Real Schur Form

Definition Let $A \in \mathbb{R}^{n \times n}$. Its characteristic polynomial has real coefficients, so conjugating the factorization of $\chi_A$ over $\mathbb{C}$ gives the same polynomial back. Hence the nonreal eigenvalues of $A$ come in pairs $\lambda$, $\overline{\lambda}$ with equal algebraic multiplicities. Eigenvectors pair up the same way. If $z \in \mathbb{C}^n$ is nonzero and $Az = \lambda z$, then conjugating every entry and using $\overline{A} = A$ gives

The pairs $(\lambda, z)$ and $(\overline{\lambda}, \overline{z})$ form a complex conjugate eigenpair. When $\lambda = a + ib$ with $a, b \in \mathbb{R}$ and $b \ne 0$, write $z = x + iy$ with $x, y \in \mathbb{R}^n$. Comparing real and imaginary parts of $A(x + iy) = (a + ib)(x + iy)$ gives

Definition Let $A \in \mathbb{R}^{n \times n}$ have a nonreal eigenvalue $\lambda = a + ib$, with $a, b \in \mathbb{R}$ and $b \ne 0$, and an eigenvector $z = x + iy$ with $x, y \in \mathbb{R}^n$. Then $x$ and $y$ are linearly independent, and the real plane

is invariant under $A$. In the ordered basis $(x, y)$ the restriction $A|_U$ has the matrix

whose eigenvalues are $a \pm ib$. Invariance is read off from $Ax = ax - by$ and $Ay = bx + ay$. For independence, the span of $x$ and $y$ is not $\{0\}$, since $z \ne 0$. If it were a line, that invariant real line would be spanned by a real eigenvector with a real eigenvalue $\mu$, so $Ax = \mu x$ and $Ay = \mu y$. That gives $Az = \mu z$, which forces $\lambda = \mu$ and contradicts $b \ne 0$. The plane $U$ contains no real eigenvector of $A$, because the restriction has no real eigenvalue. The basis $(x, y)$ need not be orthogonal, and $x$ and $y$ need not have equal lengths.

Definition The real Schur theorem says the following. Let $A \in \mathbb{R}^{n \times n}$ with $n \ge 1$. There is a real orthogonal matrix $Q$, so that $Q^{\mathsf{T}}Q = I$, such that

is real and block upper triangular, and each diagonal block $T_{jj}$ is $1 \times 1$ or $2 \times 2$. Each $1 \times 1$ block is a real eigenvalue of $A$, and each $2 \times 2$ block is a real matrix with a pair of nonreal complex conjugate eigenvalues. The matrix $T$ is a real Schur form of $A$. Since

the diagonal blocks account for every eigenvalue of $A$ with its algebraic multiplicity. For each $j$, the columns of $Q$ that belong to the first $j$ blocks span a subspace invariant under $A$. In general neither $Q$ nor $T$ is unique, and the blocks above the diagonal may be nonzero. The form is upper triangular exactly when all eigenvalues of $A$ are real.

The proof is an induction on $n$ that removes one or two dimensions at a time. Assume the theorem for every size below $n$. The fundamental theorem of algebra gives a complex eigenvalue $\lambda$ of $A$. If $\lambda$ is real, the real singular matrix $A - \lambda I$ has a nonzero real kernel vector, and normalizing it gives a unit vector $q_1$ spanning a real invariant line. If $\lambda$ is not real, the real invariant plane built from an eigenvector has an orthonormal basis $q_1, q_2$ by the Gram-Schmidt process. Completing to a real orthonormal basis of $\mathbb{R}^n$ gives an orthogonal $Q_0$ with

where $T_{11}$ is $s \times s$ with $s = 1$ or $s = 2$, and $B$ is the real $(n - s) \times (n - s)$ compression of $A$ to the orthogonal complement of the chosen line or plane. If $s = n$ the proof is complete. Otherwise the induction hypothesis gives an orthogonal $Z$ with $Z^{\mathsf{T}}BZ$ in real Schur form, and $Q = Q_0\begin{bmatrix} I_s & 0 \\ 0 & Z \end{bmatrix}$ finishes the step. Neither an invariant complement nor normality is needed.

Definition A two-by-two real Schur block is a $2 \times 2$ diagonal block $T_{jj}$ of a real Schur form. It is a real matrix

which is exactly the condition for $\chi_M(t) = t^2 - (p + s)t + (ps - qr)$ to have nonreal roots. Its eigenvalues are $\alpha \pm i\beta$, where

and they form a conjugate pair of eigenvalues of $A$. The block is the matrix of the compression of $A$ to the plane spanned by its two columns of $Q$, in that orthonormal basis. For the first block this compression is the restriction of $A$ to an invariant plane. Replacing those two columns $Q_j$ by $Q_jZ$, for any real orthogonal $2 \times 2$ matrix $Z$, replaces the block by $Z^{\mathsf{T}}T_{jj}Z$ and keeps the real Schur shape. So the entries of a block are not determined by $A$, while its eigenvalues, and with them its trace and determinant, stay the same under this replacement. In particular the entries need not follow the pattern with rows $(a, b)$ and $(-b, a)$ that the basis $(x, y)$ produces.

Definition A rotation-scaling block is a real $2 \times 2$ matrix of the form

with $a, b \in \mathbb{R}$ and $b \ne 0$, where $r = \sqrt{a^2 + b^2} > 0$ and $\theta$ is the angle with $\cos\theta = a/r$ and $\sin\theta = b/r$. Its eigenvalues are $a \pm ib$. In an orthonormal basis it rotates the plane through the angle $\theta$, measured from the first basis vector toward the second, and multiplies every length by $r$. A real $2 \times 2$ matrix $M$ with nonreal eigenvalues is a rotation-scaling block if and only if it is normal, which for a real matrix means $M^{\mathsf{T}}M = MM^{\mathsf{T}}$.

For the proof, a rotation-scaling block satisfies $M^{\mathsf{T}}M = (a^2 + b^2)I = MM^{\mathsf{T}}$. Conversely, write $M = S + K$ with $S = \tfrac{1}{2}(M + M^{\mathsf{T}})$ symmetric and $K = \tfrac{1}{2}(M - M^{\mathsf{T}})$ skew-symmetric. Expanding $M^{\mathsf{T}}M = MM^{\mathsf{T}}$ gives $SK = KS$. In two dimensions $K = cJ$ with $J$ the quarter turn with rows $(0, -1)$ and $(1, 0)$, and $c \ne 0$, since otherwise $M$ would be symmetric with real eigenvalues. With $S$ having rows $(d, e)$ and $(e, f)$,

so $SK = KS$ forces $e = 0$ and $d = f$, and $M = dI + cJ$ has rows $(d, -c)$ and $(c, d)$. An orthogonal change of basis preserves normality, because

So a block that is not normal is not a rotation-scaling block in any orthonormal basis. It still has the matrix with rows $(a, -b)$ and $(b, a)$ in the basis $(x, -y)$ built from an eigenvector $x + iy$ for $a + ib$, but for such a block that basis cannot be orthonormal.