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 $$\left[A^{-1}\right]_{ij}=\begin{cases} 2 & \text{if}\ i=j<n \\ 1 & \text{if}\ i=j=n \\ -1 & \text{if}\ \lvert i-j\rvert=1 \\ 0 & \text{otherwise}. \end{cases}$$ For example, if $$n=4$$ then $$A=\begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 2 & 2 \\ 1 & 2 & 3 & 3 \\ 1 & 2 & 3 & 4 \end{bmatrix}$$ has inverse $$A^{-1}=\begin{bmatrix} 2 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0 \\ 0 & -1 & 2 & -1 \\ 0 & 0 & -1 & 1 \end{bmatrix}$$

We can use our knowledge of $$A^{-1}$$ to eigendecompose $$A$$. To see how, let $$\{(\lambda_j,v_j)\}_{j=1}^n$$ be the eigenpairs of $$A^{-1}$$. Yueh (2005, Theorem 1) shows that the eigenvector $$v_j\in\mathbb{R}^n$$ corresponding to the $$j^\text{th}$$ eigenvalue $$\lambda_j=2\left(1+\cos\left(\frac{2j\pi}{2n+1}\right)\right)$$ has $$i^\text{th}$$ component $$[v_j]_i=\alpha\sin\left(\frac{2ij\pi}{2n+1}\right),$$ where $$\alpha\in\mathbb{R}$$ is an arbitrary scalar. This vector has length \begin{align} \lvert\vert v_j\rvert\rvert &\equiv \sqrt{\sum_{i=1}^n\left([v_j]_i\right)^2} \\ &= \sqrt{\sum_{i=1}^n\alpha^2\sin^2\left(\frac{2ij\pi}{2n+1}\right)} \\ &= \lvert\alpha\rvert\sqrt{\frac{2n+1}{4}}, \end{align} 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 $$v_1,v_2,\ldots,v_n$$ of $$A^{-1}$$ have unit length. Then, by the spectral theorem, these vectors form an orthonormal basis for $$\mathbb{R}^n$$. As a result, the $$n\times n$$ matrix $$V=\begin{bmatrix} v_1 & v_2 & \cdots & v_n\end{bmatrix}$$ 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{align} A^{-1} &= V\Lambda V^T \\ &= \sum_{j=1}^n\lambda_jv_jv_j^T \end{align} of $$A^{-1}$$. It follows from the orthogonality of $$V$$ that \begin{align} A &= \left(V\Lambda V^T\right)^{-1} \\ &= V\Lambda^{-1} V^T \\ &= \sum_{j=1}^n\frac{1}{\lambda_j}v_jv_j^T \end{align} 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 $$\mathcal{D}=\{(x_i,y_i)\}_{i=1}^n$$ generated by the process \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\E}{E} \DeclareMathOperator{\Var}{Var} \newcommand{\veps}{\sigma_\epsilon^2} \newcommand{\R}{\mathbb{R}} \renewcommand{\epsilon}{\varepsilon} \begin{align} y_i &= f(x_i)+\epsilon_i \\ \epsilon_i &\overset{\text{iid}}{\sim} \mathcal{N}(0,\veps), \end{align} where $$\{f(x)\}_{x\ge0}$$ is a sample path of a standard Wiener process and where the errors $$\epsilon_i$$ are iid normally distributed with variance $$\veps$$. I use these data to estimate $$f(x)$$ for some $$x\ge0$$.2 My estimator $$\hat{f}(x)\equiv\E[f(x)\mid\mathcal{D}]$$ has conditional variance $$\Var\left(\hat{f}(x)\mid\mathcal{D}\right)=\Var(f(x))-w^T\Sigma^{-1} w,$$ where $$w\in\R^n$$ is the vector with $$i^\text{th}$$ component $$w_i=\Cov(y_i,f(x))$$ and where $$\Sigma\in\R^{n\times n}$$ is the covariance matrix with $${ij}^\text{th}$$ entry $$\Sigma_{ij}=\Cov(y_i,y_j)$$. If $$x_i=i$$ for each $$i\in\{1,2,\ldots,n\}$$, then we can express this matrix as the sum $$\Sigma=A+\veps I,$$ 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 $$I=VV^T$$, since $$V$$ is orthogonal. It follows that \begin{align*} \Sigma^{-1} &= \left(V\Lambda^{-1}V^T+\veps VV^T\right)^{-1} \\ &= V\left(\Lambda+\frac{1}{\veps}I\right)V^T, \end{align*} from which we can derive a (relatively) closed-form expression for the conditional variance of $$\hat{f}(x)$$ given $$\mathcal{D}$$.

1. One can verify this claim by showing $$AA^{-1}$$ equals the identity matrix. ↩︎

2. I discuss this estimation problem in a recent paper. ↩︎