For a positive integer \(n\), let \(I_n\) denote the \(n\times n\) identity matrix.
Let \(A\) be a \(2\times2\) real matrix with eigenvalues \(1\) and \(-1\). Show that \(A^2=I_2\).
Find a \(3\times3\) real matrix whose only complex eigenvalues are \(1\) and \(-1\), but for which \(A^2\neq I_3\).
Let \(A\) be an \(n\times n\) real symmetric matrix whose only complex eigenvalues are \(1\) and \(-1\). Show that \(A^2=I_n\).
Solution
Since \(A\) is \(2\times 2\) and has 2 eigenvalues, noting that \(\deg \chi_A(x) = 2\), we have \(\chi_A(x) = (x-1)(x+1) = x^2 -1\). The minimal polynomial of \(A\) divides \(\chi_A(x)\), so we have \(\chi_A(A) = 0\) and thus \(A^2 - I_2 = 0 \implies A^2 = I_2\).
This will happen when \(x^2-1 = (x+1)(x-1)\) is not the minimal polynomial, and we can force the minimal polynomial to be degree 3 by inserting a nontrivial Jordan block to a diagonal matrix containing just the eigenvalues \(\pm 1\). An example that works: \[ \left(\begin{array}{rr|r} 1 & 1 & 0 \\ 0 & 1 & 0 \\ \hline 0 & 0 & -1 \end{array}\right),~ A^2 = \left(\begin{array}{rrr} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right), \quad \spec(A) = [-1, 1, 1] \]
Every symmetric matrix with \(A\) real spectrum admits a real eigendecomposition \(\Lambda D \Lambda^T\), where \(D\) is diagonal with entries the eigenvalues of \(A\) and \(\Lambda\) are orthogonal (which are also invertible). Here, we only need the fact that \(A\) is diagonalizable by invertible matrices. In our case we have \([D^2]_{ii} = (\pm 1)^2 = 1\) so \(D^2 = I_n\). Thus we have \[A^2 = (\Lambda D \inverseof{\Lambda})^2 = \Lambda D^2 \inverseof{\Lambda} = \Lambda I_n \inverseof{\Lambda} = I_n. \qed\]