We are given the following equations involving matrices $A$ and $B$:

$AB = A \quad \text{(1)}$ $BA = B \quad \text{(2)}$

We want to prove that $B^T A^T = A^T$, where $B^T$ and $A^T$ are the transposes of $B$ and $A$, respectively.

Step-by-Step Proof:

1. Start by transposing both sides of equation (2): $BA = B$ Taking the transpose of both sides: $(BA)^T = B^T$ Using the property of transposition that $(XY)^T = Y^T X^T$, we have: $A^T B^T = B^T \quad \text{(3)}$

2. Simplifying equation (3): From equation (3), we now have: $A^T B^T = B^T$ This tells us that when we multiply $A^T$ by $B^T$, the result is simply $B^T$.

3. Transpose equation (1): Now, take the transpose of equation (1): $AB = A$ Taking the transpose of both sides: $(AB)^T = A^T$ Again using the transposition property $(XY)^T = B^T A^T$, we get: $B^T A^T = A^T \quad \text{(4)}$

4. Conclusion: From equation (4), we see that $B^T A^T = A^T$, which is the result we wanted to prove.

Thus, we have shown that $B^T A^T = A^T$, as required.

Do you have any questions or would you like further details?

Here are 5 related questions you might find interesting:

1. Can we extend the result to any two matrices $A$ and $B$ with similar properties?
2. How does the proof change if $A$ or $B$ are not square matrices?
3. What other properties can be deduced about $A$ and $B$ given these conditions?
4. How does the rank of $A$ and $B$ relate to this proof?
5. Can we derive conditions for when $A$ and $B$ are invertible based on these equations?

Tip: Transposition reverses the order of matrix multiplication, so always be mindful of the order when working with transposes!