Is the set of all vectors in the span of $pmatrix{1 \ 2}$ and $pmatrix{2 \ -1}$ a subspace or not?












2












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















2












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














2












2








2





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 23:07









Wesley Strik

1,956423




1,956423










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














  • 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










1 Answer
1






active

oldest

votes


















2












$begingroup$

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




  1. 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.




  1. 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.






share|cite|improve this answer











$endgroup$









  • 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











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









2












$begingroup$

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




  1. 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.




  1. 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.






share|cite|improve this answer











$endgroup$









  • 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
















2












$begingroup$

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




  1. 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.




  1. 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.






share|cite|improve this answer











$endgroup$









  • 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














2












2








2





$begingroup$

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




  1. 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.




  1. 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.






share|cite|improve this answer











$endgroup$



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




  1. 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.




  1. 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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 10 '18 at 22:49

























answered Dec 10 '18 at 21:42









Wesley StrikWesley Strik

1,956423




1,956423








  • 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




    $begingroup$
    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


















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