 Solution

Allow an eigenvector $v\ne 0$ to have complex entries, since real matrices can initially have complex eigenvalues. Write $v^{\ast }$ for its conjugate transpose. Reality of $A$ gives $A^{\ast }=A^T$; symmetry then gives $A^{\ast }=A$. Thus $A$ is a Hermitian matrix. The scalar $v^{\ast }Av$ equals its complex conjugate, because $(v^{\ast }Av{)}^{\ast }=v^{\ast }{A}^{\ast }v=v^{\ast }Av$. If $Av=\lambda v$, thensince $v^{\ast }v>0$. This proves that all eigenvalues are real. The equality $A^{\ast }=A^T$ is precisely where reality was used; a complex symmetric matrix need not be Hermitian, as $iI$ illustrates.

If $Au=\lambda u$ and $Av=\mu v$, then $u^{\ast }Av=\mu u^{\ast }v$, but also $u^{\ast }Av=(Au{)}^{\ast }v=\overline{\lambda }{u}^{\ast }v=\lambda u^{\ast }v$. Consequently $(\mu -\lambda )u^{\ast }v=0$. Distinct eigenvalues force $u^{\ast }v=0$. For real eigenvectors this is ordinary Euclidean orthogonality. These arguments establish real symmetric spectral orthogonality without assuming a diagonalization in advance.

A real orthogonal matrix satisfies $P^{T}P=I$, equivalently $P^{-1}=P^T$. If $A$ is symmetric,so orthogonal similarity preserves symmetry. An arbitrary real invertible change of basis need not do so. For example,The final matrix is not symmetric. Orthogonal similarity preserves real symmetry; general similarity does not.