Answer to Question #121535 in Differential Geometry | Topology for Brain

Question

Define an operator T in End(F^2) by T(x,y)= (y,0) Let
U = {(x,0) | x in F}. Show that
U is invariant under T and T |U is the 0 operator on U;

Answer 1

Operator $T\in End(F^2)$ i.e $T:F^2\longrightarrow F^2$ such that $T(x,y)=(y,0)$

Let,

$$U=\{(x,0):x\in F\}$$

Note that, for any $u\in U$ ,thus $u=(x_1,0)$ for some $x_1\in F$

$T(u)=(0,0)\hspace{1cm}(\because$ By definition of $T)$ which implies $T(u)\in U$ ,Hence $T$ is invariant under $U$ .

Also note that $u$ was arbitrary and $u\in U\implies u\in F^2$ but $T|_U:U\longrightarrow U$ ,thus

$$T|_U(u)=T(u)=0\implies T|_U=0$$