How to show one set of vectors span R3 if its components do?
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The question states
Show that if a, b, c are vectors in R3, then {b+c, c+a,a+b} spans R3 iff {a,b,c} spans R3.
I've started trying Gauss-Jordan with the first set in Ax = 0 form, but I haven't got anywhere.
What am I supposed to do from here?
linear-algebra vectors
asked Nov 20 at 19:56
mathPhys
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up vote
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down vote
favorite
The question states
Show that if a, b, c are vectors in R3, then {b+c, c+a,a+b} spans R3 iff {a,b,c} spans R3.
I've started trying Gauss-Jordan with the first set in Ax = 0 form, but I haven't got anywhere.
What am I supposed to do from here?
linear-algebra vectors
asked Nov 20 at 19:56
mathPhys
11
New contributor
mathPhys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The question states
Show that if a, b, c are vectors in R3, then {b+c, c+a,a+b} spans R3 iff {a,b,c} spans R3.
I've started trying Gauss-Jordan with the first set in Ax = 0 form, but I haven't got anywhere.
What am I supposed to do from here?
linear-algebra vectors
asked Nov 20 at 19:56
mathPhys
11
New contributor
mathPhys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The question states
Show that if a, b, c are vectors in R3, then {b+c, c+a,a+b} spans R3 iff {a,b,c} spans R3.
I've started trying Gauss-Jordan with the first set in Ax = 0 form, but I haven't got anywhere.
What am I supposed to do from here?
linear-algebra vectors
linear-algebra vectors
asked Nov 20 at 19:56
mathPhys
11
New contributor
mathPhys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 20 at 19:56
mathPhys
11
New contributor
mathPhys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 20 at 19:56
mathPhys
11
New contributor
mathPhys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 20 at 19:56
mathPhys
11
asked Nov 20 at 19:56
mathPhys
11
11
New contributor
mathPhys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
mathPhys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
mathPhys is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
3 Answers
3
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oldest
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0
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HINT
Let consider
$$k_1(b+c)+k_2(c+a)+k_3(a+b)=0$$
$$(k_2+k_3)a+(k_1+k_3)b+(k_1+k_2)c=0$$
answered Nov 20 at 19:58
gimusi
87.1k74393
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Hint
$text{{a,b,c}}$ spans $mathbb R^3$ implies that $l_1a+l_2b+l_3c=0 $ for some $l_1,l_2,l_3$.
Now evaluate $l_1(b+c)+l_2(c+a)+l_3(a+b) $ and use the equality above.
answered Nov 20 at 20:06
dougle
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I'm sorry but I dont follow? Which equality, and are they the same L1, L2, L3?
– mathPhys
Nov 21 at 20:01
add a comment |
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Find a matrix $M$ such that $$begin{bmatrix}mathbf b+mathbf c & mathbf c+mathbf a & mathbf a+mathbf bend{bmatrix} = begin{bmatrix}mathbf a & mathbf b & mathbf cend{bmatrix} M$$ and use properties of determinants to argue that the columns of the matrix on the left-hand side are linearly independent iff $mathbf a$, $mathbf b$ and $mathbf c$ are.
answered Nov 20 at 23:49
amd
28.5k21049
Sorry, but could you clarify how I would go about finding M?
– mathPhys
Nov 21 at 20:03
@mathPhys Actually, I meant to post-multiply by $M$, since then you’re just taking linear combinations of the columns of $[mathbf a,mathbf b,mathbf c]$. Corrected.
– amd
Nov 21 at 20:42
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
HINT
Let consider
$$k_1(b+c)+k_2(c+a)+k_3(a+b)=0$$
$$(k_2+k_3)a+(k_1+k_3)b+(k_1+k_2)c=0$$
answered Nov 20 at 19:58
gimusi
87.1k74393
add a comment |
up vote
0
down vote
HINT
Let consider
$$k_1(b+c)+k_2(c+a)+k_3(a+b)=0$$
$$(k_2+k_3)a+(k_1+k_3)b+(k_1+k_2)c=0$$
answered Nov 20 at 19:58
gimusi
87.1k74393
add a comment |
up vote
0
down vote
up vote
0
down vote
HINT
Let consider
$$k_1(b+c)+k_2(c+a)+k_3(a+b)=0$$
$$(k_2+k_3)a+(k_1+k_3)b+(k_1+k_2)c=0$$
answered Nov 20 at 19:58
gimusi
87.1k74393
HINT
Let consider
$$k_1(b+c)+k_2(c+a)+k_3(a+b)=0$$
$$(k_2+k_3)a+(k_1+k_3)b+(k_1+k_2)c=0$$
answered Nov 20 at 19:58
gimusi
87.1k74393
answered Nov 20 at 19:58
gimusi
87.1k74393
answered Nov 20 at 19:58
gimusi
87.1k74393
answered Nov 20 at 19:58
gimusi
87.1k74393
87.1k74393
add a comment |
add a comment |
up vote
0
down vote
Hint
$text{{a,b,c}}$ spans $mathbb R^3$ implies that $l_1a+l_2b+l_3c=0 $ for some $l_1,l_2,l_3$.
Now evaluate $l_1(b+c)+l_2(c+a)+l_3(a+b) $ and use the equality above.
answered Nov 20 at 20:06
dougle
154
New contributor
dougle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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I'm sorry but I dont follow? Which equality, and are they the same L1, L2, L3?
– mathPhys
Nov 21 at 20:01
add a comment |
up vote
0
down vote
Hint
$text{{a,b,c}}$ spans $mathbb R^3$ implies that $l_1a+l_2b+l_3c=0 $ for some $l_1,l_2,l_3$.
Now evaluate $l_1(b+c)+l_2(c+a)+l_3(a+b) $ and use the equality above.
answered Nov 20 at 20:06
dougle
154
New contributor
dougle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm sorry but I dont follow? Which equality, and are they the same L1, L2, L3?
– mathPhys
Nov 21 at 20:01
add a comment |
up vote
0
down vote
up vote
0
down vote
Hint
$text{{a,b,c}}$ spans $mathbb R^3$ implies that $l_1a+l_2b+l_3c=0 $ for some $l_1,l_2,l_3$.
Now evaluate $l_1(b+c)+l_2(c+a)+l_3(a+b) $ and use the equality above.
answered Nov 20 at 20:06
dougle
154
New contributor
dougle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Hint
$text{{a,b,c}}$ spans $mathbb R^3$ implies that $l_1a+l_2b+l_3c=0 $ for some $l_1,l_2,l_3$.
Now evaluate $l_1(b+c)+l_2(c+a)+l_3(a+b) $ and use the equality above.
answered Nov 20 at 20:06
dougle
154
New contributor
dougle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Nov 20 at 20:06
dougle
154
New contributor
dougle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Nov 20 at 20:06
dougle
154
answered Nov 20 at 20:06
dougle
154
154
New contributor
dougle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
dougle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
dougle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm sorry but I dont follow? Which equality, and are they the same L1, L2, L3?
– mathPhys
Nov 21 at 20:01
add a comment |
I'm sorry but I dont follow? Which equality, and are they the same L1, L2, L3?
– mathPhys
Nov 21 at 20:01
I'm sorry but I dont follow? Which equality, and are they the same L1, L2, L3?
– mathPhys
Nov 21 at 20:01
I'm sorry but I dont follow? Which equality, and are they the same L1, L2, L3?
– mathPhys
Nov 21 at 20:01
add a comment |
up vote
0
down vote
Find a matrix $M$ such that $$begin{bmatrix}mathbf b+mathbf c & mathbf c+mathbf a & mathbf a+mathbf bend{bmatrix} = begin{bmatrix}mathbf a & mathbf b & mathbf cend{bmatrix} M$$ and use properties of determinants to argue that the columns of the matrix on the left-hand side are linearly independent iff $mathbf a$, $mathbf b$ and $mathbf c$ are.
answered Nov 20 at 23:49
amd
28.5k21049
Sorry, but could you clarify how I would go about finding M?
– mathPhys
Nov 21 at 20:03
@mathPhys Actually, I meant to post-multiply by $M$, since then you’re just taking linear combinations of the columns of $[mathbf a,mathbf b,mathbf c]$. Corrected.
– amd
Nov 21 at 20:42
add a comment |
up vote
0
down vote
Find a matrix $M$ such that $$begin{bmatrix}mathbf b+mathbf c & mathbf c+mathbf a & mathbf a+mathbf bend{bmatrix} = begin{bmatrix}mathbf a & mathbf b & mathbf cend{bmatrix} M$$ and use properties of determinants to argue that the columns of the matrix on the left-hand side are linearly independent iff $mathbf a$, $mathbf b$ and $mathbf c$ are.
answered Nov 20 at 23:49
amd
28.5k21049
Sorry, but could you clarify how I would go about finding M?
– mathPhys
Nov 21 at 20:03
@mathPhys Actually, I meant to post-multiply by $M$, since then you’re just taking linear combinations of the columns of $[mathbf a,mathbf b,mathbf c]$. Corrected.
– amd
Nov 21 at 20:42
add a comment |
up vote
0
down vote
up vote
0
down vote
Find a matrix $M$ such that $$begin{bmatrix}mathbf b+mathbf c & mathbf c+mathbf a & mathbf a+mathbf bend{bmatrix} = begin{bmatrix}mathbf a & mathbf b & mathbf cend{bmatrix} M$$ and use properties of determinants to argue that the columns of the matrix on the left-hand side are linearly independent iff $mathbf a$, $mathbf b$ and $mathbf c$ are.
answered Nov 20 at 23:49
amd
28.5k21049
Find a matrix $M$ such that $$begin{bmatrix}mathbf b+mathbf c & mathbf c+mathbf a & mathbf a+mathbf bend{bmatrix} = begin{bmatrix}mathbf a & mathbf b & mathbf cend{bmatrix} M$$ and use properties of determinants to argue that the columns of the matrix on the left-hand side are linearly independent iff $mathbf a$, $mathbf b$ and $mathbf c$ are.
answered Nov 20 at 23:49
amd
28.5k21049
edited Nov 21 at 20:42
answered Nov 20 at 23:49
amd
28.5k21049
answered Nov 20 at 23:49
amd
28.5k21049
answered Nov 20 at 23:49
amd
28.5k21049
28.5k21049
Sorry, but could you clarify how I would go about finding M?
– mathPhys
Nov 21 at 20:03
@mathPhys Actually, I meant to post-multiply by $M$, since then you’re just taking linear combinations of the columns of $[mathbf a,mathbf b,mathbf c]$. Corrected.
– amd
Nov 21 at 20:42
add a comment |
Sorry, but could you clarify how I would go about finding M?
– mathPhys
Nov 21 at 20:03
@mathPhys Actually, I meant to post-multiply by $M$, since then you’re just taking linear combinations of the columns of $[mathbf a,mathbf b,mathbf c]$. Corrected.
– amd
Nov 21 at 20:42
Sorry, but could you clarify how I would go about finding M?
– mathPhys
Nov 21 at 20:03
Sorry, but could you clarify how I would go about finding M?
– mathPhys
Nov 21 at 20:03
@mathPhys Actually, I meant to post-multiply by $M$, since then you’re just taking linear combinations of the columns of $[mathbf a,mathbf b,mathbf c]$. Corrected.
– amd
Nov 21 at 20:42
@mathPhys Actually, I meant to post-multiply by $M$, since then you’re just taking linear combinations of the columns of $[mathbf a,mathbf b,mathbf c]$. Corrected.
– amd
Nov 21 at 20:42
add a comment |
mathPhys is a new contributor. Be nice, and check out our Code of Conduct.
mathPhys is a new contributor. Be nice, and check out our Code of Conduct.
mathPhys is a new contributor. Be nice, and check out our Code of Conduct.
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