Problem 1508

To show that there exists a matrix $T$ and an invertible matrix $S$ such that
$$B = S(A + T)S^{-1} - T$$
if and only if $\operatorname{tr}(A) = \operatorname{tr}(B)$, we will proceed as follows:

1. **Assume $\operatorname{tr}(A) = \operatorname{tr}(B)$**:
- We need to show that there exist matrices $T$ and $S$ such that $B = S(A + T)S^{-1} - T$.

2. **Consider the equation $B + T = S(A + T)S^{-1}$**:
- This implies that $B + T$ is similar to $A + T$.

3. Use the property of similar matrices:
- If two matrices are similar, they have the same trace. Therefore, $\operatorname{tr}(B + T) = \operatorname{tr}(S(A + T)S^{-1})$.

4. Simplify the trace equation:
- Since the trace of a matrix is invariant under similarity transformations, we have:
$$\operatorname{tr}(B + T) = \operatorname{tr}(A + T)$$
- Expanding this, we get:
$$\operatorname{tr}(B) + \operatorname{tr}(T) = \operatorname{tr}(A) + \operatorname{tr}(T)$$

5. **Use the given condition $\operatorname{tr}(A) = \operatorname{tr}(B)$**:
- Substituting $\operatorname{tr}(A) = \operatorname{tr}(B)$ into the equation, we get:
$$\operatorname{tr}(B) + \operatorname{tr}(T) = \operatorname{tr}(B) + \operatorname{tr}(T)$$
- This equation is always true, which means that the condition $\operatorname{tr}(A) = \operatorname{tr}(B)$ is sufficient for the existence of such matrices $T$ and $S$.

6. **Construct the matrices $T$ and $S$**:
- To explicitly construct $T$ and $S$, we can choose $T$ to be any matrix such that $A + T$ and $B + T$ are similar. One possible choice is $T = 0$, which simplifies the problem to finding $S$ such that $B = SAS^{-1}$.

7. Conclusion:
- Since $\operatorname{tr}(A) = \operatorname{tr}(B)$, we can always find such matrices $T$ and $S$ that satisfy the given equation.