Showing if a vector in a span is a linear combination of the other vectors, the other vectors still span the...
Multi tool use
$begingroup$
lets say we have a subspace $W$ and that three vectors, $w1, w2$ and $w3$ span $W$.
I want to show that if $w3$ is a linear combination of $w1$ and $w2$, that $w1$ and $w2$ still span $W$
My attempt so far:
so we can write $w3$ = $(c1*w1)$ + $(c2*w2)$ for $c1,c2$ not all $0$
I'm stuck at this point and can't think of how to prove the first statement, any suggestions?
edit:
linear-algebra
$endgroup$
add a comment |
$begingroup$
lets say we have a subspace $W$ and that three vectors, $w1, w2$ and $w3$ span $W$.
I want to show that if $w3$ is a linear combination of $w1$ and $w2$, that $w1$ and $w2$ still span $W$
My attempt so far:
so we can write $w3$ = $(c1*w1)$ + $(c2*w2)$ for $c1,c2$ not all $0$
I'm stuck at this point and can't think of how to prove the first statement, any suggestions?
edit:
linear-algebra
$endgroup$
add a comment |
$begingroup$
lets say we have a subspace $W$ and that three vectors, $w1, w2$ and $w3$ span $W$.
I want to show that if $w3$ is a linear combination of $w1$ and $w2$, that $w1$ and $w2$ still span $W$
My attempt so far:
so we can write $w3$ = $(c1*w1)$ + $(c2*w2)$ for $c1,c2$ not all $0$
I'm stuck at this point and can't think of how to prove the first statement, any suggestions?
edit:
linear-algebra
$endgroup$
lets say we have a subspace $W$ and that three vectors, $w1, w2$ and $w3$ span $W$.
I want to show that if $w3$ is a linear combination of $w1$ and $w2$, that $w1$ and $w2$ still span $W$
My attempt so far:
so we can write $w3$ = $(c1*w1)$ + $(c2*w2)$ for $c1,c2$ not all $0$
I'm stuck at this point and can't think of how to prove the first statement, any suggestions?
edit:
linear-algebra
linear-algebra
edited Dec 14 '18 at 5:47
lohboys
asked Dec 14 '18 at 4:27
lohboyslohboys
9819
9819
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Take $xin W$.
Then $x=sw_1+tw_2+uw_3$ for some $s,t,u$, by assumption. So $x=sw_1+tw_2+u(c_1w_1+c_2w_2)=(s+uc_1)w_1+(t+uc_2)w_2$. Thus $operatorname{span}{w_1,w_2}=W$.
$endgroup$
$begingroup$
thanks for the reply! , now that you've shown $w1,w2$ still span $W$ how should i go about proving $w1,w2$ are linearly independent
$endgroup$
– lohboys
Dec 14 '18 at 4:46
$begingroup$
im thinking i have to show $(s+uc1)w1+(t+uc2)w2 = 0$ only when the coefficients are $0$?
$endgroup$
– lohboys
Dec 14 '18 at 4:48
$begingroup$
They may not be linearly independent.
$endgroup$
– Chris Custer
Dec 14 '18 at 5:14
1
$begingroup$
so how do i show $operatorname{dim}Wge2$ ? so that together i can show $operatorname{dim}W = 2$
$endgroup$
– lohboys
Dec 14 '18 at 5:26
1
$begingroup$
The dimension of $W$ is at least two because $v_1,v_2in W$ are linearly independent.
$endgroup$
– Chris Custer
Dec 14 '18 at 6:16
|
show 4 more comments
Your Answer
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1 Answer
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1 Answer
1
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
Take $xin W$.
Then $x=sw_1+tw_2+uw_3$ for some $s,t,u$, by assumption. So $x=sw_1+tw_2+u(c_1w_1+c_2w_2)=(s+uc_1)w_1+(t+uc_2)w_2$. Thus $operatorname{span}{w_1,w_2}=W$.
$endgroup$
$begingroup$
thanks for the reply! , now that you've shown $w1,w2$ still span $W$ how should i go about proving $w1,w2$ are linearly independent
$endgroup$
– lohboys
Dec 14 '18 at 4:46
$begingroup$
im thinking i have to show $(s+uc1)w1+(t+uc2)w2 = 0$ only when the coefficients are $0$?
$endgroup$
– lohboys
Dec 14 '18 at 4:48
$begingroup$
They may not be linearly independent.
$endgroup$
– Chris Custer
Dec 14 '18 at 5:14
1
$begingroup$
so how do i show $operatorname{dim}Wge2$ ? so that together i can show $operatorname{dim}W = 2$
$endgroup$
– lohboys
Dec 14 '18 at 5:26
1
$begingroup$
The dimension of $W$ is at least two because $v_1,v_2in W$ are linearly independent.
$endgroup$
– Chris Custer
Dec 14 '18 at 6:16
|
show 4 more comments
$begingroup$
Take $xin W$.
Then $x=sw_1+tw_2+uw_3$ for some $s,t,u$, by assumption. So $x=sw_1+tw_2+u(c_1w_1+c_2w_2)=(s+uc_1)w_1+(t+uc_2)w_2$. Thus $operatorname{span}{w_1,w_2}=W$.
$endgroup$
$begingroup$
thanks for the reply! , now that you've shown $w1,w2$ still span $W$ how should i go about proving $w1,w2$ are linearly independent
$endgroup$
– lohboys
Dec 14 '18 at 4:46
$begingroup$
im thinking i have to show $(s+uc1)w1+(t+uc2)w2 = 0$ only when the coefficients are $0$?
$endgroup$
– lohboys
Dec 14 '18 at 4:48
$begingroup$
They may not be linearly independent.
$endgroup$
– Chris Custer
Dec 14 '18 at 5:14
1
$begingroup$
so how do i show $operatorname{dim}Wge2$ ? so that together i can show $operatorname{dim}W = 2$
$endgroup$
– lohboys
Dec 14 '18 at 5:26
1
$begingroup$
The dimension of $W$ is at least two because $v_1,v_2in W$ are linearly independent.
$endgroup$
– Chris Custer
Dec 14 '18 at 6:16
|
show 4 more comments
$begingroup$
Take $xin W$.
Then $x=sw_1+tw_2+uw_3$ for some $s,t,u$, by assumption. So $x=sw_1+tw_2+u(c_1w_1+c_2w_2)=(s+uc_1)w_1+(t+uc_2)w_2$. Thus $operatorname{span}{w_1,w_2}=W$.
$endgroup$
Take $xin W$.
Then $x=sw_1+tw_2+uw_3$ for some $s,t,u$, by assumption. So $x=sw_1+tw_2+u(c_1w_1+c_2w_2)=(s+uc_1)w_1+(t+uc_2)w_2$. Thus $operatorname{span}{w_1,w_2}=W$.
answered Dec 14 '18 at 4:42
Chris CusterChris Custer
14.3k3827
14.3k3827
$begingroup$
thanks for the reply! , now that you've shown $w1,w2$ still span $W$ how should i go about proving $w1,w2$ are linearly independent
$endgroup$
– lohboys
Dec 14 '18 at 4:46
$begingroup$
im thinking i have to show $(s+uc1)w1+(t+uc2)w2 = 0$ only when the coefficients are $0$?
$endgroup$
– lohboys
Dec 14 '18 at 4:48
$begingroup$
They may not be linearly independent.
$endgroup$
– Chris Custer
Dec 14 '18 at 5:14
1
$begingroup$
so how do i show $operatorname{dim}Wge2$ ? so that together i can show $operatorname{dim}W = 2$
$endgroup$
– lohboys
Dec 14 '18 at 5:26
1
$begingroup$
The dimension of $W$ is at least two because $v_1,v_2in W$ are linearly independent.
$endgroup$
– Chris Custer
Dec 14 '18 at 6:16
|
show 4 more comments
$begingroup$
thanks for the reply! , now that you've shown $w1,w2$ still span $W$ how should i go about proving $w1,w2$ are linearly independent
$endgroup$
– lohboys
Dec 14 '18 at 4:46
$begingroup$
im thinking i have to show $(s+uc1)w1+(t+uc2)w2 = 0$ only when the coefficients are $0$?
$endgroup$
– lohboys
Dec 14 '18 at 4:48
$begingroup$
They may not be linearly independent.
$endgroup$
– Chris Custer
Dec 14 '18 at 5:14
1
$begingroup$
so how do i show $operatorname{dim}Wge2$ ? so that together i can show $operatorname{dim}W = 2$
$endgroup$
– lohboys
Dec 14 '18 at 5:26
1
$begingroup$
The dimension of $W$ is at least two because $v_1,v_2in W$ are linearly independent.
$endgroup$
– Chris Custer
Dec 14 '18 at 6:16
$begingroup$
thanks for the reply! , now that you've shown $w1,w2$ still span $W$ how should i go about proving $w1,w2$ are linearly independent
$endgroup$
– lohboys
Dec 14 '18 at 4:46
$begingroup$
thanks for the reply! , now that you've shown $w1,w2$ still span $W$ how should i go about proving $w1,w2$ are linearly independent
$endgroup$
– lohboys
Dec 14 '18 at 4:46
$begingroup$
im thinking i have to show $(s+uc1)w1+(t+uc2)w2 = 0$ only when the coefficients are $0$?
$endgroup$
– lohboys
Dec 14 '18 at 4:48
$begingroup$
im thinking i have to show $(s+uc1)w1+(t+uc2)w2 = 0$ only when the coefficients are $0$?
$endgroup$
– lohboys
Dec 14 '18 at 4:48
$begingroup$
They may not be linearly independent.
$endgroup$
– Chris Custer
Dec 14 '18 at 5:14
$begingroup$
They may not be linearly independent.
$endgroup$
– Chris Custer
Dec 14 '18 at 5:14
1
1
$begingroup$
so how do i show $operatorname{dim}Wge2$ ? so that together i can show $operatorname{dim}W = 2$
$endgroup$
– lohboys
Dec 14 '18 at 5:26
$begingroup$
so how do i show $operatorname{dim}Wge2$ ? so that together i can show $operatorname{dim}W = 2$
$endgroup$
– lohboys
Dec 14 '18 at 5:26
1
1
$begingroup$
The dimension of $W$ is at least two because $v_1,v_2in W$ are linearly independent.
$endgroup$
– Chris Custer
Dec 14 '18 at 6:16
$begingroup$
The dimension of $W$ is at least two because $v_1,v_2in W$ are linearly independent.
$endgroup$
– Chris Custer
Dec 14 '18 at 6:16
|
show 4 more comments
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