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Expanding the characteristic polynomial along the first column givesFor $s\ne 1$, solving the two nullspace equations givesThese are all the eigenvalues and eigenvectors, with nonzero vectors understood. The three displayed spanning vectors form an eigenbasis, so $A$ is diagonalizable.

At $s=1$, the sole eigenvalue is three, with algebraic multiplicity three but the same two-dimensional $E_3$. Thus $A$ is diagonalizable exactly when $s\ne 1$. Indeed, at the exceptional parameter, $A=3I+N$ with $N\ne 0$ and $N^2=0$, so its Jordan normal form has one block of size two and one of size one. This illustrates how an eigenvalue collision can destroy diagonalizability without changing the two-dimensional eigenspace.