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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 $W=span(e_1,e_2)$?

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2 Answers 2

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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.

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

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  • $\begingroup$ 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. $\endgroup$
    – RChamuel
    May 6, 2020 at 11:52
  • $\begingroup$ Which gives me $T_W$ $(e_3)=\begin{bmatrix}1\\0\\0\end{bmatrix}$ right? $\endgroup$
    – RChamuel
    May 6, 2020 at 11:58
  • $\begingroup$ Which application are you referring to with $T_{W}$? $\endgroup$ May 6, 2020 at 12:00
  • $\begingroup$ $T_W:W→W,\begin{bmatrix}s\\0\\t\end{bmatrix}→\begin{bmatrix}t\\0\\-s\end{bmatrix}$. Forgot to mention that $e_1$=$\begin{bmatrix}1\\0\\0\end{bmatrix}$ $\endgroup$
    – RChamuel
    May 6, 2020 at 12:45
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    $\begingroup$ 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 $W=span(e_1,e_3)$. $\endgroup$
    – RChamuel
    May 6, 2020 at 13:00

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