Determine whether $\{(x_1,x_2,x_3)^T \mid x_1 = x_2 = x_3\}$ is a subspace 
Determine whether $\{(x_1,x_2,x_3)^T \mid x_1 = x_2 = x_3\}$ is a subspace

I know that I must show that it is closed under addition and multiplication, but I'm confused as to how I should approach doing that.
 A: This is the set of all vectors of the form $(t,t,t)$. Note that $\lambda(t,t,t)=(\lambda t,\lambda t,\lambda t)$ and $(m,m,m)+(n,n,n)=(m+n,m+n,m+n)$ which is of the same form. Finally, note that the set is nonempty as it contains  $(0,0,0)$.
A: The notation $\{u\mid \Phi(u)\}$ is the set of those $u$ elements, which satisfy a given property $\Phi$. In your case, it is the set of somethings of the form $(x_1,x_2,x_3)^T$, such that $x_1=x_2=x_3$. Here $()^T$ means transposition, i.e. it is rather this column vector: $\begin{pmatrix} x_1\\x_2\\x_3\end{pmatrix}$, because in linear algebra one usually works with column vectors.
You have to prove that if you pick two arbitrary elements, and add them, it stays in the same set, and similarly for multiplication by an arbitrary scalar. Arbitrarity in this algebraic language means basically that you use letters instead of concrete numbers, hence will be generally true for all numbers.
A: $(0,0,0)\in W\implies W\neq\phi$
Whenever $a=(a_1,a_1,a_1)\in W$ and $b=(b_1,b_1,b_1)\in W$, then $\lambda a+\mu b=(\lambda a_1+\mu b_1,\lambda a_1+\mu b_1,\lambda a_1+\mu b_1)\in W$
Thus, $W$ is a subspace.
