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

Use the convention that the inner product is linear in its first argument, as required by the displayed representation of a linear functional. Choose an orthonormal basis $e_1,\dots ,e_n$, for instance by Gram-Schmidt orthogonalization. DefineFor $u={\sum }_j{u}_j{e}_j$, conjugate linearity in the second argument givesThusThe representation is unique as well: if two vectors represent the same functional, their difference is orthogonal to every vector and therefore to itself, so it is zero.

The adjoint operator is characterized byFor each fixed $v$, the map $u\mapsto ⟨Tu,v⟩$ is a linear functional. Part (i) gives a representing vector, which we define to be $T^{\ast }v$. Hence such a map exists on every vector of $V$. Part (b) proves that this constructed map is linear.

For scalars $\alpha ,\beta$ and vectors $v,w,u$, use the adjoint identity and conjugate linearity in the second argument:The difference of the two second-argument vectors is orthogonal to every $u$. Choosing $u$ to be that difference shows its squared norm is zero. ThereforeThe two conjugations are why the adjoint is a linear map, despite the conjugate-linear dependence of the representing functional on its second argument.

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.