= Solution
{c}
The real <spectral theorem> states that a <symmetric matrix> has an <orthonormal basis> of real <eigenvectors>, and hence is a <diagonalisable matrix>. Here is a proof in this setting.
For $n\ge1$, the <quadratic form> $q(x)=x^TAx$ attains a maximum on the <unit sphere>, by <compactness> and the <extreme value theorem>. Choose a unit maximizer $u$ and put $\lambda=u^TAu$. For any $w\perp u$, the <Rayleigh quotient> along $u+tw$ has a maximum at $t=0$, so
$$
0=\left.\frac{d}{dt}\frac{(u+tw)^TA(u+tw)}{\|u+tw\|^2}\right|_{t=0}
=2w^TAu.
$$
The last equality uses $A=A^T$ and $u^Tw=0$. Thus $Au$ is orthogonal to every $w\perp u$, so $Au=\lambda u$.
The <orthogonal complement> $u^\perp$ is an <invariant subspace>, because for $x\perp u$,
$$
u^TAx=(Au)^Tx=\lambda u^Tx=0.
$$
The <restriction> of $A$ to $u^\perp$ is again symmetric with respect to the induced <inner product>. By <mathematical induction> on <dimension>, it has an <orthonormal basis> of <eigenvectors>; adjoining $u$ gives one for $\mathbb R^n$. If $Q$ has those <basis vectors> as columns, then $Q^TAQ$ is a real <diagonal matrix> and $Q^{-1}=Q^T$.
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