Subspace of a vector space problem

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 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!

• Well if $B$ is not zero then 0 is not in the subspace. – Calvin Khor Jan 20 at 19:34
• Of course, but the OP wanted to understand that specific proof. – José Carlos Santos Jan 20 at 20:13
• 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 – 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 $$2\beta=\beta$$.
• 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. – José Carlos Santos Jan 20 at 21:34
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 $$B=0$$.