Given :

Matrices $ A_{m \times n} $, $ B_{n \times m} $, $ C_{n \times n} $, and $ D_{n \times n} $.

$ \text{tr}(\cdot) $ denotes the trace of a matrix, which is the sum of its diagonal elements.

Statement 1 :

$ \text{tr}(AB) = \text{tr}(BA) $

Analysis :

Dimensions of $ AB $ and $ BA $ :

$ A $ is $ m \times n $, $ B $ is $ n \times m $.

$ AB $ is $ m \times m $, $ BA $ is $ n \times n $.

Trace Equality for Rectangular Matrices :

Even though $ AB $ and $ BA $ are square matrices of different sizes, we can still compare their traces.

Proof :

$ \begin{align*} \text{tr}(AB) &= \sum_{i=1}^{m} (AB)_{ii} = \sum_{i=1}^{m} \sum_{k=1}^{n} A_{ik} B_{ki} \\ &= \sum_{k=1}^{n} \sum_{i=1}^{m} A_{ik} B_{ki} = \sum_{k=1}^{n} \sum_{i=1}^{m} B_{ki} A_{ik} \\ &= \sum_{k=1}^{n} (BA)_{kk} = \text{tr}(BA) \end{align*} $

Conclusion: Statement 1 is correct.

Statement 2 :

$ \text{tr}(CD) = \text{tr}(DC) $

Analysis :

Dimensions of $ C $ and $ D $ :

Both $ C $ and $ D $ are $ n \times n $ square matrices.

Trace Property for Square Matrices :

For any two $ n \times n $ matrices $ X $ and $ Y $, the trace of their product is invariant under cyclic permutations :

$ \text{tr}(XY) = \text{tr}(YX) $

Conclusion : Statement 2 is correct.

Final Answer :

Both Statement 1 and Statement 2 are correct.

Answer : Option C