To prove that $S = \{ v \in V \mid f(v) \in U \}$ is a subspace of $V$, let's proceed by verifying that $S$ meets the necessary conditions for being a subspace.

Given:

1. $V$ and $W$ are vector spaces over a field $\mathbb{K}$.
2. $U$ is a subspace of $W$.
3. $f: V \rightarrow W$ is a linear map.

To Prove:

We need to show that $S = \{ v \in V \mid f(v) \in U \}$ is a subspace of $V$.

Proof:

To confirm that $S$ is a subspace, we must check the following properties:

1. The Zero Vector is in $S$:

   Since $f$ is a linear map, $f(0) = 0$. Given that $U$ is a subspace of $W$, it contains the zero vector of $W$, which means $0 \in U$. Therefore, $f(0) \in U$, implying $0 \in S$. Thus, $S$ contains the zero vector of $V$.

2. Closure Under Addition:

   Take any $v_1, v_2 \in S$. By definition of $S$, we have $f(v_1) \in U$ and $f(v_2) \in U$.

   Since $U$ is a subspace of $W$, it is closed under addition, so $f(v_1) + f(v_2) \in U$.

   By the linearity of $f$, we have: $f(v_1 + v_2) = f(v_1) + f(v_2).$ Thus, $f(v_1 + v_2) \in U$, which implies $v_1 + v_2 \in S$. Therefore, $S$ is closed under addition.

3. Closure Under Scalar Multiplication:

   Take any $v \in S$ and any scalar $\alpha \in \mathbb{K}$. Since $v \in S$, we have $f(v) \in U$.

   Because $U$ is a subspace of $W$, it is closed under scalar multiplication, so $\alpha f(v) \in U$.

   By the linearity of $f$, we have: $f(\alpha v) = \alpha f(v).$ Thus, $f(\alpha v) \in U$, which implies $\alpha v \in S$. Therefore, $S$ is closed under scalar multiplication.

Conclusion:

Since $S$ contains the zero vector, is closed under addition, and is closed under scalar multiplication, $S$ is a subspace of $V$.

Would you like more details or have any questions on specific steps?

Here are some related questions you might find useful:

1. How would this proof change if $f$ were not a linear map?
2. What are some examples of vector spaces and subspaces in practice?
3. How does the subspace test work in general for subsets of vector spaces?
4. What if $f$ was a bijective map? How would that impact $S$?
5. Can you think of another approach to demonstrate that $S$ is a subspace?

Tip: When proving a subset is a subspace, always verify the subspace conditions: zero vector inclusion, closure under addition, and closure under scalar multiplication.