31.7. Examples

Example 2: Diagonalization Using Similarity Let A = $\begin{bmatrix} 4 & 1 \\ 0 & 3 \end{bmatrix}$ Find if A is diagonalizable. Solution: • Characteristic equation: $\text{det}(A - \lambda I) = (4 - \lambda)(3 - \lambda)$ • Eigenvalues: $\lambda_1 = 4, \lambda_2 = 3$ • Since eigenvalues are distinct, A is diagonalizable. • Find eigenvectors: – For $\lambda = 4:$ $(A - 4I)x = 0$ – For $\lambda = 3:$ $(A - 3I)x = 0$ • Construct matrix P with eigenvectors, compute D = P^{-1}AP Thus, A is similar to D, confirming diagonalization.

In this example, we are determining if matrix A is diagonalizable. To do this, we first find the characteristic equation by taking the determinant of $A - \lambda I$. The calculation results in the eigenvalues 4 and 3. Since these are distinct, we know that A can have a complete set of linearly independent eigenvectors, hence A is diagonalizable. Next, we find the eigenvectors for each eigenvalue. By forming a matrix P from these eigenvectors, we can compute the diagonal matrix D. This process of diagonalization confirms that A is indeed similar to a diagonal matrix D, reducing complexity in computations.

Imagine a high school where students are organized by class based on their subjects. If each student (student being an eigenvector) can belong to their distinct classes (eigenvalues), we can arrange them neatly into separate groups without confusion (diagonalization). On the other hand, if there are too many mixed classes with overlapping subjects (like matrix A with insufficient distinct linearly independent eigenvectors), we cannot neatly group them by subjects (indicating non-diagonalizability). Diagonalization here is akin to creating a class schedule that clearly defines who belongs to which subject, making it easy to manage.

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