Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2011 / Algebra / 2 / c

The finite-dimensional spectral theorem says that a self-adjoint operator on a complex inner-product space has an orthonormal basis of eigenvectors and all its eigenvalues are real. Equivalently, its matrix is unitarily diagonalisable with real diagonal entries.

For $a\in \mathbb{R}$, the matrix is equal to its conjugate transpose:Thus it represents a self-adjoint operator for the standard inner product on ${\mathbb{C}}^3$. The theorem applies directly, proving

Set $N=M_i-I$. Direct multiplication giveswith $N\ne 0$. If $M_i$ were diagonalisable, then subtracting the identity in the same basis would diagonalise $N$. A diagonal matrix whose cube is zero has every diagonal entry zero and is itself zero. Similarity would then imply $N=0$, a contradiction. HenceThis proves failure directly; merely observing that it is not self-adjoint would not suffice, because non-self-adjoint matrices can still be diagonalisable.