Here's a detailed and educational rewrite of the solution:

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Understanding the Matrices and Key Properties

The problem involves matrix operations, specifically powers of matrices and similarity transformations. Let's first analyze the given matrices:

1. Matrix P: P = \left[ {\matrix{ {{{\sqrt 3 } \over 2}} & {{1 \over 2}} \cr { - {1 \over 2}} & {{{\sqrt 3 } \over 2}} \cr } } \right] We can recognize this matrix as a rotation matrix. A general 2D rotation matrix for a counter-clockwise rotation by an angle $\theta$ is given by R(\theta) = \left[ {\matrix{ {\cos \theta} & {-\sin \theta} \cr {\sin \theta} & {\cos \theta} \cr } } \right]. However, our matrix $P$ is of the form \left[ {\matrix{ {\cos \theta} & {\sin \theta} \cr {-\sin \theta} & {\cos \theta} \cr } } \right]. Comparing the elements: $\cos \theta = \frac{\sqrt{3}}{2}$ and $\sin \theta = \frac{1}{2}$. This corresponds to $\theta = 30^\circ$ or $\frac{\pi}{6}$ radians. So, $P$ represents a rotation matrix. An important property of rotation matrices is that they are orthogonal matrices. For an orthogonal matrix, its transpose is equal to its inverse, i.e., $P^T = P^{-1}$. This implies that $P P^T = P^T P = I$, where $I$ is the identity matrix.

   Let's find $P^T$: P^T = \left[ {\matrix{ {{{\sqrt 3 } \over 2}} & { - {1 \over 2}} \cr {{1 \over 2}} & {{{\sqrt 3 } \over 2}} \cr } } \right] Now, let's verify $P P^T$: P P^T = \left[ {\matrix{ {{{\sqrt 3 } \over 2}} & {{1 \over 2}} \cr { - {1 \over 2}} & {{{\sqrt 3 } \over 2}} \cr } } \right] \left[ {\matrix{ {{{\sqrt 3 } \over 2}} & { - {1 \over 2}} \cr {{1 \over 2}} & {{{\sqrt 3 } \over 2}} \cr } } \right] = \left[ {\matrix{ {{{\left( {{{\sqrt 3 } \over 2}} \right)}^2} + {{\left( {{1 \over 2}} \right)}^2}} & {{{\sqrt 3 } \over 2}\left( { - {1 \over 2}} \right) + {1 \over 2}\left( {{{\sqrt 3 } \over 2}} \right)} \cr { - {1 \over 2}\left( {{{\sqrt 3 } \over 2}} \right) + {{\sqrt 3 } \over 2}\left( {{1 \over 2}} \right)} & {{{\left( { - {1 \over 2}} \right)}^2} + {{\left( {{{\sqrt 3 } \over 2}} \right)}^2}} \cr } } \right] = \left[ {\matrix{ {{3 \over 4} + {1 \over 4}} & {{ - \sqrt 3 } \over 4} + {{\sqrt 3 } \over 4}} \cr {{ - \sqrt 3 } \over 4} + {{\sqrt 3 } \over 4}} & {{1 \over 4} + {3 \over 4}} \cr } } \right] = \left[ {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right] = I Similarly, $P^T P = I$. This property is crucial for simplifying the expression.

2. Matrix A: A = \left[ {\matrix{ 1 & 1 \cr 0 & 1 \cr } } \right] This is an upper triangular matrix with 1s on the diagonal. We will need to find its powers later.

3. Matrix Q: $Q = P A P^T$ This form $P A P^T$ is known as a similarity transformation. It essentially transforms matrix $A$ into a new basis defined by $P$.

Simplifying the Expression $P^T Q^{2015} P$

Our goal is to evaluate $P^T Q^{2015} P$. Let's substitute the definition of $Q$ and use the property $P^T P = I$.

The expression is $P^T Q^{2015} P$. We can expand $Q^{2015}$ as $Q \cdot Q \cdot \dots \cdot Q$ (2015 times). Let's look at the structure: $P^T Q^{2015} P = P^T (P A P^T)^{2015} P$.

A common trick for such expressions is to analyze $Q^n$. $Q^2 = (P A P^T)(P A P^T) = P A (P^T P) A P^T$. Since $P^T P = I$, we have: $Q^2 = P A I A P^T = P A^2 P^T$.

Similarly, $Q^3 = Q^2 Q = (P A^2 P^T)(P A P^T) = P A^2 (P^T P) A P^T = P A^2 I A P^T = P A^3 P^T$.

By induction, we can generalize this pattern: $Q^n = P A^n P^T$ This identity is very powerful and simplifies the problem significantly.

Now, we can apply this identity for $n=2015$: $Q^{2015} = P A^{2015} P^T$ Substitute this back into the expression we want to evaluate: $P^T Q^{2015} P = P^T (P A^{2015} P^T) P$ Now, we can use the associative property of matrix multiplication and the identity $P^T P = I$: $P^T Q^{2015} P = (P^T P) A^{2015} (P^T P)$ $P^T Q^{2015} P = I A^{2015} I$