Diagonalize the following matrix: \[ A = \begin{pmatrix} 2 & 1 \newline 1 & 2 \end{pmatrix} \]
Answer
We can now write the matrix \(A\) as \(A = PDP^{-1}\), where \(P\) is the matrix whose columns are the eigenvectors, \(D\) is the diagonal matrix whose entries are the eigenvalues, and \(P^{-1}\) is the inverse of \(P\). In this case, \(P = \begin{pmatrix} 1 & 1 \newline -1 & 1 \end{pmatrix}\), \(D = \begin{pmatrix} 1 & 0 \newline 0 & 3 \end{pmatrix}\), and \(P^{-1} = \frac{1}{2}\begin{pmatrix} 1 & -1 \newline 1 & 1 \end{pmatrix}\).
Step 1 ：Find the eigenvalues of the matrix. This can be done by solving the characteristic equation, \(\text{det}(A-\lambda I) = 0\), where \(I\) is the identity matrix and \(\lambda\) are the eigenvalues.
Step 2 ：The characteristic equation is \(\text{det}\left(\begin{pmatrix} 2-\lambda & 1 \newline 1 & 2-\lambda \end{pmatrix}\right) = 0\), which simplifies to \((2-\lambda)^2 - 1 = 0\). Solving this equation gives \(\lambda = 1, 3\).
Step 3 ：Find the eigenvectors corresponding to each eigenvalue. For \(\lambda = 1\), we solve \( (A - \lambda I)X = 0 \) for \(X\), giving the equation \(\begin{pmatrix} 1 & 1 \newline 1 & 1 \end{pmatrix}X = 0\). The solution to this is \(X = \begin{pmatrix} 1 \newline -1 \end{pmatrix}\). For \(\lambda = 3\), we solve the same equation, giving the equation \(\begin{pmatrix} -1 & 1 \newline 1 & -1 \end{pmatrix}X = 0\). The solution to this is \(X = \begin{pmatrix} 1 \newline 1 \end{pmatrix}\).
Step 4 ：We can now write the matrix \(A\) as \(A = PDP^{-1}\), where \(P\) is the matrix whose columns are the eigenvectors, \(D\) is the diagonal matrix whose entries are the eigenvalues, and \(P^{-1}\) is the inverse of \(P\). In this case, \(P = \begin{pmatrix} 1 & 1 \newline -1 & 1 \end{pmatrix}\), \(D = \begin{pmatrix} 1 & 0 \newline 0 & 3 \end{pmatrix}\), and \(P^{-1} = \frac{1}{2}\begin{pmatrix} 1 & -1 \newline 1 & 1 \end{pmatrix}\).
