If $$P = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}$$, $$A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$$ and $$Q = PAP^T$$, then $$P^T Q^{2015} P$$ is:

We have $$P=\begin{bmatrix}\dfrac{\sqrt3}{2}&\dfrac12\\[4pt]-\dfrac12&\dfrac{\sqrt3}{2}\end{bmatrix}$$ and $$P^T=\begin{bmatrix}\dfrac{\sqrt3}{2}&-\dfrac12\\[4pt]\dfrac12&\dfrac{\sqrt3}{2}\end{bmatrix}$$. Because the columns of $$P$$ are orthonormal, the fundamental relation $$P^T P=I$$ holds, where $$I$$ is the identity matrix. Hence $$P^T=P^{-1}$$, that is, $$P$$ is an orthogonal (rotation) matrix.

The matrix $$A=\begin{bmatrix}1&1\\0&1\end{bmatrix}$$ is upper-triangular with both diagonal entries equal to $$1$$, so its determinant is $$1$$ and it is nonsingular.

Now we form $$Q=PAP^T$$. Because $$P$$ is orthogonal, this construction makes $$Q$$ similar to $$A$$. Similar matrices share all algebraic properties such as eigenvalues and, crucially for us, their powers relate in a very simple way.

To find an explicit relation, we first square $$Q$$:

$$Q^2=(PAP^T)(PAP^T)=PA(P^TP)AP^T=PAIA P^T=PA^2P^T.$$

Here we inserted $$P^T P=I$$ between the two occurrences of $$A$$, then simplified. Repeating the same idea gives the general rule

$$Q^n=P\,A^n\,P^T\qquad\text{for every positive integer }n.$$

This can be proved rigorously by mathematical induction, but the key point is that every time two successive factors $$P^T P$$ meet, they collapse to the identity and disappear.

The quantity required in the question is $$P^T Q^{2015} P$$. Using the formula we just obtained, we substitute $$n=2015$$ and obtain

$$P^T Q^{2015} P=P^T\bigl(PA^{2015}P^T\bigr)P=(P^TP)\,A^{2015}\,(P^TP)=I\,A^{2015}\,I=A^{2015}.$$

So the whole problem reduces to finding the $$2015^{\text{th}}$$ power of the simple matrix $$A$$.

At this stage we recall (and will verify) a well-known fact: for the Jordan block $$A=\begin{bmatrix}1&1\\0&1\end{bmatrix}$$, the general power is

$$A^n=\begin{bmatrix}1&n\\0&1\end{bmatrix}\qquad(n\ge 1).$$

We establish this formula by induction on $$n$$. For $$n=1$$ it is trivially true. Assume it holds for some $$n=k$$, i.e. $$A^k=\begin{bmatrix}1&k\\0&1\end{bmatrix}$$. Then

$$A^{k+1}=A^k A=\begin{bmatrix}1&k\\0&1\end{bmatrix}\begin{bmatrix}1&1\\0&1\end{bmatrix}=\begin{bmatrix}1&k+1\\0&1\end{bmatrix}.$$

Thus the pattern continues, completing the proof. Putting $$n=2015$$ gives immediately

$$A^{2015}=\begin{bmatrix}1&2015\\0&1\end{bmatrix}.$$

Since we have already shown that $$P^T Q^{2015} P=A^{2015}$$, we conclude

$$P^T Q^{2015} P=\begin{bmatrix}1&2015\\0&1\end{bmatrix}.$$

Comparing this matrix with the options supplied, we see it matches exactly with Option C.

Hence, the correct answer is Option C.