Suppose that $(u_1,ldots, u_m, v_1, ldots, v_n)$ is linearly independent. Prove that $W_1 ∩ W_2 = {0}$.
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Let $V$ be a real vector space and $W_1$ and $W_2$ two finite dimensional subspaces. Let
$(u_1,ldots, u_m)$ be a basis for $W_1$ and $(v_1,ldots, v_n)$ be a basis for $W2$.
Suppose that $(u_1,ldots, u_m, v_1,ldots, v_n)$ is linearly independent. Prove that $W_1 ∩ W_2 = {0}$.
I know that if $w ∈ W_1 ∩ W_2$, then $w$ is a linear combination of the $u_i$ and also a linear combination of the $v_j$). I am however not sure how to proceed next.
linear-algebra vector-spaces
edited Nov 16 at 9:49
José Carlos Santos
140k19111204
asked Nov 16 at 9:40
Lowly0palace
11
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up vote
0
down vote
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Let $V$ be a real vector space and $W_1$ and $W_2$ two finite dimensional subspaces. Let
$(u_1,ldots, u_m)$ be a basis for $W_1$ and $(v_1,ldots, v_n)$ be a basis for $W2$.
Suppose that $(u_1,ldots, u_m, v_1,ldots, v_n)$ is linearly independent. Prove that $W_1 ∩ W_2 = {0}$.
I know that if $w ∈ W_1 ∩ W_2$, then $w$ is a linear combination of the $u_i$ and also a linear combination of the $v_j$). I am however not sure how to proceed next.
linear-algebra vector-spaces
edited Nov 16 at 9:49
José Carlos Santos
140k19111204
asked Nov 16 at 9:40
Lowly0palace
11
New contributor
Lowly0palace 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
Let $V$ be a real vector space and $W_1$ and $W_2$ two finite dimensional subspaces. Let
$(u_1,ldots, u_m)$ be a basis for $W_1$ and $(v_1,ldots, v_n)$ be a basis for $W2$.
Suppose that $(u_1,ldots, u_m, v_1,ldots, v_n)$ is linearly independent. Prove that $W_1 ∩ W_2 = {0}$.
I know that if $w ∈ W_1 ∩ W_2$, then $w$ is a linear combination of the $u_i$ and also a linear combination of the $v_j$). I am however not sure how to proceed next.
linear-algebra vector-spaces
edited Nov 16 at 9:49
José Carlos Santos
140k19111204
asked Nov 16 at 9:40
Lowly0palace
11
New contributor
Lowly0palace is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Let $V$ be a real vector space and $W_1$ and $W_2$ two finite dimensional subspaces. Let
$(u_1,ldots, u_m)$ be a basis for $W_1$ and $(v_1,ldots, v_n)$ be a basis for $W2$.
Suppose that $(u_1,ldots, u_m, v_1,ldots, v_n)$ is linearly independent. Prove that $W_1 ∩ W_2 = {0}$.
I know that if $w ∈ W_1 ∩ W_2$, then $w$ is a linear combination of the $u_i$ and also a linear combination of the $v_j$). I am however not sure how to proceed next.
linear-algebra vector-spaces
linear-algebra vector-spaces
edited Nov 16 at 9:49
José Carlos Santos
140k19111204
asked Nov 16 at 9:40
Lowly0palace
11
New contributor
Lowly0palace is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 16 at 9:49
José Carlos Santos
140k19111204
asked Nov 16 at 9:40
Lowly0palace
11
New contributor
Lowly0palace is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 16 at 9:49
José Carlos Santos
140k19111204
edited Nov 16 at 9:49
José Carlos Santos
140k19111204
edited Nov 16 at 9:49
José Carlos Santos
140k19111204
140k19111204
asked Nov 16 at 9:40
Lowly0palace
11
New contributor
Lowly0palace is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 16 at 9:40
Lowly0palace
11
asked Nov 16 at 9:40
Lowly0palace
11
11
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Lowly0palace is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
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1 Answer
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Since $w$ is a linear combination of the $u_i$'s and also of the $v_j$'s, then you have$$w=alpha_1u_1+cdots+alpha_mu_m=beta_1v_1+cdots+beta_nv_n.$$But then$$alpha_1u_1+cdots+alpha_mu_m-beta_1v_1-cdots-beta_nv_n=0.$$Therefore, the linear independence implies that all coefficients are $0$. In particular, $w=0$.
answered Nov 16 at 9:47
José Carlos Santos
140k19111204
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Since $w$ is a linear combination of the $u_i$'s and also of the $v_j$'s, then you have$$w=alpha_1u_1+cdots+alpha_mu_m=beta_1v_1+cdots+beta_nv_n.$$But then$$alpha_1u_1+cdots+alpha_mu_m-beta_1v_1-cdots-beta_nv_n=0.$$Therefore, the linear independence implies that all coefficients are $0$. In particular, $w=0$.
answered Nov 16 at 9:47
José Carlos Santos
140k19111204
add a comment |
up vote
2
down vote
Since $w$ is a linear combination of the $u_i$'s and also of the $v_j$'s, then you have$$w=alpha_1u_1+cdots+alpha_mu_m=beta_1v_1+cdots+beta_nv_n.$$But then$$alpha_1u_1+cdots+alpha_mu_m-beta_1v_1-cdots-beta_nv_n=0.$$Therefore, the linear independence implies that all coefficients are $0$. In particular, $w=0$.
answered Nov 16 at 9:47
José Carlos Santos
140k19111204
add a comment |
up vote
2
down vote
up vote
2
down vote
Since $w$ is a linear combination of the $u_i$'s and also of the $v_j$'s, then you have$$w=alpha_1u_1+cdots+alpha_mu_m=beta_1v_1+cdots+beta_nv_n.$$But then$$alpha_1u_1+cdots+alpha_mu_m-beta_1v_1-cdots-beta_nv_n=0.$$Therefore, the linear independence implies that all coefficients are $0$. In particular, $w=0$.
answered Nov 16 at 9:47
José Carlos Santos
140k19111204
Since $w$ is a linear combination of the $u_i$'s and also of the $v_j$'s, then you have$$w=alpha_1u_1+cdots+alpha_mu_m=beta_1v_1+cdots+beta_nv_n.$$But then$$alpha_1u_1+cdots+alpha_mu_m-beta_1v_1-cdots-beta_nv_n=0.$$Therefore, the linear independence implies that all coefficients are $0$. In particular, $w=0$.
answered Nov 16 at 9:47
José Carlos Santos
140k19111204
answered Nov 16 at 9:47
José Carlos Santos
140k19111204
answered Nov 16 at 9:47
José Carlos Santos
140k19111204
answered Nov 16 at 9:47
José Carlos Santos
140k19111204
140k19111204
add a comment |
add a comment |
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