 Solution

A complex $3\times 3$ 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 $A=PI{P}^{-1}=I$. Therefore the specified matrix, which is not a scalar identity matrix, is not diagonalizable.

For the displayed real symmetric matrix, choose the orthogonal matrixIts columns form an orthonormal eigenbasis, andIn the orthogonal coordinates $u=(x+y)/\sqrt{2}$, $v=(x-y)/\sqrt{2}$, 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 $1/\sqrt{3}$. The figure gives its principal-axis sketch.

A diagonalizable matrix can certainly have exactly two distinct eigenvalues: $\mathrm{diag}(1,1,2)$ is already diagonal and has an eigenbasis despite its repeated eigenvalue.