Find an invertible real matrix \(P\) such that \(P^{-1}AP\) and \(P^{-1}BP\) are both diagonal, where \(A\) and \(B\) are the real matrices
\[ A = \begin{bmatrix} 1 & 2 \\ 0 & 2 \end{bmatrix}, \quad B = \begin{bmatrix} 3 & -8 \\ 0 & -1 \end{bmatrix} \]
\[ A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}, \qquad B = \begin{bmatrix} 1 & a \\ a & 1 \end{bmatrix}. \]
Solution
(a).
1. \(A\) has eigenvalues \(1\) and \(2\) (it is upper-triangular), with eigenvectors \(e_1 = (1,0)\) and \((2,1)\).
\(A e_1 = e_1\), and \(A(2,1) = (4, 2) = 2(2,1)\).
2. \(B\) has eigenvalues \(3\) and \(-1\), with eigenvectors \(e_1 = (1,0)\) and \((2,1)\).
\(B e_1 = 3e_1\), and \(B(2,1) = (-2, -1) = -1(2,1)\).
3. Hence \(A\) and \(B\) are simultaneously diagonalized by \(P = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\) (whose columns are the common eigenvectors).
1 and 2 (both matrices share the eigenvectors \(e_1\) and \((2,1)\)).
4. Indeed \(P^{-1}AP = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}\) and \(P^{-1}BP = \begin{pmatrix} 3 & 0 \\ 0 & -1 \end{pmatrix}\).
3.
(b).
1. \(A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\) has eigenvalues \(0\) and \(2\), with eigenvectors \((1,-1)\) and \((1,1)\).
\(A(1,-1) = (0,0)\) and \(A(1,1) = (2,2) = 2(1,1)\).
2. \(B = \begin{pmatrix} 1 & a \\ a & 1 \end{pmatrix}\) has eigenvalues \(1 + a\) and \(1 - a\), with eigenvectors \((1,1)\) and \((1,-1)\).
\(B(1,1) = (1+a, 1+a) = (1+a)(1,1)\) and \(B(1,-1) = (1-a, a-1) = (1-a)(1,-1)\).
3. Hence \(A\) and \(B\) are simultaneously diagonalized by \(P = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}\) (columns are the common eigenvectors \((1,-1)\) and \((1,1)\)).
1 and 2.
4. Indeed \(P^{-1}AP = \begin{pmatrix} 0 & 0 \\ 0 & 2 \end{pmatrix}\) and \(P^{-1}BP = \begin{pmatrix} 1-a & 0 \\ 0 & 1+a \end{pmatrix}\).
3.
5. Q.E.D.
4 (a) and 4 (b).