Decomposing matrices of pairwise minima

Let $A$ be the $n\times n$ matrix with ${ij}^{\text{th}}$ entry $A_{ij}=min\{i,j\}$. From a previous post, we know $A$ has a tridiagonal inverse $A^{-1}$ with ${ij}^{\text{th}}$ entry¹ $${\left[A^{-1}\right]}_{ij}=\{\begin{array}{cc}2& \text{if}i=j<n\\ 1& \text{if}i=j=n\\ -1& \text{if}|i-j|=1\\ 0& \text{otherwise}.\end{array}$$ For example, if $n=4$ then $$A=\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& 2& 2& 2\\ 1& 2& 3& 3\\ 1& 2& 3& 4\end{array}\right]$$ has inverse $$A^{-1}=\left[\begin{array}{cccc}2& -1& 0& 0\\ -1& 2& -1& 0\\ 0& -1& 2& -1\\ 0& 0& -1& 1\end{array}\right]$$

We can use our knowledge of $A^{-1}$ to eigendecompose $A$. To see how, let $\left\{\right({\lambda }_j,v_j){\}}_{j=1}^n$ be the eigenpairs of $A^{-1}$. Yueh (2005, Theorem 1) shows that the eigenvector $v_j\in R^n$ corresponding to the $j^{\text{th}}$ eigenvalue $${\lambda }_j=2(1+\cos \left(\frac{2j\pi }{2n+1}\right))$$ has $i^{\text{th}}$ component $$[v_j{]}_i=\alpha \sin \left(\frac{2ij\pi }{2n+1}\right),$$ where $\alpha \in R$ is an arbitrary scalar. This vector has length $$\begin{array}{cc}\left|\right|v_j\left|\right|& \equiv \sqrt{\sum _{i=1}^n{\left(\right[v_j{]}_i)}^2}\\ & =\sqrt{\sum _{i=1}^n{\alpha }^2{\sin }^2\left(\frac{2ij\pi }{2n+1}\right)}\\ & =\left|\alpha \right|\sqrt{\frac{2n+1}{4}},\end{array}$$ where the last equality can be verified using Wolfram Alpha and proved using complex analysis. So choosing $\alpha =2/\sqrt{2n+1}$ ensures that the eigenvectors $v_1,v_2,\dots ,v_n$ of $A^{-1}$ have unit length. Then, by the spectral theorem, these vectors form an orthonormal basis for $R^n$. As a result, the $n\times n$ matrix $$V=\left[\begin{array}{cccc}{v}_1& v_2& \cdots & v_n\end{array}\right]$$ with ${ij}^{\text{th}}$ entry $V_{ij}=[v_j{]}_i$ is orthogonal. Moreover, letting $\Lambda$ be the $n\times n$ diagonal matrix with ${ii}^{\text{th}}$ entry ${\Lambda }_{ii}={\lambda }_i$ yields the eigendecomposition $$\begin{array}{cc}{A}^{-1}& =V\Lambda V^T\\ & =\sum _{j=1}^n{\lambda }_j{v}_j{v}_j^T\end{array}$$ of $A^{-1}$. It follows from the orthogonality of $V$ that $$\begin{array}{cc}A& ={\left(V\Lambda V^T\right)}^{-1}\\ & =V{\Lambda }^{-1}{V}^T\\ & =\sum _{j=1}^n\frac{1}{{\lambda }_j}{v}_j{v}_j^T\end{array}$$ is the eigendecomposition of $A$. Thus $A$ and $A^{-1}$ have the same eigenvectors, but the eigenvalues of $A$ are the reciprocated eigenvalues of $A^{-1}$.

Here’s one scenario in which this decomposition is useful: Suppose I observe data $D=\left\{\right(x_i,y_i){\}}_{i=1}^n$ generated by the process $$\begin{array}{cc}{y}_i& =f\left(x_i\right)+{\epsilon }_i\\ {\epsilon }_i& \stackrel{\text{iid}}{\sim }N(0,{\sigma }_{\epsilon }^2),\end{array}$$ where $\left\{f\right(x){\}}_{x\ge 0}$ is a sample path of a standard Wiener process and where the errors ${\epsilon }_i$ are iid normally distributed with variance ${\sigma }_{\epsilon }^2$. I use these data to estimate $f\left(x\right)$ for some $x\ge 0$.² My estimator $\hat{f}\left(x\right)\equiv E\left[f\right(x)\mid D]$ has conditional variance $$\mathrm{Var}\left(\hat{f}\right(x)\mid D)=\mathrm{Var}\left(f\right(x\left)\right)-w^T{\Sigma }^{-1}w,$$ where $w\in R^n$ is the vector with $i^{\text{th}}$ component $w_i=\mathrm{Cov}(y_i,f(x\left)\right)$ and where $\Sigma \in R^{n\times n}$ is the covariance matrix with ${ij}^{\text{th}}$ entry ${\Sigma }_{ij}=\mathrm{Cov}(y_i,y_j)$. If $x_i=i$ for each $i\in \{1,2,\dots ,n\}$, then we can express this matrix as the sum $$\Sigma =A+{\sigma }_{\epsilon }^{2}I,$$ where $A$ is the $n\times n$ matrix defined above and where $I$ is the $n\times n$ identity matrix. But we know $A=V{\Lambda }^{-1}{V}^T$. We also know $I=V{V}^T$, since $V$ is orthogonal. It follows that $$\begin{array}{cc}{\Sigma }^{-1}& ={(V{\Lambda }^{-1}{V}^T+{\sigma }_{\epsilon }^{2}V{V}^T)}^{-1}\\ & =V(\Lambda +\frac{1}{{\sigma }_{\epsilon }^2}I)V^T,\end{array}$$ from which we can derive a (relatively) closed-form expression for the conditional variance of $\hat{f}\left(x\right)$ given $D$.