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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2023

Paper 2

Using <Euler's formula>,
$$
\cos(\pi x)=\frac12e^{i\pi x}+\frac12e^{-i\pi x},
$$
$$
-\frac12\sin(\pi x)
=\frac i4e^{i\pi x}-\frac i4e^{-i\pi x}.
$$
Thus
$$
\widehat c_0=2,
\qquad
\widehat c_1=\frac12+\frac i4,
\qquad
\widehat c_{-1}=\frac12-\frac i4.
$$
Ordering the modes as $(-1,0,1)$, let $a=1/2+i/4$. Then
$$
C=\begin{pmatrix}
2&\overline a&0\\
a&2&\overline a\\
0&a&2
\end{pmatrix},
\qquad
D=\begin{pmatrix}-1&0&0\\0&0&0\\0&0&1\end{pmatrix}.
$$
Consequently
$$
\boxed{
B=-CD=
\begin{pmatrix}
2&0&0\\
\frac12+\frac i4&0&-\frac12+\frac i4\\
0&0&-2
\end{pmatrix}.}
$$
This matrix is <triangular matrix>[triangular], so its eigenvalues are its diagonal entries:
$$
\boxed{\operatorname{spec}(B)=\{-2,0,2\}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 41C / d / Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 41C / d / Solution

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Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 41C / d / Solution

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