= Solution
A complex $3\times3$ <matrix> is <diagonalizable> exactly when it has a basis of three linearly independent <eigenvectors>. If all its <eigenvalues> are one, a diagonalization would have $D=I$ and hence $A=PIP^{-1}=I$. Therefore the specified matrix, which is not a scalar <identity matrix>, is \b[not diagonalizable].
For the displayed real symmetric matrix, choose the <orthogonal matrix>
$$
Q=\begin{pmatrix}1/\sqrt2&1/\sqrt2&0\\1/\sqrt2&-1/\sqrt2&0\\0&0&1\end{pmatrix}.
$$
Its columns form an <orthonormal eigenbasis>, and
$$
\boxed{Q^TAQ=\operatorname{diag}(1,3,3).}
$$
In the <orthogonal coordinates> $u=(x+y)/\sqrt2$, $v=(x-y)/\sqrt2$, the <quadric surface> has equation $u^2+3v^2+3z^2=1$. It is a <prolate spheroid>, with long semiaxis one in direction $(1,1,0)$ and two equal short semiaxes $1/\sqrt3$. The figure gives its principal-axis sketch.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2008/paper-1-principal-axis-spheroid.png]
{title=The diagonalized quadric is a prolate spheroid with semiaxes one and one over the square root of three}
A diagonalizable matrix can certainly have exactly two distinct <eigenvalues>: $\operatorname{diag}(1,1,2)$ is already diagonal and has an eigenbasis despite its repeated eigenvalue.
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