 Solution

Write the given matrix as $A$. Expanding its characteristic polynomial along the middle column givesThus the eigenvalues are $1,-2,-2$, counted with algebraic multiplicity.

For $\lambda =1$, solving $(A-I)v=0$ gives $z=2x$ and $y=-x$. For $\lambda =-2$, solving $(A+2I)v=0$ gives $z=3x$, with $y$ free. A convenient eigenbasis is thereforeThe last two eigenvectors span the two-dimensional eigenspace for $-2$. The three columns have determinant $-1$, so they are linearly independent and form a basis of ${\mathbb{R}}^3$. In particular the repeated eigenvalue causes no failure of diagonalizability here.