Show that for $T:V \to W$ if $H$ is a subspace of $W$, then $T^{-1}(H)$ is a subspace of $V$

Question

I'm trying to show that even if there isn't a function $T^{-1}$ for $T:V \to W$ a linear transformation, if $H$ is a subspace of $W$ then $T^{-1}(H)$ is a subspace of $V$

What I know is that $T^{-1}(H)$ is defined as: $$T^{-1}(H) = \{v \in V \mid T(v) \in H\}$$

It seems perfectly sensible to me that $T^{-1}(H)$ is a subspace of $W$. I know that $H$ is a subspace of $W$, thus it has a spanning set of $0 \leq n < dim(V)$ vectors. Because I don't know anything about $T$ (if it is injective, surjective, etc...) I can't make any claims that I can think of. I'm pretty much stuck here...

Any help is greatly appreciated!

Answer 1

Since $T^{-1}(H)$ is a subset of the vector space $V$, you only need to show that $T^{-1}(H)$ is a subspace of $V$ (using the subspace test). The subspace test consists of two tests

• If $v_1$ and $v_2$ are in $T^{-1}(H)$, then you need to show that $v_1+v_2$ is in $T^{-1}(H)$.

• If $v_1$ is in $T^{-1}(H)$ and $c$ is a scalar, then you need to show that $cv_1$ is in $T^{-1}(H)$.

I'll show how to check the first one:

Suppose that $v_1$ and $v_2$ are in $T^{-1}(H)$. Then, by definition, $T(v_1)$ and $T(v_2)$ are in $H$. Since $H$ is closed under addition, $T(v_1)+T(v_2)$ is also in $H$. Since $T$ is a linear map, $T(v_1+v_2)=T(v_1)+T(v_2)$. Therefore, $T(v_1+v_2)$ is in $H$. Therefore, by the definition of $T^{-1}(H)$, $v_1+v_2$ is in $T^{-1}(H)$ since its image is in $H$.

Answer 2

Let $v,w\in T^{-1}(H)$ and $c\in F$ the base field. Then it suffices to show that $v+cw\in T^{-1}(H)$. Pass to the image and use linearity, then go back to the preimage.

Answer 3

Does $0$ belong to $T^{-1}(H)= Yes, because $T(0)=0$.

If $v,w\in T^{-1}(H)$, and $\alpha,\beta\in\mathbb R$, is it true that $\alpha v+\beta w\in T^{-1}(H)$? Yes, because$$T(\alpha v+\beta w)=\alpha T(v)+\beta T(w)\in H.$$