How to find eigenvalues and eigenvectors of a 3x3 matrix?

For the matrix: $$A = \begin{pmatrix} 2 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 2 \end{pmatrix}$$ Step 1: Compute $A - \lambda I$: $$\begin{pmatrix} 2-\lambda & 1 & 0 \\ 1 & 2-\lambda & 1 \\ 0 & 1 & 2-\lambda \end{pmatrix}$$ Step 2: Compute determinant. For a tridiagonal matrix, expansion along the first row gives: $$\det = (2-\lambda)\begin{vmatrix} 2-\lambda & 1 \\ 1 & 2-\lambda \end{vmatrix} - 1\begin{vmatrix} 1 & 1 \\ 0 & 2-\lambda \end{vmatrix}$$ $$= (2-\lambda)[(2-\lambda)^2 - 1] - 1[(2-\lambda) - 0]$$ $$= (2-\lambda)(\lambda^2 - 4\lambda + 3) - (2-\lambda)$$ $$= (2-\lambda)(\lambda^2 - 4\lambda + 2)$$ Step 3: Solve $(2-\lambda)(\lambda^2 - 4\lambda + 2) = 0$, giving: $$\lambda = 2, \quad \lambda = 2 \pm \sqrt{2}$$ Step 4: Find eigenvectors by solving $(A - \lambda I)\mathbf{v} = 0$ for each eigenvalue. **Trick:** For symmetric tridiagonal matrices like this, there's a pattern: eigenvalues are $\lambda_k = 2 + 2\cos\left(\frac{k\pi}{n+1}\right)$ for $k = 1, \ldots, n$, where $n = 3$. So $\lambda_k = 2 + 2\cos(k\pi/4)$, giving $2 + \sqrt{2}, 2, 2 - \sqrt{2}$.