The <identity operator> is positive and self-adjoint, but on the infinite-dimensional <Hilbert space> $L^2[0,1]$ it has eigenvalue one with infinite multiplicity. Consequently
$$
\operatorname{tr}I=\sum_{k=1}^{\infty}1=\infty.
$$
A square-integrable Hilbert-space random variable has a trace-class <covariance operator> with trace $\mathbb E\lVert X\rVert^2<\infty$. The identity therefore cannot be such a covariance operator.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 225 / 1 / a / ii / Solution

{title2=$I$}

The identity operator maps every vector to itself. Every nonzero vector is an <eigenvector> with <eigenvalue> one.

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