Key Concepts and Formulas

1. Orthogonal Matrices: A square matrix $P$ is orthogonal if its transpose is equal to its inverse, i.e., $P^T = P^{-1}$. This property implies $P^T P = P P^T = I$, where $I$ is the identity matrix. Rotation matrices are a classic example of orthogonal matrices.
2. Similar Matrices: Two square matrices $A$ and $B$ are similar if there exists an invertible matrix $P$ such that $B = P A P^{-1}$. A crucial property for similar matrices is that if $B = P A P^{-1}$, then $B^k = P A^k P^{-1}$ for any positive integer $k$. Similar matrices share many properties, including their trace and eigenvalues.
3. Powers of Triangular Matrices: For an upper (or lower) triangular matrix $M$, the diagonal elements of its $k$-th power, $M^k$, are simply the $k$-th powers of its original diagonal elements. For example, if $M = \begin{bmatrix} a & b \\ 0 & d \end{bmatrix}$, then $M^k = \begin{bmatrix} a^k & \star \\ 0 & d^k \end{bmatrix}$, where $\star$ is some off-diagonal element.
4. Trace of a Matrix: The trace of a square matrix is the sum of its diagonal elements. For a matrix $M = \begin{bmatrix} m_{11} & m_{12} \\ m_{21} & m_{22} \end{bmatrix}$, $\text{Tr}(M) = m_{11} + m_{22}$.

Step-by-Step Solution

Step 1: Analyze Matrix $P$ and its Properties We are given the matrix $P$: $P = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$ This is a standard 2D rotation matrix. To determine its properties, we first find its transpose, $P^T$: $P^T = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$ Next, we compute the product $P^T P$: $P^T P = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$ $P^T P = \begin{bmatrix} \cos^2 \theta + \sin^2 \theta & -\cos \theta \sin \theta + \sin \theta \cos \theta \\ -\sin \theta \cos \theta + \cos \theta \sin \theta & \sin^2 \theta + \cos^2 \theta \end{bmatrix}$ Using the identity $\cos^2 \theta + \sin^2 \theta = 1$: $P^T P = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$ Why this step? The result $P^T P = I$ confirms that $P$ is an orthogonal matrix. A key property of orthogonal matrices is that their inverse is equal to their transpose, i.e., $P^{-1} = P^T$. This simplification is fundamental for efficiently manipulating the given matrix expressions.

Step 2: Simplify the Expression for $C$ using Similarity Transformation We are given $B = PAP^T$ and $C = P^T B^{10} P$. From Step 1, we know that $P^T = P^{-1}$. Substituting this into the expression for $B$: $B = PAP^{-1}$ This shows that matrix $B$ is similar to matrix $A$. A crucial property of similar matrices is how their powers are related. If $B = PAP^{-1}$, then for any positive integer $k$: $B^k = (PAP^{-1})(PAP^{-1})\dots(PAP^{-1}) \quad (\text{k times})$ Due to the cancellation of $P^{-1}P = I$ in the middle terms: $B^k = P A (P^{-1}P) A (P^{-1}P) \dots (P^{-1}P) A P^{-1} = P A^k P^{-1}$ Applying this property for $k=10$: $B^{10} = P A^{10} P^{-1}$ Now, substitute $P^{-1} = P^T$ back into this expression: $B^{10} = P A^{10} P^T$ Finally, substitute this expression for $B^{10}$ into the definition of $C$: $C = P^T B^{10} P$ $C = P^T (P A^{10} P^T) P$ Again, using the property $P^T P = I$: $C = (P^T P) A^{10} (P^T P)$ $C = I A^{10} I$ $C = A^{10}$ Why this step? This is the most critical simplification. By leveraging the orthogonal nature of $P$ and the properties of similar matrices, we've reduced the complex problem of calculating $B^{10}$ and then $C$ to simply calculating the 10th power of matrix $A$. This avoids direct computation of $B$, which would be cumbersome due to the $\theta$ dependency.

Step 3: Calculate $A^{10}$ Now we need to compute $A^{10}$. The given matrix $A$ is: $A = \begin{bmatrix} \frac{1}{\sqrt{2}} & -2 \\ 0 & 1 \end{bmatrix}$ We observe that $A$ is an upper triangular matrix. Why this observation? For any triangular matrix (upper or lower), the diagonal elements of its $k$-th power are simply the $k$-th powers of its original diagonal elements. This property significantly simplifies the calculation of $A^{10}$, as we only need the diagonal elements to find the trace.

Let the diagonal elements of $A$ be $a_{11} = \frac{1}{\sqrt{2}}$ and $a_{22} = 1$. Then, the diagonal elements of $A^{10}$ will be $a_{11}^{10}$ and $a_{22}^{10}$. Calculate these values: $a_{11}^{10} = \left(\frac{1}{\sqrt{2}}\right)^{10} = \frac{1^{10}}{(\sqrt{2})^{10}} = \frac{1}{2^{10/2}} = \frac{1}{2^5} = \frac{1}{32}$ $a_{22}^{10} = 1^{10} = 1$ So, the matrix $A^{10}$ (which is equal to $C$) will have the form: $C = A^{10} = \begin{bmatrix} \frac{1}{32} & \star \\ 0 & 1 \end{bmatrix}$ where $\star$ represents the off-diagonal element, which is not required for finding the trace.

Step 4: Calculate the Sum of Diagonal Elements of $C$ and find $m+n$ The problem asks for the sum of the diagonal elements of $C$, which is $\text{Tr}(C)$. From Step 3, the diagonal elements of $C$ are $\frac{1}{32}$ and $1$. $\text{Tr}(C) = \frac{1}{32} + 1$ $\text{Tr}(C) = \frac{1+32}{32} = \frac{33}{32}$ We are given that the sum of the diagonal elements of $C$ is $\frac{m}{n}$, where $\operatorname{gcd}(m, n)=1$. Comparing $\frac{33}{32}$ with $\frac{m}{n}$, we have $m=33$ and $n=32$. Let's verify the greatest common divisor: $33 = 3 \times 11$ $32 = 2^5$ The greatest common divisor of $33$ and $32$ is $1$, so $\operatorname{gcd}(33, 32)=1$ is satisfied.

Finally, we need to find $m+n$: $m+n = 33 + 32 = 65$ Why this step? This is the final calculation as per the problem statement, providing the required integer value.

Common Mistakes & Tips

• Forgetting $P^T = P^{-1}$: A common error is not recognizing $P$ as an orthogonal matrix, leading to much more complicated calculations involving $P^T$ and $P^{-1}$ explicitly. Always check for special matrix types like orthogonal or symmetric matrices.
• Direct Calculation of $B^{10}$: Attempting to compute $B = PAP^T$ explicitly and then raising it to the 10th power would be extremely time-consuming and prone to errors. The similarity transformation $B^k = P A^k P^{-1}$ is a critical shortcut.
• Incorrectly Powering Matrices: For a general matrix, the diagonal elements of $M^k$ are NOT simply the $k$-th powers of the original diagonal elements. This simplification only applies to triangular (or diagonal) matrices. Ensure to identify matrix types before applying such shortcuts.

Summary

This problem effectively tests the understanding of matrix properties and transformations. The solution began by identifying the rotation matrix $P$ as an orthogonal matrix, which allowed us to use the property $P^T = P^{-1}$. This was crucial for simplifying the expression for $B$ to $B = PAP^{-1}$, demonstrating that $A$ and $B$ are similar matrices. Leveraging the similarity property, we simplified $C = P^T B^{10} P$ down to $C = A^{10}$. Finally, recognizing that $A$ is an upper triangular matrix allowed for a straightforward calculation of the diagonal elements of $A^{10}$ by simply raising the original diagonal elements to the power of 10. Summing these diagonal elements yielded $\frac{33}{32}$, from which $m=33$ and $n=32$ were identified, leading to $m+n = 65$.

The final answer is $\boxed{65}$, which corresponds to option (A).