State the Spectral Theorem (over \(\mathbb{R}\)).
Let \(A = \begin{bmatrix} 2 & -2 \\ -2 & 5 \end{bmatrix}\). Find an orthogonal matrix \(P\) for which \(P^{-1}AP\) is diagonal.
Solution
Part (a).
1. Spectral Theorem (over \(\RR\)): if \(A\) is a real symmetric matrix, then \(A\) is orthogonally diagonalizable; i.e. there is an orthogonal matrix \(P\) (with \(P^{-1} = P^T\)) such that \(P^{-1} A P\) is diagonal, and the diagonal entries are the (real) eigenvalues of \(A\).
statement of the theorem.
Part (b).
1. The eigenvalues of \(A\) are \(1\) and \(6\).
\(\det(A - \lambda I) = (2-\lambda)(5-\lambda) - 4 = \lambda^2 - 7\lambda + 6 = (\lambda - 1)(\lambda - 6)\).
2. An eigenvector for \(\lambda = 1\) is \(\begin{bmatrix} 2 \\ 1 \end{bmatrix}\).
\((A - I)\begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ -2 & 4 \end{bmatrix}\begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}\).
3. An eigenvector for \(\lambda = 6\) is \(\begin{bmatrix} -1 \\ 2 \end{bmatrix}\).
\((A - 6I)\begin{bmatrix} -1 \\ 2 \end{bmatrix} = \begin{bmatrix} -4 & -2 \\ -2 & -1 \end{bmatrix}\begin{bmatrix} -1 \\ 2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}\).
4. The eigenvectors \(\begin{bmatrix} 2 \\ 1 \end{bmatrix}\) and \(\begin{bmatrix} -1 \\ 2 \end{bmatrix}\) are orthogonal.
\(2(-1) + 1(2) = 0\).
5. Normalize them to get the orthogonal matrix\[P = \frac{1}{\sqrt{5}}\begin{bmatrix} 2 & -1 \\ 1 & 2 \end{bmatrix}.\]
each eigenvector has norm \(\sqrt{5}\).
6. Then \(P^{-1} A P = \begin{bmatrix} 1 & 0 \\ 0 & 6 \end{bmatrix}\).
\(P\) is orthogonal (so \(P^{-1} = P^T\)) and its columns are eigenvectors of \(A\) with eigenvalues \(1\) and \(6\).
7. Q.E.D.
6.