What are the properties of symmetric, anti-symmetric, and diagonal matrices

In general, given matrices $A,B$ appropriately sized so that $AB$ is defined, we also know that $B^\dagger A^\dagger$ is defined, and in particular that $B^\dagger A^\dagger=(AB)^\dagger.$ (By $\dagger$ I denote transpose.)

Now, for any square matrix $A$ and any integer $n$ for which $A^n$ is defined (negative $n$ make sense if and only if $A$ is invertible, while nonnegative $n$ always make sense), it follows that $\left(A^\dagger\right)^n$ is defined, and that $\left(A^\dagger\right)^n=(A^n)^\dagger.$ (Why?)

From there, we can readily see that (defined) even powers of antisymmetric matrices are symmetric, as are all (defined) integer powers of symmetric matrices.

Since a sum of symmetric matrices of the same size is again symmetric (why?), then it follows that $Q^{2012}+D^{2013}$ is symmetric. (Why?)

For the second, keep in mind that for any matrix $A$ and any constant $c,$ we have $\left(cA\right)^\dagger=cA^\dagger.$ This, together with the above observations, will allow us to conclude (after some manipulation) that $(P+Q)(P-Q)$ is symmetric.

It should be observed at the start that the conditions on the matrices $P$, $Q$, and $D$, that they are respectively symmetric, anti-symmetric, and diagonal, implies they are all square. Since they are given to be all the same size, the operations of matrix addition and multiplication between them are defined and permissible.

If $Q$ is an anti- or skew-symmetric matrix, $Q^T = -Q$, whence $Q^2 = -Q^TQ$; for square matrices $A$, we have $(A^TA)^T = A^T(A^T)^T =A^TA$, hence $A^TA$ is symmetric for any square matrix $A$. From this we see that $Q^2$ is symmetric, hence $Q^{2m} = (Q^2)^m$ is also symmetric for any positive integer $m$; this follows from the elementary fact that $(A^T)^m = (A^m)^T$ which is easily seen by a simple application of the two-matrix rule $(AB)^T = B^TA^T$. We also incidentally see that $Q^{2m + 1}$ is skew, $(Q^{2m + 1})^T = -Q^{2m + 1}$: $(Q^{2m + 1})^T = (Q^T)^{2m + 1} = (-Q)^{2m + 1} = (-Q)Q^{2m} = -Q^{2m + 1}$. So, for a skew-symmetric matrix, the even powers are symmetric and the odd powers are skew.

To paraphrase Tom Jefferson, We hold this truth to be (nearly) self-evident: that $D^m$ is diagonal for any diagonal matrix $D$. Nearly self-evident though this assertion may be, it can be easily validated by showing that the product of two diagonal matrices is diagonal, which itself follows from a straightforward application of the definition of matrix multiplication. Since a diagonal matrix is clearly symmetric, we see that for any non-negative integers $m, n$, $Q^{2m + 1} + D^n$, being the sum of two symmetric matrices, is symmetric. In particular, $Q^{2012} + D^{2013}$ is symmetric.

It is a bit harder to see that $(P + Q)(P - Q)$ is skew, because it is in fact symmetric. This follows easily from the symmetry of $A^TA$ for any square $A$: we have $(P - Q)^T = P^T - Q^T = P + Q$ whence $(P + Q)(P - Q) = (P - Q)^T(P - Q)$ is indeed symmetric.

And so it goes . . .

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!