During the lecture I've missed out on, there was the following problem:
Let $\mathcal{T}:\mathbb{R}^4\to\mathbb{R}^4$, where linear operator $\mathcal{T}$ is defined by $\mathcal{T}(a,b,c,d)=(a+b+2c-d,b+d,2c-d,c+d)$, and let $W=\{(t,s,0,0)|t,s\in\mathbb{R}\}$. Check whether $W$ is invariant subspace of $\mathbb{R}^4$ with respect to linear operator $\mathcal{T}$.
Solution from the lecture:
We need to prove that $\forall w\in W$ it stands that $\mathcal{T}(w)\in W$, from the definition of invariant subspace. Let's take arbitrary $(t,s,0,0)\in W$. Then $\mathcal{T}(t,s,0,0)=(t+s,s,0,0)\in W$, hence $W$ is invariant subspace of $\mathbb{R}^4$.
I don't understand how they just arrived to the conclusion $(t+s,s,0,0)\in W$. Why is this true when we know that vectors $(t,s,0,0)$ and $(t+s,s,0,0)$ are generated by linear combinations of two different vectors?
$$(t,s,0,0)=t(1,0,0,0)+s(0,1,0,0)$$
and
$$(t+s,s,0,0)=(t+s)(1,0,0,0)+s(0,1,0,0)$$
so both them are generated by the same span$\{(1,0,0,0),(0,1,0,0)\}$