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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2017

Paper 1

Componentwise,
$$
[(AB)^T]_{ij}=(AB)_{ji}=\sum_kA_{jk}B_{ki}
=\sum_k(B^T)_{ik}(A^T)_{kj}=(B^TA^T)_{ij},
$$
so
$$
\boxed{(AB)^T=B^TA^T}.
$$
Repeated application gives $(A^k)^T=(A^T)^k$. Transposing the <matrix exponential> term by term therefore yields
$$
\boxed{(e^A)^T=e^{A^T}}.
$$
Multiplication of the two power series gives
$$
e^{tA}e^{tA^T}
=I+t(A+A^T)
+t^2\left(\frac12A^2+AA^T+\frac12(A^T)^2\right)+O(t^3).
$$
Hence
$$
\boxed{Q_0=I,\qquad Q_1=A+A^T,\qquad
Q_2=\frac12A^2+AA^T+\frac12(A^T)^2}.
$$
If $e^{tA}$ is an <orthogonal matrix> for every real $t$, then
$$
e^{tA}(e^{tA})^T=e^{tA}e^{tA^T}=I.
$$
Its coefficient of $t$ must vanish, so
$$
\boxed{A^T=-A};
$$
that is, $A$ is a <skew-symmetric matrix>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 1 / 2C / Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 1 / 2C / Solution

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