How to show that vector $b$ is in the vector space $V$? Announcing the arrival of Valued...
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How to show that vector $b$ is in the vector space $V$?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Showing vectors span a vector space by definitionLinear independence of vectors $w_1,w_2,w_3$How to show that $T$ is linearly independent?Proof to show that sums of vectors spanning a vector space also span a vector spaceShow linear independence of vector additionspan between vector and set of vectorsSuppose that $v_1,v_2,v_3,v_4$ spans $V$. Prove that the list $v_1 - v_2, v_2 - v_3, v_3-v_4,v_4$ also spans $V$.Simplified vector equation and geometric representationFind the dual space and dual basis of basis $beta$ in vector space $mathbb{C^3}$Span of a Vector Space
$begingroup$
How do I show that vector $b$ is in the vector space $V$?
vector $b = (0, 4, 7)$, $V = operatorname{span} {(1, 2, 2), (1, 1, 1), (-1, 0, 1)}$
Do I use the equation $b = c_1 v_1 +c_2 v_2 +c_3 v_3$ to solve for each $v_1$, $v_2$, $v_3$? If I do so, what do I do from then?
linear-algebra
edited Mar 23 at 22:00
Rócherz
3,0263823
asked Mar 23 at 21:37
devil0108devil0108
174
$endgroup$
add a comment |
$begingroup$
How do I show that vector $b$ is in the vector space $V$?
vector $b = (0, 4, 7)$, $V = operatorname{span} {(1, 2, 2), (1, 1, 1), (-1, 0, 1)}$
Do I use the equation $b = c_1 v_1 +c_2 v_2 +c_3 v_3$ to solve for each $v_1$, $v_2$, $v_3$? If I do so, what do I do from then?
linear-algebra
edited Mar 23 at 22:00
Rócherz
3,0263823
asked Mar 23 at 21:37
devil0108devil0108
174
$endgroup$
1
$begingroup$
You mean, you solve for $c_1, c_2, c_3$.
$endgroup$
– lcv
Mar 23 at 21:40
add a comment |
$begingroup$
How do I show that vector $b$ is in the vector space $V$?
vector $b = (0, 4, 7)$, $V = operatorname{span} {(1, 2, 2), (1, 1, 1), (-1, 0, 1)}$
Do I use the equation $b = c_1 v_1 +c_2 v_2 +c_3 v_3$ to solve for each $v_1$, $v_2$, $v_3$? If I do so, what do I do from then?
linear-algebra
edited Mar 23 at 22:00
Rócherz
3,0263823
asked Mar 23 at 21:37
devil0108devil0108
174
$endgroup$
How do I show that vector $b$ is in the vector space $V$?
vector $b = (0, 4, 7)$, $V = operatorname{span} {(1, 2, 2), (1, 1, 1), (-1, 0, 1)}$
Do I use the equation $b = c_1 v_1 +c_2 v_2 +c_3 v_3$ to solve for each $v_1$, $v_2$, $v_3$? If I do so, what do I do from then?
linear-algebra
linear-algebra
edited Mar 23 at 22:00
Rócherz
3,0263823
asked Mar 23 at 21:37
devil0108devil0108
174
edited Mar 23 at 22:00
Rócherz
3,0263823
asked Mar 23 at 21:37
devil0108devil0108
174
edited Mar 23 at 22:00
Rócherz
3,0263823
edited Mar 23 at 22:00
Rócherz
3,0263823
edited Mar 23 at 22:00
Rócherz
3,0263823
3,0263823
asked Mar 23 at 21:37
devil0108devil0108
174
asked Mar 23 at 21:37
devil0108devil0108
174
asked Mar 23 at 21:37
devil0108devil0108
174
174
1
$begingroup$
You mean, you solve for $c_1, c_2, c_3$.
$endgroup$
– lcv
Mar 23 at 21:40
add a comment |
1
$begingroup$
You mean, you solve for $c_1, c_2, c_3$.
$endgroup$
– lcv
Mar 23 at 21:40
1
1
$begingroup$
You mean, you solve for $c_1, c_2, c_3$.
$endgroup$
– lcv
Mar 23 at 21:40
$begingroup$
You mean, you solve for $c_1, c_2, c_3$.
$endgroup$
– lcv
Mar 23 at 21:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You're in the right direction, but you need to solve for $c_1, c_2, c_3$ and not $v_1, v_2, v_3$. For $b$ to be in $V$, it needs to be a linear combination of the given vectors, that is, $b=c_1(1,2,2)+c_2(1,1,1)+c_3(-1,0,1)$ for some real $c_1, c_2, c_3$, or equivalently $(0,4,7)=(c_1+c_2-c_3,2c_1+c_2,2c_1+c_2+c_3)$. This is equivalent to the system of equations:
$$
begin{cases}
c_1+c_2-c_3=0 \
2c_1+c_2=4 \
2c_1+c_2+c_3=7
end{cases}
$$
Can you proceed from here? Solve this system (or show it has a solution).
answered Mar 23 at 21:41
Yuval GatYuval Gat
9741213
$endgroup$
$begingroup$
Yes, I can do this from here. Essentially, is this question the same as writing vector b as a linear combination of vector V? How do I justify the answers?
$endgroup$
– devil0108
Mar 23 at 21:48
$begingroup$
@devil0108 It’s the same as writing the vector $b$ as a linear combination of the given three vectors. The span of a set of vectors is precisely, by definition, the set of all linear combinations of those vectors. Therefore, the existence of a solution to the system above is equivalent to $b$ being in the space $V$, which is spanned by three vectors we started with.
$endgroup$
– Yuval Gat
Mar 23 at 21:54
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You're in the right direction, but you need to solve for $c_1, c_2, c_3$ and not $v_1, v_2, v_3$. For $b$ to be in $V$, it needs to be a linear combination of the given vectors, that is, $b=c_1(1,2,2)+c_2(1,1,1)+c_3(-1,0,1)$ for some real $c_1, c_2, c_3$, or equivalently $(0,4,7)=(c_1+c_2-c_3,2c_1+c_2,2c_1+c_2+c_3)$. This is equivalent to the system of equations:
$$
begin{cases}
c_1+c_2-c_3=0 \
2c_1+c_2=4 \
2c_1+c_2+c_3=7
end{cases}
$$
Can you proceed from here? Solve this system (or show it has a solution).
answered Mar 23 at 21:41
Yuval GatYuval Gat
9741213
$endgroup$
$begingroup$
Yes, I can do this from here. Essentially, is this question the same as writing vector b as a linear combination of vector V? How do I justify the answers?
$endgroup$
– devil0108
Mar 23 at 21:48
$begingroup$
@devil0108 It’s the same as writing the vector $b$ as a linear combination of the given three vectors. The span of a set of vectors is precisely, by definition, the set of all linear combinations of those vectors. Therefore, the existence of a solution to the system above is equivalent to $b$ being in the space $V$, which is spanned by three vectors we started with.
$endgroup$
– Yuval Gat
Mar 23 at 21:54
add a comment |
$begingroup$
You're in the right direction, but you need to solve for $c_1, c_2, c_3$ and not $v_1, v_2, v_3$. For $b$ to be in $V$, it needs to be a linear combination of the given vectors, that is, $b=c_1(1,2,2)+c_2(1,1,1)+c_3(-1,0,1)$ for some real $c_1, c_2, c_3$, or equivalently $(0,4,7)=(c_1+c_2-c_3,2c_1+c_2,2c_1+c_2+c_3)$. This is equivalent to the system of equations:
$$
begin{cases}
c_1+c_2-c_3=0 \
2c_1+c_2=4 \
2c_1+c_2+c_3=7
end{cases}
$$
Can you proceed from here? Solve this system (or show it has a solution).
answered Mar 23 at 21:41
Yuval GatYuval Gat
9741213
$endgroup$
$begingroup$
Yes, I can do this from here. Essentially, is this question the same as writing vector b as a linear combination of vector V? How do I justify the answers?
$endgroup$
– devil0108
Mar 23 at 21:48
$begingroup$
@devil0108 It’s the same as writing the vector $b$ as a linear combination of the given three vectors. The span of a set of vectors is precisely, by definition, the set of all linear combinations of those vectors. Therefore, the existence of a solution to the system above is equivalent to $b$ being in the space $V$, which is spanned by three vectors we started with.
$endgroup$
– Yuval Gat
Mar 23 at 21:54
add a comment |
$begingroup$
You're in the right direction, but you need to solve for $c_1, c_2, c_3$ and not $v_1, v_2, v_3$. For $b$ to be in $V$, it needs to be a linear combination of the given vectors, that is, $b=c_1(1,2,2)+c_2(1,1,1)+c_3(-1,0,1)$ for some real $c_1, c_2, c_3$, or equivalently $(0,4,7)=(c_1+c_2-c_3,2c_1+c_2,2c_1+c_2+c_3)$. This is equivalent to the system of equations:
$$
begin{cases}
c_1+c_2-c_3=0 \
2c_1+c_2=4 \
2c_1+c_2+c_3=7
end{cases}
$$
Can you proceed from here? Solve this system (or show it has a solution).
answered Mar 23 at 21:41
Yuval GatYuval Gat
9741213
$endgroup$
You're in the right direction, but you need to solve for $c_1, c_2, c_3$ and not $v_1, v_2, v_3$. For $b$ to be in $V$, it needs to be a linear combination of the given vectors, that is, $b=c_1(1,2,2)+c_2(1,1,1)+c_3(-1,0,1)$ for some real $c_1, c_2, c_3$, or equivalently $(0,4,7)=(c_1+c_2-c_3,2c_1+c_2,2c_1+c_2+c_3)$. This is equivalent to the system of equations:
$$
begin{cases}
c_1+c_2-c_3=0 \
2c_1+c_2=4 \
2c_1+c_2+c_3=7
end{cases}
$$
Can you proceed from here? Solve this system (or show it has a solution).
answered Mar 23 at 21:41
Yuval GatYuval Gat
9741213
answered Mar 23 at 21:41
Yuval GatYuval Gat
9741213
answered Mar 23 at 21:41
Yuval GatYuval Gat
9741213
answered Mar 23 at 21:41
Yuval GatYuval Gat
9741213
9741213
$begingroup$
Yes, I can do this from here. Essentially, is this question the same as writing vector b as a linear combination of vector V? How do I justify the answers?
$endgroup$
– devil0108
Mar 23 at 21:48
$begingroup$
@devil0108 It’s the same as writing the vector $b$ as a linear combination of the given three vectors. The span of a set of vectors is precisely, by definition, the set of all linear combinations of those vectors. Therefore, the existence of a solution to the system above is equivalent to $b$ being in the space $V$, which is spanned by three vectors we started with.
$endgroup$
– Yuval Gat
Mar 23 at 21:54
add a comment |
$begingroup$
Yes, I can do this from here. Essentially, is this question the same as writing vector b as a linear combination of vector V? How do I justify the answers?
$endgroup$
– devil0108
Mar 23 at 21:48
$begingroup$
@devil0108 It’s the same as writing the vector $b$ as a linear combination of the given three vectors. The span of a set of vectors is precisely, by definition, the set of all linear combinations of those vectors. Therefore, the existence of a solution to the system above is equivalent to $b$ being in the space $V$, which is spanned by three vectors we started with.
$endgroup$
– Yuval Gat
Mar 23 at 21:54
$begingroup$
Yes, I can do this from here. Essentially, is this question the same as writing vector b as a linear combination of vector V? How do I justify the answers?
$endgroup$
– devil0108
Mar 23 at 21:48
$begingroup$
Yes, I can do this from here. Essentially, is this question the same as writing vector b as a linear combination of vector V? How do I justify the answers?
$endgroup$
– devil0108
Mar 23 at 21:48
$begingroup$
@devil0108 It’s the same as writing the vector $b$ as a linear combination of the given three vectors. The span of a set of vectors is precisely, by definition, the set of all linear combinations of those vectors. Therefore, the existence of a solution to the system above is equivalent to $b$ being in the space $V$, which is spanned by three vectors we started with.
$endgroup$
– Yuval Gat
Mar 23 at 21:54
$begingroup$
@devil0108 It’s the same as writing the vector $b$ as a linear combination of the given three vectors. The span of a set of vectors is precisely, by definition, the set of all linear combinations of those vectors. Therefore, the existence of a solution to the system above is equivalent to $b$ being in the space $V$, which is spanned by three vectors we started with.
$endgroup$
– Yuval Gat
Mar 23 at 21:54
add a comment |
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$begingroup$
You mean, you solve for $c_1, c_2, c_3$.
$endgroup$
– lcv
Mar 23 at 21:40