So the task says:

for $\alpha_1\ldots,\alpha_n, \beta\in \mathbb R$, we define $U = \{(x_1\ldots,x_n) ∈ \mathbb{R}^n \,|\, \sum_{i=1}^n \alpha_ix_i = \beta\}$. 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.

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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!

  • 2
    $\begingroup$ Well if $B$ is not zero then 0 is not in the subspace. $\endgroup$ – Calvin Khor Jan 20 at 19:34
  • $\begingroup$ Of course, but the OP wanted to understand that specific proof. $\endgroup$ – José Carlos Santos Jan 20 at 20:13
  • $\begingroup$ 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 Jan 20 at 20:35

Because if $\beta\neq0$, then you take, say, $\lambda=\mu=1$, and then it will be false that $\lambda\beta+\mu\beta=\beta$, since then this would mean that $2\beta=\beta$.

  • $\begingroup$ 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 Jan 20 at 20:05
  • $\begingroup$ If $\mu=\lambda=\frac12$, you deduce nothing. But the proof says that for all real $\mu$ and $\lambda$ we have $\mu\beta+\lambda\beta=\beta$. So, I used $\mu=\lambda=1$ to reach a contradiction. That's all. $\endgroup$ – José Carlos Santos Jan 20 at 21:34
  • $\begingroup$ Ok thank you! I understand it now. $\endgroup$ – Petar Jan 20 at 21:51
  • $\begingroup$ If my answer was useful, perhaps that you could mark it as the accepted one. $\endgroup$ – José Carlos Santos Jan 20 at 21:53
  • $\begingroup$ yes i did it now ,tnx $\endgroup$ – Petar Jan 20 at 22:09

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 $\mathbf x\in U\leftrightarrow\mathbf x\cdot\mathbf a=B$. Now, $\mathbf x,\mathbf y\in U\implies\mathbf x+\mathbf y\in U\implies(\mathbf x+\mathbf y)\cdot \mathbf a=\mathbf{x\cdot a}+\mathbf{y\cdot a}=B+B=B$, that implies $B=0$.


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