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QB_EM_ADV_020
Question Bank • QB_EM_ADV_020
NAT • 2 Marks
Section 1: Engineering Mathematics
Linear Algebra
Subtopic: Eigen values and Eigen vectors
Question Statement:
Consider the real symmetric tridiagonal matrix:
$A = \begin{pmatrix} 2 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 2 \end{pmatrix}$
The maximum eigenvalue $\lambda_{\max}$ of $A$ is ________ (round off to two decimal places).
Step-by-Step Solution & Explanation
**Method 1: Characteristic Equation**
$\det(A - \lambda I) = \begin{vmatrix} 2 - \lambda & 1 & 0 \\ 1 & 2 - \lambda & 1 \\ 0 & 1 & 2 - \lambda \end{vmatrix} = 0$
Expanding along the first row:
$(2 - \lambda)[(2 - \lambda)^2 - 1] - 1[1(2 - \lambda) - 0] = 0$
$(2 - \lambda)[(2 - \lambda)^2 - 2] = 0$
Roots:
1. $2 - \lambda = 0 \implies \lambda = 2$
2. $(2 - \lambda)^2 = 2 \implies \lambda = 2 \pm \sqrt{2}$
$\lambda_1 = 2 - \sqrt{2} \approx 0.59, \quad \lambda_2 = 2, \quad \lambda_3 = 2 + \sqrt{2} \approx 3.4142$
The maximum eigenvalue is $\lambda_{\max} = 2 + \sqrt{2} \approx 3.41$.
**Method 2: Analytical Tridiagonal Toeplitz Formula**
For an $n \times n$ Toeplitz matrix with diagonal $a=2$ and off-diagonal $b=1$:
$\lambda_k = a + 2b \cos\left(\frac{k\pi}{n+1}\right), \quad k = 1, \dots, n$
For $n = 3$, the largest eigenvalue occurs at $k = 1$:
$\lambda_1 = 2 + 2(1) \cos\left(\frac{\pi}{4}\right) = 2 + 2\left(\frac{\sqrt{2}}{2}\right) = 2 + \sqrt{2} \approx 3.41$
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