= Solution
{c}
Suppose $A^T=A$. For <eigenvectors> $v_i,v_j$ with distinct real <eigenvalues>, symmetry gives
$$
\lambda_i v_i^Tv_j=(Av_i)^Tv_j=v_i^TA^Tv_j=v_i^TAv_j
=\lambda_jv_i^Tv_j.
$$
Thus $(\lambda_i-\lambda_j)v_i^Tv_j=0$, 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 $P^TP=I$, so it is an <orthogonal matrix> and $P^{-1}=P^T$. As in part (a),
$$
\boxed{P^TAP=\operatorname{diag}(\lambda_1,\ldots,\lambda_n).}
$$
This proves the needed <orthogonal diagonalization> from symmetry and distinct <eigenvalues>, without invoking an unproved <spectral theorem>.
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