Let's go through the proof step-by-step.

Part 1: Prove that if $Ax = 0$ for all $n \times 1$ column matrices $x$, then $A$ is a zero matrix.

1. Assumption: We are given that $Ax = 0$ for all $n \times 1$ column vectors $x$.

2. Implication: The equation $Ax = 0$ for all $x$ implies that every linear transformation applied by $A$ results in the zero vector. This property holds for any arbitrary vector $x$ in $\mathbb{R}^n$.

3. Conclusion: To show that $A$ is a zero matrix, consider the standard basis vectors $e_1, e_2, \dots, e_n$ of $\mathbb{R}^n$, where each $e_i$ has a 1 in the $i$-th position and 0 in all other positions.

   Since $Ax = 0$ for all $x$, applying $A$ to each basis vector $e_i$ must also yield the zero vector. Therefore: $A e_i = 0 \quad \text{for each } i = 1, 2, \dots, n.$ This means that each column of $A$ must be zero because each $e_i$ corresponds to selecting a specific column of $A$. Thus, $A$ is a zero matrix.

Part 2: Prove that if $Ax = Bx$ for all $x$, then $A = B$.

1. Assumption: Suppose $Ax = Bx$ for all $x \in \mathbb{R}^n$.

2. Reformulate the Equation: We can rewrite this as: $(A - B)x = 0 \quad \text{for all } x.$

3. Application of Part 1: From Part 1, we know that if $(A - B)x = 0$ for all $x$, then $A - B$ must be the zero matrix. Thus: $A - B = 0,$ which implies that $A = B$.

Conclusion

The proof is complete. We have shown that:

1. If $Ax = 0$ for all $x$, then $A$ is a zero matrix.
2. If $Ax = Bx$ for all $x$, then $A = B$.

Would you like further clarification or have any additional questions?

Here are some follow-up questions you might find helpful:

1. What is the significance of proving $Ax = 0$ implies $A$ is a zero matrix?
2. Can this method be applied to non-square matrices $A$ and $B$?
3. How does this proof relate to the concept of linear independence?
4. Can you provide an example of matrices $A$ and $B$ to illustrate this proof?
5. How does the result change if $x$ is restricted to specific types of vectors?

Tip: When proving results involving matrices, using basis vectors or arbitrary vectors in $\mathbb{R}^n$ is often a powerful technique for showing properties that hold for all vectors.