Let \(A\) be an \(n \times n\) matrix with real entries such that \(A^2 + I = 0\). Prove that \(n\) is even, and if \(n = 2k\), then \(A\) is similar over the field of real numbers to a matrix of the block form \[ \begin{bmatrix} 0 & -I \\ I & 0 \end{bmatrix} \] where \(I\) is the \(k \times k\) identity matrix.
Solution
1. The minimal polynomial of \(A\) is \(x^2+1\).
The relation \[ A^2+I=0 \] shows that the minimal polynomial \(m_A\) divides \(x^2+1\). Over \(\mathbb R\), \(x^2+1\) is irreducible. The minimal polynomial cannot be constant, so \[ \boxed{m_A(x)=x^2+1}. \]
2. Rational form forces \(n\) to be even.
Let \[ p_r\mid p_{r-1}\mid\cdots\mid p_1=m_A \] be the invariant factors of \(A\). Every nonconstant \(p_i\) divides \(m_A=x^2+1\). Irreducibility therefore forces \[ p_i=x^2+1 \] for every invariant factor. Their product is the characteristic polynomial, whose degree is \(n\) [@HK71]. If there are \(k\) invariant factors, then \[ n=\sum_{i=1}^k\deg p_i=2k. \] Thus \(n\) is even.
3. The rational form is a direct sum of standard two-dimensional rotation blocks.
For \(p(x)=x^2+1\), Hoffman–Kunze’s companion matrix is \[ C(p)=\begin{pmatrix}0&-1\\1&0\end{pmatrix}. \] Therefore \(A\) is similar over \(\mathbb R\) to the direct sum of \(k\) copies of \(C(p)\). In the corresponding basis write the vectors of the \(j\)th companion block as \(u_j,v_j\), so \[ Au_j=v_j, \qquad Av_j=-u_j. \] Reorder the basis as \[ u_1,\ldots,u_k,v_1,\ldots,v_k. \] Then the matrix becomes \[ \boxed{ \begin{pmatrix} 0&-I_k\\ I_k&0 \end{pmatrix}.} \] This is exactly the required real similarity form.