 22:49
answered Dec 10 '18 at 21:42
Wesley StrikWesley Strik
1,858423
1,858423
1
$begingroup$
Right on, thank you!
$endgroup$
– Paul Romano
Dec 11 '18 at 16:49
$begingroup$
No problem ;) it's what we're here for.
$endgroup$
– Wesley Strik
Dec 11 '18 at 17:01
add a comment |
1
$begingroup$
Right on, thank you!
$endgroup$
– Paul Romano
Dec 11 '18 at 16:49
$begingroup$
No problem ;) it's what we're here for.
$endgroup$
– Wesley Strik
Dec 11 '18 at 17:01
1
1
$begingroup$
Right on, thank you!
$endgroup$
– Paul Romano
Dec 11 '18 at 16:49
$begingroup$
Right on, thank you!
$endgroup$
– Paul Romano
Dec 11 '18 at 16:49
$begingroup$
No problem ;) it's what we're here for.
$endgroup$
– Wesley Strik
Dec 11 '18 at 17:01
$begingroup$
No problem ;) it's what we're here for.
$endgroup$
– Wesley Strik
Dec 11 '18 at 17:01
add a comment |
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$begingroup$
Explicitly writing down the definition of the span of a set of vectors will help a ton!
$endgroup$
– wilkersmon
Dec 10 '18 at 21:35
$begingroup$
I'm guessing here you mean it is a subspace of $mathbb{R}^2$ where we consider the standard canonical basis ${e_1,e_2}$ such that by $v= pmatrix{1 \ 2}$ you mean $v= e_1 + 2 e_2$
$endgroup$
– Wesley Strik
Dec 10 '18 at 22:47