Let \(k\in\mathbb R\), let \(n\ge2\), and let \(A=(a_{ij})\) be the \(n\times n\) matrix with \[ a_{ii}=k, \qquad a_{i,i+1}=a_{i+1,i}=1, \] and all other entries equal to \(0\).
Let \(\lambda_{\min}\) and \(\lambda_{\max}\) be the smallest and largest eigenvalues of \(A\). Show that \[ \lambda_{\min}\le k-1 \qquad\text{and}\qquad \lambda_{\max}\ge k+1. \]
Solution
1. For every nonzero vector \(v\in\RR^n\),\[ \lambda_{\min} \leq \frac{v^TAv}{v^Tv} \leq \lambda_{\max}. \]
The matrix \(A\) is real symmetric, so by the spectral theorem it has an orthonormal eigenbasis \[ u_1,\ldots,u_n \] with corresponding real eigenvalues \[ \lambda_1,\ldots,\lambda_n \] lying between \(\lambda_{\min}\) and \(\lambda_{\max}\). Write \[ v=\sum_{j=1}^n c_ju_j. \] Then \[ \frac{v^TAv}{v^Tv} = \frac{\sum_j\lambda_j c_j^2}{\sum_jc_j^2}, \] which is a weighted average of the eigenvalues. Hence it lies between their minimum and maximum.
2. One has\[ \lambda_{\min}\leq k-1. \]
Take \[ v=e_1-e_2. \] Only the upper-left \(2\times2\) block of \(A\) contributes to \(v^TAv\), so \[ v^TAv = \begin{pmatrix}1&-1\end{pmatrix} \begin{pmatrix}k&1\\1&k\end{pmatrix} \begin{pmatrix}1\\-1\end{pmatrix} = 2k-2. \] Since \(v^Tv=2\), \[ \frac{v^TAv}{v^Tv}=k-1. \] Step 1 therefore gives \(\lambda_{\min}\leq k-1\).
3. One has\[ \lambda_{\max}\geq k+1. \]
Take \[ w=e_1+e_2. \] Again using the upper-left \(2\times2\) block, \[ w^TAw = \begin{pmatrix}1&1\end{pmatrix} \begin{pmatrix}k&1\\1&k\end{pmatrix} \begin{pmatrix}1\\1\end{pmatrix} = 2k+2. \] Since \(w^Tw=2\), \[ \frac{w^TAw}{w^Tw}=k+1. \] Step 1 therefore gives \(\lambda_{\max}\geq k+1\).
4. Q.E.D.
Steps 2 and 3 are the required inequalities.