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. ↩︎