# Invariant subspace of $R^3$

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 $$W=span(e_1,e_2)$$?

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.

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 $$(T(W)\subseteq W)$$.

• So, $T_W$ $(e_1)=\begin{bmatrix}0\\0\\-1\end{bmatrix}$ right? If this statement is true then I need evaluate $T_W$ $(e_3)$ like the same way. May 6, 2020 at 11:52
• Which gives me $T_W$ $(e_3)=\begin{bmatrix}1\\0\\0\end{bmatrix}$ right? May 6, 2020 at 11:58
• Which application are you referring to with $T_{W}$? May 6, 2020 at 12:00
• $T_W:W→W,\begin{bmatrix}s\\0\\t\end{bmatrix}→\begin{bmatrix}t\\0\\-s\end{bmatrix}$. Forgot to mention that $e_1$=$\begin{bmatrix}1\\0\\0\end{bmatrix}$ May 6, 2020 at 12:45
• Thanks a lot! In the class we did a example like this: hizliresim.com/VtnxGu. But in the example it shown as $W=span(e_1,e_2)$. I'm bit concerned about the changes between $W=span(e_1,e_2)$ to $W=span(e_1,e_3)$. If I'm getting it right $T_W:W→W$ will be $\begin{bmatrix}s\\0\\t\end{bmatrix}$→$\begin{bmatrix}t\\0\\-s\end{bmatrix}$ in $W=span(e_1,e_3)$. May 6, 2020 at 13:00