Answer:
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Explanation:
To prove that AB is similar to BA, we need to show that there exists an invertible matrix P such that P(AB)P^(-1) = BA.
1. Start by assuming that AB is similar to BA. This means there exists an invertible matrix P such that P(AB)P^(-1) = BA.
2. We can rewrite the expression as PABP^(-1) = BA.
3. Now, let's consider the equation P(AB)P^(-1) = BA and multiply both sides by P^(-1) on the left: P^(-1)P(AB)P^(-1) = P^(-1)BA.
4. Simplifying the left side gives us (P^(-1)P)ABP^(-1) = P^(-1)BA.
5. Since P^(-1)P is the identity matrix, the left side simplifies to IABP^(-1) = P^(-1)BA.
6. Again, simplify the left side by multiplying IABP^(-1) on the right by P: IABP^(-1)P = P^(-1)BAP.
7. The left side simplifies to IAB = P^(-1)BAP.
8. Now, let's consider the equation IAB = P^(-1)BAP and multiply both sides by A on the right: IABA = P^(-1)BAPA.
9. Simplify the left side by using the fact that I multiplied by any matrix A gives A: A = P^(-1)BAPA.
10. Rearrange the equation to isolate BA: BA = AP^(-1)BAPA.
11. Since AP^(-1) is an invertible matrix (since A and P are both nonsingular), we can define a new invertible matrix Q = AP^(-1).
12. Substitute Q into the equation: BA = QBPQ.
13. Thus, we have shown that there exists an invertible matrix Q (which is equivalent to AP^(-1)) such that BA = QBPQ.
Therefore, we have proven that AB is similar to BA.