Let $V$ be finite dimensional vector space. $T$ is a linear transformation from $V$ into $V$ and $E$ is a subspace of $V$. Define

Question

Define $T^{-1}(E)={\{u\in V|T(u)\in E}\}$

Prove that $T^{-1}(E)$ is a subspace of $V$

I know the three rules that define a subspace, but I am not sure how to apply on this statement.

so far I did:

take $U$1,$U$2 from $V$, since $V$ is linear,

$T(U1+U2)=T(U1)+T(U2)$

$T(kU)=kT(U)$

Answer 1

Let $x,y \in T^{-1}E$ then $T(x+y)=T(x)+T(y)\in E.$ Therefore $x+y \in T^{-1}E.$ Let $r\in F$ and $x\in T^{-1}E$ then $T(rx)=rT(x)\in E.$

Answer 2

Since $E$ is a vector space, and $T(U_1),T(U_2)\in E$, $T(U_1+U_2)\in E$ too. This proves that $U_1,U_2\in T^{-1}(E)\implies U_1,U_2\in T^{-1}(E)$.

And since $E$ is a vector space, and $T(U)\in E$, $kT(U)\in E$ too. This proves that $U\in T^{-1}(E)\implies kU\in T^{-1}(E)$.

Finally, $T^{-1}(E)\neq\emptyset$, since $0\in T^{-1}(E)$.

Answer 3

We might want to show

• $T^{-1}(E)$ is not empty
• If $v,w\in T^{-1}(U)$, then $v+w\in T^{-1}(U)$
• If $v\in T^{-1}(U)$ and $c\in K$, then $cv\in T^{-1}(U)$

For the first, we know that $T(0)=0$ because $T$ is linear, and we know that $0\in E$ because $E$ is a subspace. Hence $0\in T^{-1}(E)$

For the second, assume $v,w\in T^{-1}(E)$. Then by definition of $T^{-1}(E)$, $T(v)\in E$ and $T(w)\in E$. We want to know if $u+v\in T^{–1}(E)$, i.e., if $T(u+v)\in E$. But $T(u+v)=T(u)+T(v)$ by linearity of $T$. And as $E$ is a subspace, it is closed under addition, i.e., $T(v)+T(w)\in E$.

Let $v\in T {-1}(U)$ and $c\in K$. By definition, this means $T(v)\in E$. And we want to show that $cv\in T^{-1}(E)$, i.e., that $T(cv)\in E$. By linearity of $T$, we have $T(cv)=cT(v)$. And as $E$ is a subspace, $T(v)\in E$ also implies $cT(v)\in E$.