Past exam of the mathematical methods I course of the University of Oxford / 2021 / 9

The matrix exponential power series is absolutely convergent for a finite matrix, so taking the conjugate transpose term by term is valid. Since $M^{†}=M$,The absolute convergence also justifies the Cauchy product. In $e^{-iM}{e}^{iM}$, the coefficient of $M^n$ is proportional to ${\sum }_{k=0}^n(-1{)}^k(\genfrac{}{}{0ex}{}{n}{k})$, which is zero for $n>0$ and one for $n=0$, by the binomial theorem. Consequently $U^{†}U=I$; the reversed product is also $I$. This proves that the exponential is unitary directly from the power series, without invoking diagonalization.

First check that the proposed space is closed under the stated operations: $|\alpha a_i+b_i{|}^2\le 2{\alpha }^2{a}_i^2+2b_i^2$, and summing gives a finite result. For any finite $N$, the Cauchy-Schwarz inequality givesThe left side is an increasing, bounded sequence of real numbers. Its limit exists, proving absolute convergence of ${\sum }_i{a}_i{b}_i$. This justifies termwise addition and scalar multiplication in the defining sum. Symmetry and bilinearity follow, while $⟨A,A⟩={\sum }_i{a}_i^2\ge 0$, with equality only when every $a_i=0$. Thus it is a real inner product on the real l2 sequence space.

The state update matrix isIts eigenvalues are $1$ and $-1/2$, with eigenvectors $(1,1{)}^T$ and $(1,-2{)}^T$. The initial column $(G_1,G_0{)}^T=(1,0{)}^T$ decomposes as $\frac{2}{3}(1,1{)}^T+\frac{1}{3}(1,-2{)}^T$. Applying $A^i$ and reading the second component givesThe alternating term decays because its eigenvalue has modulus below one. The convergence of an adjacent-term averaging recurrence gives the same result by a first-order recurrence for successive differences.