To prove that $(ST)^2 = 0$, we need to use the given information that $\text{range}(S) \subseteq \text{null}(T)$. We will proceed step-by-step:

Step 1: Definition of $(ST)^2$

We need to show that $(ST)^2 = 0$. This expression expands as: $(ST)^2 = STST$ So, we need to show that $STST = 0$ for any vector $v \in V$.

Step 2: Using the assumption about the range of $S$

From the assumption $\text{range}(S) \subseteq \text{null}(T)$, we know that for any vector $v \in V$, the vector $S(v)$ is in the null space of $T$. This means that: $T(S(v)) = 0 \quad \text{for all} \quad v \in V.$ In other words, applying $T$ to any vector in the range of $S$ gives 0.

Step 3: Applying $STST$ to an arbitrary vector $v$

Now, let's compute $STST(v)$ for an arbitrary vector $v \in V$: $STST(v) = ST(S(T(v))).$ First, apply $T$ to $v$, which gives $T(v) \in V$. Then apply $S$ to $T(v)$, which gives $S(T(v))$.

Now, since $S(T(v))$ is in the range of $S$ and by our assumption $\text{range}(S) \subseteq \text{null}(T)$, we have: $T(S(T(v))) = 0.$

Thus, $STST(v) = S(0) = 0.$

Step 4: Conclusion

Since $STST(v) = 0$ for all $v \in V$, it follows that: $(ST)^2 = 0.$

Thus, we have proved that $(ST)^2 = 0$.

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Would you like further clarification on any step, or details on how these concepts apply in linear algebra?

Here are 5 related questions to deepen your understanding:

1. What is the null space of a linear transformation?
2. Can you explain the relationship between a linear map's kernel and image?
3. How does the property $\text{range}(S) \subseteq \text{null}(T)$ affect the behavior of the product $ST$?
4. What is the significance of nilpotent operators in linear algebra?
5. How would the proof change if $\text{range}(S)$ were not a subspace of $\text{null}(T)$?

Tip: The null space (or kernel) of a linear transformation consists of all vectors that the transformation maps to the zero vector. Understanding its geometric and algebraic properties is key to solving many linear algebra problems.