Discussion
Let \(A = \begin{pmatrix} 1 & 3 & 5 \\ -6 & 8 & 3 \\ -4 & 6 & 5 \end{pmatrix}\) and \(P = \frac{1}{2}(A + A^{T})\). Then
Answer: A
The relation is \(P^{T} = P\).
Transposition distributes over a sum and leaves a scalar factor alone, and \((A^{T})^{T} = A\).
\[ P^{T} = \frac{1}{2}(A^{T} + A) = P \]
So P is the symmetric part of A, whatever A is; the entries of A play no role. \(P^{T} = -P\) describes the skew-symmetric part \(\frac{1}{2}(A - A^{T})\), and the multiples 2P, -2P and 3P cannot hold for a non-zero P.
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