Diagonalize the matrix A : 2 2 1 3

solution :

\(P = \begin{pmatrix} 2 & 1 \\ -1 & 1 \end{pmatrix}\), \(D = P^{-1}AP = \begin{pmatrix} 1 & 0 \\ 0 & 4 \end{pmatrix}\)

\[A = \begin{pmatrix} 2 & 2 \\ 1 & 3 \end{pmatrix}\]

Step 1: Characteristic equation

\[|A - \lambda I| = (2-\lambda)(3-\lambda) - 2 = 0\]

\[\lambda^2 - 5\lambda + 4 = 0\]

\[\lambda = 1,\ 4\]

Step 2: Eigenvector for \(\lambda = 1\)

\((A - I)X = 0\): \(x_1 + 2x_2 = 0\), so \(X_1 = \begin{pmatrix} 2 \\ -1 \end{pmatrix}\).

Step 3: Eigenvector for \(\lambda = 4\)

\((A - 4I)X = 0\): \(-2x_1 + 2x_2 = 0\), so \(x_1 = x_2\) and \(X_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}\).

Step 4: Modal matrix and its inverse

\[P = \begin{pmatrix} 2 & 1 \\ -1 & 1 \end{pmatrix},\ |P| = 3\]

\[P^{-1} = \frac{1}{3}\begin{pmatrix} 1 & -1 \\ 1 & 2 \end{pmatrix}\]

Step 5: Check

\[AP = \begin{pmatrix} 2 & 4 \\ -1 & 4 \end{pmatrix}\]

\[P^{-1}AP = \frac{1}{3}\begin{pmatrix} 3 & 0 \\ 0 & 12 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 4 \end{pmatrix}\]

So A is diagonalised to \(D = \text{diag}(1, 4)\); the eigenvalues appear on the diagonal in the same order as their eigenvectors in P.