Because every column of $X$ has <sample mean> zero, every column $u_i$ of $U=XV$ also has sample mean zero. The <spectral theorem for real symmetric matrices> gives
$$
U^TU=V^TX^TXV=V^T\Sigma V=\Lambda.
$$
Hence for $i\ne j$ the <sample covariance> is $n^{-1}u_i^Tu_j=0$, while the <sample variance> of $u_i$ is
$$
\frac1n u_i^Tu_i=\frac{\Lambda_{ii}}n.
$$
Thus the $u_i$ are pairwise uncorrelated <sample principal components>, ordered by decreasing sample variance.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 29J / c / Solution

Let the columns of a centered <design matrix> $X\in\mathbb R^{n\times p}$ be variables and write $X^TX=V\Lambda V^T$ by the <spectral theorem for real symmetric matrices>. The columns $u_i$ of $U=XV$ are the sample principal-component score vectors. They satisfy $u_i^Tu_j=\Lambda_{ij}$, so distinct score vectors have zero <sample covariance> and $u_i$ has <sample variance> $\Lambda_{ii}/n$ under the divisor-$n$ convention.

Sample principal component

For paired scalar observations $(x_i,y_i)$, their sample covariance is $n^{-1}\sum_i(x_i-\bar x)(y_i-\bar y)$, or the same sum divided by $n-1$ under the unbiased convention. It is an off-diagonal entry of the <sample covariance matrix> of the paired vectors.

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