Invariant subspace of $R^3$

Question

Let $T:R^3→R^3$ be the linear operator defined by $$T(\begin{bmatrix}a\\b\\c\end{bmatrix})=\begin{bmatrix}b+c\\2b\\a-b+c\end{bmatrix}$$

Show that $W=span(e_1,e_3)$ is a T-invariant subspace of $R^3$.

Let $\alpha={e_1,e_3} $ be ordered basis for W and $\beta={e_1,e_2,e_3}$ be ordered basis for $R^3=V$.

(In my textbook's example, W was $W=span({e_1,e_2})$ and the $T_W:W→W,\begin{bmatrix}s\\t\\0\end{bmatrix}→\begin{bmatrix}t\\-s\\0\end{bmatrix}$) So my question is how can I show that W is a T-invariant subspace? And also how can I write matrices like $W=span(e_1,e_2)$?

Answer 1

You see that, being $a,c\in \mathbb{R}$, $$T(a\cdot e_1 + c\cdot e_3) = \begin{pmatrix}c\\0\\a+c\end{pmatrix}$$ Clearly, $$\begin{pmatrix}c\\0\\a+c\end{pmatrix}\in \text{span}(e_1,e_3)=W$$ because it equals $(c\cdot e_1 + (a+c)\cdot e_3)$, so it's a $T$-invariant subspace by the definition of invariant subspace $(T(W)\subseteq W)$.

Answer 2

It is easy to see that $T(e_1)=e_3 \in W$ and that $T(e_3)=e_1+e_3 \in W.$

This gives

$$T(W)=W,$$

hence $W$ is a $T-$ invariant subspace.