Finding the span and basis of V.
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
edited Nov 28 '18 at 5:47
Chinnapparaj R
5,2481828
asked Nov 28 '18 at 4:39
Cheryl Cheryl
755
add a comment |
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
edited Nov 28 '18 at 5:47
Chinnapparaj R
5,2481828
asked Nov 28 '18 at 4:39
Cheryl Cheryl
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
add a comment |
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
edited Nov 28 '18 at 5:47
Chinnapparaj R
5,2481828
asked Nov 28 '18 at 4:39
Cheryl Cheryl
755
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
edited Nov 28 '18 at 5:47
Chinnapparaj R
5,2481828
asked Nov 28 '18 at 4:39
Cheryl Cheryl
755
edited Nov 28 '18 at 5:47
Chinnapparaj R
5,2481828
asked Nov 28 '18 at 4:39
Cheryl Cheryl
755
edited Nov 28 '18 at 5:47
Chinnapparaj R
5,2481828
edited Nov 28 '18 at 5:47
Chinnapparaj R
5,2481828
edited Nov 28 '18 at 5:47
Chinnapparaj R
5,2481828
5,2481828
asked Nov 28 '18 at 4:39
Cheryl Cheryl
755
asked Nov 28 '18 at 4:39
Cheryl Cheryl
755
asked Nov 28 '18 at 4:39
Cheryl 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
add a comment |
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
add a comment |
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 $2$.
answered Nov 28 '18 at 5:14
mathnoobmathnoob
1,795422
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
add a comment |
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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 $2$.
answered Nov 28 '18 at 5:14
mathnoobmathnoob
1,795422
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
add a comment |
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 $2$.
answered Nov 28 '18 at 5:14
mathnoobmathnoob
1,795422
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
add a comment |
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 $2$.
answered Nov 28 '18 at 5:14
mathnoobmathnoob
1,795422
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 $2$.
answered Nov 28 '18 at 5:14
mathnoobmathnoob
1,795422
edited Nov 28 '18 at 6:48
answered Nov 28 '18 at 5:14
mathnoobmathnoob
1,795422
answered Nov 28 '18 at 5:14
mathnoobmathnoob
1,795422
answered Nov 28 '18 at 5:14
mathnoobmathnoob
1,795422
1,795422
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
add a comment |
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
add a comment |
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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