Properties of Sum of Transformed Matrices Using Permutation Matrices

To determine which options are correct, let us analyze the given matrix:

$$A = \begin{bmatrix} 2 & 1 & 3 \\ 1 & 0 & 2 \\ 3 & 2 & 1 \end{bmatrix}$$

and the sum: $X = \sum_{k=1}^6 P_k A P_k^T$ where $P_1, P_2, \dots, P_6$ are the six $3 \times 3$ permutation matrices.

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Option B: Symmetry of $X$

Notice that the given matrix $A$ is symmetric since $A^T = A$:

$$A^T = \begin{bmatrix} 2 & 1 & 3 \\ 1 & 0 & 2 \\ 3 & 2 & 1 \end{bmatrix}^T = \begin{bmatrix} 2 & 1 & 3 \\ 1 & 0 & 2 \\ 3 & 2 & 1 \end{bmatrix} = A$$

Taking the transpose of $X$: $X^T = \left(\sum_{k=1}^6 P_k A P_k^T\right)^T = \sum_{k=1}^6 (P_k^T)^T A^T P_k^T = \sum_{k=1}^6 P_k A P_k^T = X$ Thus, $X$ is a symmetric matrix.
Option (B) is correct.

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Option A: Eigenvector and Eigenvalue $\alpha$

Let $V = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$. Since each $P_k$ is a permutation matrix, it simply permutes the components of any vector it multiplies. Hence: $P_k^T V = V \quad \text{for each } k = 1, 2, \dots, 6$

Now, compute $AV$:

$$AV = \begin{bmatrix} 2 & 1 & 3 \\ 1 & 0 & 2 \\ 3 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 2+1+3 \\ 1+0+2 \\ 3+2+1 \end{bmatrix} = \begin{bmatrix} 6 \\ 3 \\ 6 \end{bmatrix}$$

Multiplying $X$ by $V$:

$$XV = \sum_{k=1}^6 P_k A P_k^T V = \sum_{k=1}^6 P_k (AV) = \sum_{k=1}^6 P_k \begin{bmatrix} 6 \\ 3 \\ 6 \end{bmatrix}$$

As $P_1, P_2, \dots, P_6$ represent all $3! = 6$ permutations of the vector $\begin{bmatrix} 6 \\ 3 \\ 6 \end{bmatrix}$, the entries of each row of the sum will consist of the number $3$ appearing $2$ times and the number $6$ appearing $4$ times.
Therefore, each component in the resulting sum vector is: $2 \times 3 + 4 \times 6 = 6 + 24 = 30$

Thus:

$$XV = \begin{bmatrix} 30 \\ 30 \\ 30 \end{bmatrix} = 30 \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = 30 V$$

Hence, $\alpha = 30$.
Option (A) is correct.

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Option C: Trace of $X$

The sum of diagonal entries of a matrix is its trace. Using the cyclic property of the trace ($\text{Tr}(AB) = \text{Tr}(BA)$) and the fact that $P_k^T P_k = I$: $\text{Tr}(X) = \text{Tr}\left(\sum_{k=1}^6 P_k A P_k^T\right) = \sum_{k=1}^6 \text{Tr}(P_k A P_k^T) = \sum_{k=1}^6 \text{Tr}(A P_k^T P_k) = \sum_{k=1}^6 \text{Tr}(A) = 6 \cdot \text{Tr}(A)$

The trace of matrix $A$ is: $\text{Tr}(A) = 2 + 0 + 1 = 3$

Thus: $\text{Tr}(X) = 6 \times 3 = 18$ Option (C) is correct.

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Option D: Invertibility of $X - 30I$

From Option (A), we have: $XV = 30V \implies (X - 30I)V = \mathbf{0}$ Since $V = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \neq \mathbf{0}$, the matrix equation $(X - 30I)V = \mathbf{0}$ has a non-trivial solution, which implies: $\det(X - 30I) = 0$ Hence, $X - 30I$ is a singular (non-invertible) matrix.
Option (D) is incorrect.

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Conclusion:

The correct options are A, B, and C.