Subspace of a vector space problem












2












$begingroup$


So the task says:



for $alpha_1ldots,alpha_n, betain mathbb R$, we define
$U = {(x_1ldots,x_n) ∈ mathbb{R}^n ,|, sum_{i=1}^n alpha_ix_i = beta}$.
When is the $U$ subspace?



I know that if subspace is a subspace only if $x,y∈ U$ and $A,C ∈ mathbb R$ this is true:



$Ax + Cy ∈ U$,



I could not find the answer so i checked the answer.



enter image description here



I understand everything in explanation except the last part where it says that B needs to be 0 can someone explain to me why $beta$ needs to be 0.



Thank you!






linear-algebra







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Well if $B$ is not zero then 0 is not in the subspace.
    $endgroup$
    – Calvin Khor
    5 hours ago










  • $begingroup$
    Of course, but the OP wanted to understand that specific proof.
    $endgroup$
    – José Carlos Santos
    4 hours ago










  • $begingroup$
    Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
    $endgroup$
    – Petar
    4 hours ago
















2












$begingroup$


So the task says:



for $alpha_1ldots,alpha_n, betain mathbb R$, we define
$U = {(x_1ldots,x_n) ∈ mathbb{R}^n ,|, sum_{i=1}^n alpha_ix_i = beta}$.
When is the $U$ subspace?



I know that if subspace is a subspace only if $x,y∈ U$ and $A,C ∈ mathbb R$ this is true:



$Ax + Cy ∈ U$,



I could not find the answer so i checked the answer.



enter image description here



I understand everything in explanation except the last part where it says that B needs to be 0 can someone explain to me why $beta$ needs to be 0.



Thank you!






linear-algebra







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Well if $B$ is not zero then 0 is not in the subspace.
    $endgroup$
    – Calvin Khor
    5 hours ago










  • $begingroup$
    Of course, but the OP wanted to understand that specific proof.
    $endgroup$
    – José Carlos Santos
    4 hours ago










  • $begingroup$
    Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
    $endgroup$
    – Petar
    4 hours ago














2












2








2





$begingroup$


So the task says:



for $alpha_1ldots,alpha_n, betain mathbb R$, we define
$U = {(x_1ldots,x_n) ∈ mathbb{R}^n ,|, sum_{i=1}^n alpha_ix_i = beta}$.
When is the $U$ subspace?



I know that if subspace is a subspace only if $x,y∈ U$ and $A,C ∈ mathbb R$ this is true:



$Ax + Cy ∈ U$,



I could not find the answer so i checked the answer.



enter image description here



I understand everything in explanation except the last part where it says that B needs to be 0 can someone explain to me why $beta$ needs to be 0.



Thank you!






linear-algebra







share|cite|improve this question











$endgroup$




So the task says:



for $alpha_1ldots,alpha_n, betain mathbb R$, we define
$U = {(x_1ldots,x_n) ∈ mathbb{R}^n ,|, sum_{i=1}^n alpha_ix_i = beta}$.
When is the $U$ subspace?



I know that if subspace is a subspace only if $x,y∈ U$ and $A,C ∈ mathbb R$ this is true:



$Ax + Cy ∈ U$,



I could not find the answer so i checked the answer.



enter image description here



I understand everything in explanation except the last part where it says that B needs to be 0 can someone explain to me why $beta$ needs to be 0.



Thank you!






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 5 hours ago









José Carlos Santos

155k22124227




155k22124227










asked 5 hours ago









PetarPetar

182




182








  • 1




    $begingroup$
    Well if $B$ is not zero then 0 is not in the subspace.
    $endgroup$
    – Calvin Khor
    5 hours ago










  • $begingroup$
    Of course, but the OP wanted to understand that specific proof.
    $endgroup$
    – José Carlos Santos
    4 hours ago










  • $begingroup$
    Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
    $endgroup$
    – Petar
    4 hours ago














  • 1




    $begingroup$
    Well if $B$ is not zero then 0 is not in the subspace.
    $endgroup$
    – Calvin Khor
    5 hours ago










  • $begingroup$
    Of course, but the OP wanted to understand that specific proof.
    $endgroup$
    – José Carlos Santos
    4 hours ago










  • $begingroup$
    Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
    $endgroup$
    – Petar
    4 hours ago








1




1




$begingroup$
Well if $B$ is not zero then 0 is not in the subspace.
$endgroup$
– Calvin Khor
5 hours ago




$begingroup$
Well if $B$ is not zero then 0 is not in the subspace.
$endgroup$
– Calvin Khor
5 hours ago












$begingroup$
Of course, but the OP wanted to understand that specific proof.
$endgroup$
– José Carlos Santos
4 hours ago




$begingroup$
Of course, but the OP wanted to understand that specific proof.
$endgroup$
– José Carlos Santos
4 hours ago












$begingroup$
Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
$endgroup$
– Petar
4 hours ago




$begingroup$
Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
$endgroup$
– Petar
4 hours ago










2 Answers
2






active

oldest

votes


















3












$begingroup$

Because if $betaneq0$, then you take, say, $lambda=mu=1$, and then it will be false that $lambdabeta+mubeta=beta$, since then this would mean that $2beta=beta$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
    $endgroup$
    – Petar
    5 hours ago










  • $begingroup$
    If $mu=lambda=frac12$, you deduce nothing. But the proof says that for all real $mu$ and $lambda$ we have $mubeta+lambdabeta=beta$. So, I used $mu=lambda=1$ to reach a contradiction. That's all.
    $endgroup$
    – José Carlos Santos
    3 hours ago












  • $begingroup$
    Ok thank you! I understand it now.
    $endgroup$
    – Petar
    3 hours ago










  • $begingroup$
    If my answer was useful, perhaps that you could mark it as the accepted one.
    $endgroup$
    – José Carlos Santos
    3 hours ago










  • $begingroup$
    yes i did it now ,tnx
    $endgroup$
    – Petar
    2 hours ago



















2












$begingroup$

Notice that $sum_1^na_ix_i$ is merely the dot product of $mathbf x=(x_1,x_2,...,x_n)in U$ with the vector $mathbf a=(a_1,a_2,...,a_n)$. Given that vector $mathbf xin Uleftrightarrowmathbf xcdotmathbf a=B$. Now, $mathbf x,mathbf yin Uimpliesmathbf x+mathbf yin Uimplies(mathbf x+mathbf y)cdot mathbf a=mathbf{xcdot a}+mathbf{ycdot a}=B+B=B$, that implies $B=0$.






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Because if $betaneq0$, then you take, say, $lambda=mu=1$, and then it will be false that $lambdabeta+mubeta=beta$, since then this would mean that $2beta=beta$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
      $endgroup$
      – Petar
      5 hours ago










    • $begingroup$
      If $mu=lambda=frac12$, you deduce nothing. But the proof says that for all real $mu$ and $lambda$ we have $mubeta+lambdabeta=beta$. So, I used $mu=lambda=1$ to reach a contradiction. That's all.
      $endgroup$
      – José Carlos Santos
      3 hours ago












    • $begingroup$
      Ok thank you! I understand it now.
      $endgroup$
      – Petar
      3 hours ago










    • $begingroup$
      If my answer was useful, perhaps that you could mark it as the accepted one.
      $endgroup$
      – José Carlos Santos
      3 hours ago










    • $begingroup$
      yes i did it now ,tnx
      $endgroup$
      – Petar
      2 hours ago
















    3












    $begingroup$

    Because if $betaneq0$, then you take, say, $lambda=mu=1$, and then it will be false that $lambdabeta+mubeta=beta$, since then this would mean that $2beta=beta$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
      $endgroup$
      – Petar
      5 hours ago










    • $begingroup$
      If $mu=lambda=frac12$, you deduce nothing. But the proof says that for all real $mu$ and $lambda$ we have $mubeta+lambdabeta=beta$. So, I used $mu=lambda=1$ to reach a contradiction. That's all.
      $endgroup$
      – José Carlos Santos
      3 hours ago












    • $begingroup$
      Ok thank you! I understand it now.
      $endgroup$
      – Petar
      3 hours ago










    • $begingroup$
      If my answer was useful, perhaps that you could mark it as the accepted one.
      $endgroup$
      – José Carlos Santos
      3 hours ago










    • $begingroup$
      yes i did it now ,tnx
      $endgroup$
      – Petar
      2 hours ago














    3












    3








    3





    $begingroup$

    Because if $betaneq0$, then you take, say, $lambda=mu=1$, and then it will be false that $lambdabeta+mubeta=beta$, since then this would mean that $2beta=beta$.






    share|cite|improve this answer









    $endgroup$



    Because if $betaneq0$, then you take, say, $lambda=mu=1$, and then it will be false that $lambdabeta+mubeta=beta$, since then this would mean that $2beta=beta$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 5 hours ago









    José Carlos SantosJosé Carlos Santos

    155k22124227




    155k22124227












    • $begingroup$
      Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
      $endgroup$
      – Petar
      5 hours ago










    • $begingroup$
      If $mu=lambda=frac12$, you deduce nothing. But the proof says that for all real $mu$ and $lambda$ we have $mubeta+lambdabeta=beta$. So, I used $mu=lambda=1$ to reach a contradiction. That's all.
      $endgroup$
      – José Carlos Santos
      3 hours ago












    • $begingroup$
      Ok thank you! I understand it now.
      $endgroup$
      – Petar
      3 hours ago










    • $begingroup$
      If my answer was useful, perhaps that you could mark it as the accepted one.
      $endgroup$
      – José Carlos Santos
      3 hours ago










    • $begingroup$
      yes i did it now ,tnx
      $endgroup$
      – Petar
      2 hours ago


















    • $begingroup$
      Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
      $endgroup$
      – Petar
      5 hours ago










    • $begingroup$
      If $mu=lambda=frac12$, you deduce nothing. But the proof says that for all real $mu$ and $lambda$ we have $mubeta+lambdabeta=beta$. So, I used $mu=lambda=1$ to reach a contradiction. That's all.
      $endgroup$
      – José Carlos Santos
      3 hours ago












    • $begingroup$
      Ok thank you! I understand it now.
      $endgroup$
      – Petar
      3 hours ago










    • $begingroup$
      If my answer was useful, perhaps that you could mark it as the accepted one.
      $endgroup$
      – José Carlos Santos
      3 hours ago










    • $begingroup$
      yes i did it now ,tnx
      $endgroup$
      – Petar
      2 hours ago
















    $begingroup$
    Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
    $endgroup$
    – Petar
    5 hours ago




    $begingroup$
    Ok thank you! But what if µ = 1/2 and λ = 1/2 than if B is diffrent than 0, (1/2 * B) + (1/2*B) = B
    $endgroup$
    – Petar
    5 hours ago












    $begingroup$
    If $mu=lambda=frac12$, you deduce nothing. But the proof says that for all real $mu$ and $lambda$ we have $mubeta+lambdabeta=beta$. So, I used $mu=lambda=1$ to reach a contradiction. That's all.
    $endgroup$
    – José Carlos Santos
    3 hours ago






    $begingroup$
    If $mu=lambda=frac12$, you deduce nothing. But the proof says that for all real $mu$ and $lambda$ we have $mubeta+lambdabeta=beta$. So, I used $mu=lambda=1$ to reach a contradiction. That's all.
    $endgroup$
    – José Carlos Santos
    3 hours ago














    $begingroup$
    Ok thank you! I understand it now.
    $endgroup$
    – Petar
    3 hours ago




    $begingroup$
    Ok thank you! I understand it now.
    $endgroup$
    – Petar
    3 hours ago












    $begingroup$
    If my answer was useful, perhaps that you could mark it as the accepted one.
    $endgroup$
    – José Carlos Santos
    3 hours ago




    $begingroup$
    If my answer was useful, perhaps that you could mark it as the accepted one.
    $endgroup$
    – José Carlos Santos
    3 hours ago












    $begingroup$
    yes i did it now ,tnx
    $endgroup$
    – Petar
    2 hours ago




    $begingroup$
    yes i did it now ,tnx
    $endgroup$
    – Petar
    2 hours ago











    2












    $begingroup$

    Notice that $sum_1^na_ix_i$ is merely the dot product of $mathbf x=(x_1,x_2,...,x_n)in U$ with the vector $mathbf a=(a_1,a_2,...,a_n)$. Given that vector $mathbf xin Uleftrightarrowmathbf xcdotmathbf a=B$. Now, $mathbf x,mathbf yin Uimpliesmathbf x+mathbf yin Uimplies(mathbf x+mathbf y)cdot mathbf a=mathbf{xcdot a}+mathbf{ycdot a}=B+B=B$, that implies $B=0$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Notice that $sum_1^na_ix_i$ is merely the dot product of $mathbf x=(x_1,x_2,...,x_n)in U$ with the vector $mathbf a=(a_1,a_2,...,a_n)$. Given that vector $mathbf xin Uleftrightarrowmathbf xcdotmathbf a=B$. Now, $mathbf x,mathbf yin Uimpliesmathbf x+mathbf yin Uimplies(mathbf x+mathbf y)cdot mathbf a=mathbf{xcdot a}+mathbf{ycdot a}=B+B=B$, that implies $B=0$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Notice that $sum_1^na_ix_i$ is merely the dot product of $mathbf x=(x_1,x_2,...,x_n)in U$ with the vector $mathbf a=(a_1,a_2,...,a_n)$. Given that vector $mathbf xin Uleftrightarrowmathbf xcdotmathbf a=B$. Now, $mathbf x,mathbf yin Uimpliesmathbf x+mathbf yin Uimplies(mathbf x+mathbf y)cdot mathbf a=mathbf{xcdot a}+mathbf{ycdot a}=B+B=B$, that implies $B=0$.






        share|cite|improve this answer









        $endgroup$



        Notice that $sum_1^na_ix_i$ is merely the dot product of $mathbf x=(x_1,x_2,...,x_n)in U$ with the vector $mathbf a=(a_1,a_2,...,a_n)$. Given that vector $mathbf xin Uleftrightarrowmathbf xcdotmathbf a=B$. Now, $mathbf x,mathbf yin Uimpliesmathbf x+mathbf yin Uimplies(mathbf x+mathbf y)cdot mathbf a=mathbf{xcdot a}+mathbf{ycdot a}=B+B=B$, that implies $B=0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 5 hours ago









        Shubham JohriShubham Johri

        4,780717




        4,780717






























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            南乌拉尔铁路局

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            南乌拉尔铁路局 Южно-Уральская железная дорога 成立 1934年1月4日 国家   蘇聯（1934年—1991年）   俄羅斯（1991年—） 总部 车里雅宾斯克 隶属 俄罗斯铁路股份公司 路局代码 80 铁路里程 4,934.9公里 奖项 网址 http://yuzd.rzd.ru/ 查 论 编 南乌拉尔铁路局 （俄语： Южно-Уральская железная дорога ， 羅馬化： Yuzhno-Uralskaya zheleznaya doroga ，简称 ЮУЖД ）是俄罗斯铁路股份公司属下的铁路管理局之一，总部位于车里雅宾斯克，负责管辖俄罗斯联邦乌拉尔南部地区的铁路系统，范围包括奥伦堡州、车里雅宾斯克州、库尔干州的大部分铁路，以及萨马拉州、斯维尔德洛夫斯克州、巴什科尔托斯坦共和国的少部分铁路，并有部分铁路延伸至或穿过哈萨克斯坦境内。南乌拉尔铁路局的营业里程为4,934.9公里，与伏尔加铁路局、古比雪夫铁路局、斯维尔德洛夫斯克铁路局、西西伯利亚铁路局相邻，并与南方的哈萨克斯坦铁路连接。 外部链接 （俄文） Южно-Уральская железная дорога — филиал ОАО РЖД （俄文） 南乌拉尔铁路局管辖路线图（一） （俄文） 南乌拉尔铁路局管辖路线图（二） 查 论 编 俄罗斯铁路（集团）股份公司 十月铁路局 · 莫斯科铁路局 · 北高加索铁路局 · 高尔基铁路局 · 古比雪夫铁路局 · 北方铁路局 · 伏尔加铁路局 · 东南铁路局 · 斯维尔德洛夫斯克铁路局 · 南乌拉尔铁路局 · 西西伯利亚铁路局 · 克拉斯诺亚尔斯克铁路局 · 东西伯利亚铁路局 · 后贝加尔铁路局 · 远东铁路局 · 加里宁格勒铁路局 This page is only for reference, If you need detailed information, please check here
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