Past exam of the mathematical methods I course of the University of Oxford / 2016 / 11

The transpose of a matrix product reverses factor order:For the requested two-dimensional check, put $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ and $B=\left(\begin{array}{cc}e& f\\ g& h\end{array}\right)$. Both sides areMore generally, the entry formula $(AB{)}_{ji}={\sum }_k{A}_{jk}{B}_{ki}$ is exactly $(B^T{A}^T{)}_{ij}$.

The conjugate transpose also reverses factor order. If $U$ and $V$ are unitary matrices,Thus their product is unitary. This is closure under multiplication in the unitary group.

For a Hermitian matrix $H$, $H^{†}=H$. Reversing the product of $n$ identical factors has no effect, soHence every positive integer power is Hermitian. The identical-factor condition matters; a product of different Hermitian matrices need not be Hermitian.

If $A$ and $B$ are Hermitian matrices, then $(AB{)}^{†}=B^{†}{A}^{†}=BA$. ThereforeThe required condition is that the matrices commute.

At $t=0$ the power series gives $\mathbf{v}(0)={\mathbf{v}}_0$. Equate the coefficient of $t^n$ after differentiating the series and substituting into the linear system of ordinary differential equations:Mathematical induction yields ${\mathbf{v}}_n=(-iH{)}^n{\mathbf{v}}_0/n!$. Summing gives the matrix exponential solution $\mathbf{v}(t)=e^{-iHt}{\mathbf{v}}_0$.

For the specified diagonal Hermitian matrix, its powers and exponential act separately on the two components. ThusThis unitary matrix preserves ${\mathbf{v}}^{†}\mathbf{v}$: the two components acquire opposite phases of unit modulus. The coefficient recurrence applies to any constant finite square matrix $H$; unitarity here additionally uses its Hermitian property.