= Solution
A real <orthogonal matrix> satisfies $Q^TQ=I$; a complex <unitary matrix> satisfies $U^\dagger U=I$. Their columns are orthonormal in the corresponding real or complex <inner product>.
Adopt the convention that a <complex inner product> is linear in its second argument, conjugate symmetric, and positive definite: $\langle u,u\rangle\geq0$, with equality exactly when $u=0$. Conjugate symmetry and second-argument linearity then imply antilinearity in the first:
$$
\langle\alpha u_1+\beta u_2,v\rangle=\overline{\langle v,\alpha u_1+\beta u_2\rangle}=\overline{\alpha\langle v,u_1\rangle+\beta\langle v,u_2\rangle}=\bar\alpha\langle u_1,v\rangle+\bar\beta\langle u_2,v\rangle.
$$
For the specified real symmetric matrix use $b=a^2$. Its upper block is the <outer product> $(1,a)^T(1,a)$, so it has one zero eigenvalue and one eigenvalue $1+a^2$. A complete <orthonormal eigenbasis> is
$$
\boxed{\lambda_0=0,\quad v_0=\frac{(-a,1,0)^T}{\sqrt{1+a^2}};\qquad \lambda_+=1+a^2=1+b,\quad v_+=\frac{(1,a,0)^T}{\sqrt{1+a^2}};\qquad \lambda_c=c,\quad v_c=(0,0,1)^T.}
$$
If $c=1+b$, the last two eigenvalues coincide and their <eigenspace> is the span of $v_+,v_c$; every nonzero vector in that span is an eigenvector. Since $c>0$, the zero eigenspace remains one-dimensional.
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