Past exam of the mathematical methods I course of the University of Oxford / 2021 resit / 8

Differentiation satisfies $D(af+bg)=aD(f)+bD(g)$ for constant scalars $a,b$, so it is a linear map on suitable vector spaces of differentiable functions. An arbitrary set $S$ need not even be a vector space, and an endomorphism interpretation also requires $D(S)\subseteq S$.

For a noninvertible example, take $S=\mathrm{span}\{1,x\}$. Then $D(a+bx)=b$ and the kernel of a linear map is the nonzero linear span of $1$, so $D:S\to S$ is not injective. The matrix in this basis is $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$.

For an invertible example, take $S=\mathrm{span}\{e^x,e^{2x}\}$. In this basis, $D$ has matrix $\mathrm{diag}(1,2)$, andThus differentiation is not always invertible; the domain and codomain determine the answer.

For an invertible matrix $P$ whose columns give the new basis in old coordinates, the new matrix is $B=P^{-1}AP$. Multiplicativity of the determinant givessince $\det (P^{-1})\det P=1$.

To justify the cyclic property of the trace directly, for two square matrices $R,S$,Apply it with $R=P^{-1}A$ and $S=P$:Thus both the determinant and trace are invariant under change of basis.

Write the ordered basis as $\mathcal{B}=(1,t,e^t,t{e}^t)$. Differentiation givesThe corresponding coordinate vectors form the columns of the representing matrix:This is an example of differentiation on an exponential-polynomial space.

Choose $\boxed{v_4=e^t(1-t)}$, and denote the new ordered basis by $\mathcal{C}=(b_1,b_2,b_3,b_4)$. The change of basis matrix, with columns expressed in $\mathcal{B}$, isThis nonzero determinant proves that the proposed fourth function completes a basis. Express the derivatives in that basis:Reading off columns therefore gives

Yes. The old matrix is triangular, with determinant zero and trace $0+0+1+1=2$. The upper-left two-by-two block of the new matrix has determinant zero, so the whole determinant remains zero. Its trace isThusThis verifies both change of basis invariants from part (b).