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
    4 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
    2 hours ago










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









$endgroup$













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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
      4 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
      2 hours ago










    • $begingroup$
      If my answer was useful, perhaps that you could mark it as the accepted one.
      $endgroup$
      – José Carlos Santos
      2 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
      4 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
      2 hours ago










    • $begingroup$
      If my answer was useful, perhaps that you could mark it as the accepted one.
      $endgroup$
      – José Carlos Santos
      2 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
      4 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
      2 hours ago










    • $begingroup$
      If my answer was useful, perhaps that you could mark it as the accepted one.
      $endgroup$
      – José Carlos Santos
      2 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
      4 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
      2 hours ago










    • $begingroup$
      If my answer was useful, perhaps that you could mark it as the accepted one.
      $endgroup$
      – José Carlos Santos
      2 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
    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












    $begingroup$
    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
    2 hours ago




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












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




    $begingroup$
    If my answer was useful, perhaps that you could mark it as the accepted one.
    $endgroup$
    – José Carlos Santos
    2 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 4 hours ago









        Shubham JohriShubham Johri

        4,780717




        4,780717






























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