Statement 1: $\operatorname{tr}(A B)=\operatorname{tr}(B A)$
Trace: sum of eigen value
$$
A_{m \times n} \&B_{n \times m}=A B_{m \times m}
$$
$$B_{n \times m} \& A_{m \times n}=B A_{n \times n}$$
So, whatever size of matrix $A B$ and $B A$.
$\Rightarrow$ Non-zero eigen value of $A B \& B A$ are same and other eigen value are zero.
example:
suppose. $A B_{4 \times 4}$ has 4 eigen value
$$
\left(\lambda_1=2, \lambda_2=1, \lambda_3=7, \lambda_4=8\right)
$$
and $\mathrm{BA}_{7 \times 7}$ has 7 eigen value
$\Rightarrow$ out of 7,4 eigen value of $B A$ must same cs $A B$. and rest are zero.
$$
\begin{aligned}
& \left.\lambda_1=2\right] \operatorname{}\;\;\;\;\; \lambda_5=0 \\
& \left.\begin{array}{l}
\lambda_2=1 \\
\lambda_3=7
\end{array}\right\} \begin{array}{ll}
\text { } & \lambda_6=0 \\
& \lambda_7=0
\end{array} \\
& \lambda_4=8 \\
&
\end{aligned}
$$
Statement 2: $\operatorname{tr}(C D)=\operatorname{tr}(D C)$
it's very easy Both matrices have the same non-zero eigenvalues and their sum is also the same.
Both statements are true.
option C Correct