Finding the span and basis of V.












1















Let $V ={(x, y, z, w)|x=y+z, w=2y}$ be a subspace of $Bbb{R}^4$.



(i) Write down an explicit form of a general vector of $V$.



(ii) Express $V$ in linear span form.



(iii) Write down a basis for $V$ and find $dim V$.




To answer (i), I managed to get $Big{l(1,0,1,0) + t(1/2,1/2,0,1)|t, l in Bbb{R}Big}$. Which brings me to the next part (ii) and (iii), is it right to say that both the basis and span of $V$ is



$V= {(1,0,1,0),(1/2,1/2,0,1)}$?






linear-algebra matrices change-of-basis







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  • 2




    The span is not the basis, but all linear combinations of the basis (sums of constant times basis element).
    – coffeemath
    Nov 28 '18 at 4:47










  • $left{begin{bmatrix}1\0end{bmatrix}, , begin{bmatrix}0\1end{bmatrix}right}$ is a basis of $Bbb{R}^2$ but the span of these vectors is entire $Bbb{R}^2$. So basis and span are not the same thing.
    – Anurag A
    Nov 28 '18 at 4:58
















1















Let $V ={(x, y, z, w)|x=y+z, w=2y}$ be a subspace of $Bbb{R}^4$.



(i) Write down an explicit form of a general vector of $V$.



(ii) Express $V$ in linear span form.



(iii) Write down a basis for $V$ and find $dim V$.




To answer (i), I managed to get $Big{l(1,0,1,0) + t(1/2,1/2,0,1)|t, l in Bbb{R}Big}$. Which brings me to the next part (ii) and (iii), is it right to say that both the basis and span of $V$ is



$V= {(1,0,1,0),(1/2,1/2,0,1)}$?






linear-algebra matrices change-of-basis







share|cite|improve this question




















  • 2




    The span is not the basis, but all linear combinations of the basis (sums of constant times basis element).
    – coffeemath
    Nov 28 '18 at 4:47










  • $left{begin{bmatrix}1\0end{bmatrix}, , begin{bmatrix}0\1end{bmatrix}right}$ is a basis of $Bbb{R}^2$ but the span of these vectors is entire $Bbb{R}^2$. So basis and span are not the same thing.
    – Anurag A
    Nov 28 '18 at 4:58














1












1








1


0






Let $V ={(x, y, z, w)|x=y+z, w=2y}$ be a subspace of $Bbb{R}^4$.



(i) Write down an explicit form of a general vector of $V$.



(ii) Express $V$ in linear span form.



(iii) Write down a basis for $V$ and find $dim V$.




To answer (i), I managed to get $Big{l(1,0,1,0) + t(1/2,1/2,0,1)|t, l in Bbb{R}Big}$. Which brings me to the next part (ii) and (iii), is it right to say that both the basis and span of $V$ is



$V= {(1,0,1,0),(1/2,1/2,0,1)}$?






linear-algebra matrices change-of-basis







share|cite|improve this question
















Let $V ={(x, y, z, w)|x=y+z, w=2y}$ be a subspace of $Bbb{R}^4$.



(i) Write down an explicit form of a general vector of $V$.



(ii) Express $V$ in linear span form.



(iii) Write down a basis for $V$ and find $dim V$.




To answer (i), I managed to get $Big{l(1,0,1,0) + t(1/2,1/2,0,1)|t, l in Bbb{R}Big}$. Which brings me to the next part (ii) and (iii), is it right to say that both the basis and span of $V$ is



$V= {(1,0,1,0),(1/2,1/2,0,1)}$?






linear-algebra matrices change-of-basis




linear-algebra matrices change-of-basis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Nov 28 '18 at 5:47









Chinnapparaj R

5,3051826




5,3051826










asked Nov 28 '18 at 4:39









Cheryl

755




755








  • 2




    The span is not the basis, but all linear combinations of the basis (sums of constant times basis element).
    – coffeemath
    Nov 28 '18 at 4:47










  • $left{begin{bmatrix}1\0end{bmatrix}, , begin{bmatrix}0\1end{bmatrix}right}$ is a basis of $Bbb{R}^2$ but the span of these vectors is entire $Bbb{R}^2$. So basis and span are not the same thing.
    – Anurag A
    Nov 28 '18 at 4:58














  • 2




    The span is not the basis, but all linear combinations of the basis (sums of constant times basis element).
    – coffeemath
    Nov 28 '18 at 4:47










  • $left{begin{bmatrix}1\0end{bmatrix}, , begin{bmatrix}0\1end{bmatrix}right}$ is a basis of $Bbb{R}^2$ but the span of these vectors is entire $Bbb{R}^2$. So basis and span are not the same thing.
    – Anurag A
    Nov 28 '18 at 4:58








2




2




The span is not the basis, but all linear combinations of the basis (sums of constant times basis element).
– coffeemath
Nov 28 '18 at 4:47




The span is not the basis, but all linear combinations of the basis (sums of constant times basis element).
– coffeemath
Nov 28 '18 at 4:47












$left{begin{bmatrix}1\0end{bmatrix}, , begin{bmatrix}0\1end{bmatrix}right}$ is a basis of $Bbb{R}^2$ but the span of these vectors is entire $Bbb{R}^2$. So basis and span are not the same thing.
– Anurag A
Nov 28 '18 at 4:58




$left{begin{bmatrix}1\0end{bmatrix}, , begin{bmatrix}0\1end{bmatrix}right}$ is a basis of $Bbb{R}^2$ but the span of these vectors is entire $Bbb{R}^2$. So basis and span are not the same thing.
– Anurag A
Nov 28 '18 at 4:58










1 Answer
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Let $y=t,z=s$, then we can then deduce that vectors in $V$ are of this form: $$Big{(t+s,t,s,2t)=t(1,1,0,2)+s(1,0,1,0);:;s,tin mathbb{R}Big}$$ From that the two vectors $(1,1,0,2),(1,0,1,0)$ are linearly independent and form a basis for $V$. So dimension of $V$ is $2$.






share|cite|improve this answer























  • I would replace the "we can see" part by "since the two vectors are linearly independent, they form a basis of V".
    – Dirk
    Nov 28 '18 at 5:57










  • yes, I edited it!
    – mathnoob
    Nov 28 '18 at 6:48











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1 Answer
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Let $y=t,z=s$, then we can then deduce that vectors in $V$ are of this form: $$Big{(t+s,t,s,2t)=t(1,1,0,2)+s(1,0,1,0);:;s,tin mathbb{R}Big}$$ From that the two vectors $(1,1,0,2),(1,0,1,0)$ are linearly independent and form a basis for $V$. So dimension of $V$ is $2$.






share|cite|improve this answer























  • I would replace the "we can see" part by "since the two vectors are linearly independent, they form a basis of V".
    – Dirk
    Nov 28 '18 at 5:57










  • yes, I edited it!
    – mathnoob
    Nov 28 '18 at 6:48
















1














Let $y=t,z=s$, then we can then deduce that vectors in $V$ are of this form: $$Big{(t+s,t,s,2t)=t(1,1,0,2)+s(1,0,1,0);:;s,tin mathbb{R}Big}$$ From that the two vectors $(1,1,0,2),(1,0,1,0)$ are linearly independent and form a basis for $V$. So dimension of $V$ is $2$.






share|cite|improve this answer























  • I would replace the "we can see" part by "since the two vectors are linearly independent, they form a basis of V".
    – Dirk
    Nov 28 '18 at 5:57










  • yes, I edited it!
    – mathnoob
    Nov 28 '18 at 6:48














1












1








1






Let $y=t,z=s$, then we can then deduce that vectors in $V$ are of this form: $$Big{(t+s,t,s,2t)=t(1,1,0,2)+s(1,0,1,0);:;s,tin mathbb{R}Big}$$ From that the two vectors $(1,1,0,2),(1,0,1,0)$ are linearly independent and form a basis for $V$. So dimension of $V$ is $2$.






share|cite|improve this answer














Let $y=t,z=s$, then we can then deduce that vectors in $V$ are of this form: $$Big{(t+s,t,s,2t)=t(1,1,0,2)+s(1,0,1,0);:;s,tin mathbb{R}Big}$$ From that the two vectors $(1,1,0,2),(1,0,1,0)$ are linearly independent and form a basis for $V$. So dimension of $V$ is $2$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 28 '18 at 6:48

























answered Nov 28 '18 at 5:14









mathnoob

1,797422




1,797422












  • I would replace the "we can see" part by "since the two vectors are linearly independent, they form a basis of V".
    – Dirk
    Nov 28 '18 at 5:57










  • yes, I edited it!
    – mathnoob
    Nov 28 '18 at 6:48


















  • I would replace the "we can see" part by "since the two vectors are linearly independent, they form a basis of V".
    – Dirk
    Nov 28 '18 at 5:57










  • yes, I edited it!
    – mathnoob
    Nov 28 '18 at 6:48
















I would replace the "we can see" part by "since the two vectors are linearly independent, they form a basis of V".
– Dirk
Nov 28 '18 at 5:57




I would replace the "we can see" part by "since the two vectors are linearly independent, they form a basis of V".
– Dirk
Nov 28 '18 at 5:57












yes, I edited it!
– mathnoob
Nov 28 '18 at 6:48




yes, I edited it!
– mathnoob
Nov 28 '18 at 6:48


















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