 Solution

Suppose $A^T=A$. For eigenvectors $v_i,v_j$ with distinct real eigenvalues, symmetry givesThus $({\lambda }_i-{\lambda }_j)v_i^T{v}_j=0$, and the eigenvectors are orthogonal. Divide each by its nonzero Euclidean norm to obtain an orthonormal basis. The matrix $P$ having these normalized columns satisfies $P^{T}P=I$, so it is an orthogonal matrix and $P^{-1}=P^T$. As in part (a),This proves the needed orthogonal diagonalization from symmetry and distinct eigenvalues, without invoking an unproved spectral theorem.