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.






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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







    share|cite|improve this question









    New contributor




    Lowly0palace is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      up vote
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      down vote

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      up vote
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      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







      share|cite|improve this question









      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






      share|cite|improve this question









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      Lowly0palace is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      edited Nov 16 at 9:49









      José Carlos Santos

      140k19111204




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      asked Nov 16 at 9:40









      Lowly0palace

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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, $w=0$.






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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, $w=0$.






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              up vote
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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, $w=0$.






              share|cite|improve this answer























                up vote
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                up vote
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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, $w=0$.






                share|cite|improve this answer












                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, $w=0$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 16 at 9:47









                José Carlos Santos

                140k19111204




                140k19111204






















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