Find a \(2 \times 2\) matrix \(P\) such that \(P \begin{pmatrix} 1 & 2 \\ \frac{1}{2} & 1 \end{pmatrix} P^{-1}\) is diagonal.
Solution
1. Let \(A = \begin{pmatrix} 1 & 2 \\ 1/2 & 1 \end{pmatrix}\); its eigenvalues are \(0\) and \(2\).
\(\det(A - \lambda I) = (1-\lambda)^2 - 1 = \lambda^2 - 2\lambda = \lambda(\lambda - 2)\).
2. An eigenvector for \(\lambda = 0\) is \(\begin{pmatrix} -2 \\ 1 \end{pmatrix}\).
\(A\begin{pmatrix} -2 \\ 1 \end{pmatrix} = \begin{pmatrix} -2 + 2 \\ -1 + 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\).
3. An eigenvector for \(\lambda = 2\) is \(\begin{pmatrix} 2 \\ 1 \end{pmatrix}\).
\(A\begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 + 2 \\ 1 + 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \end{pmatrix} = 2\begin{pmatrix} 2 \\ 1 \end{pmatrix}\).
4. For \(P A P^{-1}\) to be diagonal, the rows of \(P\) must be left eigenvectors of \(A\) (i.e. eigenvectors of \(A^T\)).
\(P A P^{-1}\) is diagonal iff the rows of \(P\) are left eigenvectors of \(A\).
5. Left eigenvectors: for \(\lambda = 0\), solve \((x, y)A = 0\), i.e. \(x + y/2 = 0\) and \(2x + y = 0\), giving \((x, y) = (-1, 2)\); for \(\lambda = 2\), solve \((x,y)A = 2(x,y)\), giving \((x, y) = (1, 2)\).
direct computation.
6. Let \(P = \begin{pmatrix} -1 & 2 \\ 1 & 2 \end{pmatrix}\) (rows are the left eigenvectors).
definition.
7. Then \(P A P^{-1} = \begin{pmatrix} 0 & 0 \\ 0 & 2 \end{pmatrix}\).
the rows of \(P\) are left eigenvectors with eigenvalues \(0\) and \(2\), so \(P A P^{-1}\) is diagonal with diagonal entries \(0\) and \(2\).
8. Q.E.D.
7.