IMC 1996 · Day 1 · P3

(i) Let $B = \frac{1}{2}(A + E)$. Then $$B^2 = \frac{1}{4} (A^2 + 2AE + E) = \frac{1}{4} (2AE + 2E) = \frac{1}{2} (A + E) = B.$$ Hence $B$ is a projection. Thus there exists a basis of eigenvectors for $B$, and the matrix of $B$ in this basis is of the form $\operatorname{diag}(1, \dots, 1, 0, \dots, 0)$.

Since $A = 2B - E$ the eigenvalues of $A$ are $\pm 1$ only.

(ii) Let $\{ A_i : i \in I \}$ be a set of commuting diagonalizable operators on $V$, and let $A_1$ be one of these operators. Choose an eigenvalue $\lambda$ of $A_1$ and denote $V_\lambda = \{ v \in V : A_1 v = \lambda v \}$. Then $V_\lambda$ is a subspace of $V$, and since $A_1 A_i = A_i A_1$ for each $i \in I$ we obtain that $V_\lambda$ is invariant under each $A_i$. If $V_\lambda = V$ then $A_1$ is either $E$ or $-E$, and we can start with another operator $A_i$. If $V_\lambda \ne V$ we proceed by induction on $\dim V$ in order to find a common eigenvector for all $A_i$. Therefore $\{ A_i : i \in I \}$ are simultaneously diagonalizable.

If they are involutions then $|I| \le 2^n$ since the diagonal entries may equal 1 or $-1$ only.