Complex Eigenvalues and Real Rotations

Definition The quarter-turn rotation matrix is

the standard matrix of the counterclockwise rotation of $\mathbb{R}^2$ about the origin through $90^\circ$. It sends $(x, y)$ to $(-y, x)$, preserves lengths, and satisfies $R^2 = -I$ and $R^4 = I$. Its characteristic polynomial is

Every real $t$ gives $t^2 + 1 \ge 1$, so $R$ has no real eigenvalue and no nonzero real eigenvector. Preserving lengths is not the reason. The reflection with diagonal entries $1$ and $-1$ also preserves lengths, and it has the real eigenlines spanned by $\mathbf{e}_1$ and $\mathbf{e}_2$.

Definition For a positive integer $n$, the complex coordinate space $\mathbb{C}^n$ is the set of columns with $n$ complex entries, with entrywise addition and multiplication by complex scalars. As a complex vector space it has dimension $n$, with the standard basis $\mathbf{e}_1, \dots, \mathbf{e}_n$. The real space $\mathbb{R}^n$ sits inside it as the columns with real entries. A real or complex $m \times n$ matrix multiplies a column of $\mathbb{C}^n$ by the usual row by column rule, now with complex arithmetic, so a real $n \times n$ matrix also acts on $\mathbb{C}^n$. Every $\mathbf{z}$ in $\mathbb{C}^n$ splits uniquely as $\mathbf{z} = \mathbf{x} + i\mathbf{y}$ with $\mathbf{x}$ and $\mathbf{y}$ in $\mathbb{R}^n$, so one vector of $\mathbb{C}^n$ carries $2n$ real numbers. Two uses of complex numbers must be kept apart. A real pair $(x, y)$ can be encoded as the single complex number $x + iy$ in the complex plane, where multiplication by $i$ is the quarter-turn. A vector of $\mathbb{C}^2$ is a column of two complex numbers and carries four real numbers. It is an arrow in the real plane $\mathbb{R}^2$ only when both of its entries are real.

Definition Let $A$ be an $n \times n$ matrix with real or complex entries. A complex number $\lambda$ is an eigenvalue of $A$ over $\mathbb{C}$ if there is a nonzero vector $\mathbf{v}$ in $\mathbb{C}^n$ with $A\mathbf{v} = \lambda\mathbf{v}$, and such a $\mathbf{v}$ is a complex eigenvector for $\lambda$. As over $\mathbb{R}$, the eigenvalues are exactly the roots of $\chi_A(t) = \det(tI - A)$, now taken in $\mathbb{C}$, and the eigenspace of $\lambda$ is the null space of $A - \lambda I$ in $\mathbb{C}^n$, found by elimination with complex arithmetic. By the fundamental theorem of algebra, $\chi_A$ factors into linear factors over $\mathbb{C}$, so $A$ has $n$ complex eigenvalues counted with algebraic multiplicity. Their sum is the trace of $A$ and their product is $\det A$. If $A$ is real and $\lambda$ is not real, the eigenspace of $\lambda$ contains no nonzero real vector, because $A\mathbf{w} = \lambda\mathbf{w}$ with $\mathbf{w}$ real and nonzero would force $\lambda$ to be real.

Definition An $n \times n$ matrix $A$ with real or complex entries is diagonalizable over $\mathbb{C}$ if $A = PDP^{-1}$ for an invertible complex $n \times n$ matrix $P$ and a complex diagonal matrix $D$. As over $\mathbb{R}$, an invertible $P$ and a diagonal $D$ give $A = PDP^{-1}$ exactly when the columns of $P$ form an eigenbasis of $\mathbb{C}^n$ whose eigenvalues are the diagonal entries of $D$ in matching order. So $A$ is diagonalizable over $\mathbb{C}$ exactly when its complex eigenspaces have dimensions adding up to $n$. Since $\chi_A$ always factors over $\mathbb{C}$, an equivalent test is that every eigenvalue has geometric multiplicity equal to its algebraic multiplicity, and $n$ distinct complex eigenvalues are enough. A real matrix that is diagonalizable over $\mathbb{R}$ is also diagonalizable over $\mathbb{C}$, but not conversely, so a claim that a matrix is diagonalizable should name its scalars. If $A$ and its eigenvalue $\lambda$ are both real, elimination on $A - \lambda I$ uses only real numbers, so the complex eigenspace of $\lambda$ has the same dimension as the real one. When the factorization exists, $A^k = PD^kP^{-1}$ for every integer $k \ge 0$.

Definition The conjugate of a complex vector or matrix is taken entry by entry. Let $A$ be an $n \times n$ matrix with real entries. If $A\mathbf{v} = \lambda\mathbf{v}$ for a nonzero $\mathbf{v}$ in $\mathbb{C}^n$, then

so $\overline{\lambda}$ is also an eigenvalue of $A$, with eigenvector $\overline{\mathbf{v}}$. The proof conjugates both sides of $A\mathbf{v} = \lambda\mathbf{v}$. Conjugation passes through sums and products, $\overline{A} = A$ because the entries are real, and $\overline{\mathbf{v}} \ne \mathbf{0}$ because $\mathbf{v} \ne \mathbf{0}$. So the nonreal eigenvalues of a real matrix come in pairs $\lambda$ and $\overline{\lambda}$, and the two members of a pair have the same algebraic multiplicity because $\chi_A$ has real coefficients. In particular a real matrix of odd size has at least one real eigenvalue, because its $n$ eigenvalues, counted with multiplicity, include an even number of nonreal ones. A real eigenvalue is its own conjugate, so the argument gives no new eigenvalue for it. The hypothesis that $A$ is real is essential. The $1 \times 1$ matrix with entry $i$ has the eigenvalue $i$ but not $-i$.

Definition The conjugate transpose of an $m \times n$ complex matrix $A$ is the $n \times m$ matrix $A^{*} = \overline{A}^{\top}$, whose entry in row $j$ and column $k$ is $\overline{a_{kj}}$. For a real matrix the conjugate transpose is the ordinary transpose, $A^{*} = A^{\top}$. It satisfies $(AB)^{*} = B^{*}A^{*}$ and $(A^{*})^{*} = A$. For a column $\mathbf{v}$ in $\mathbb{C}^n$,

a real number that is positive for every $\mathbf{v} \ne \mathbf{0}$. That is why the standard length and inner product on $\mathbb{C}^n$ use the conjugate transpose, with $\lVert \mathbf{v} \rVert = \sqrt{\mathbf{v}^{*}\mathbf{v}}$ and $\langle \mathbf{x}, \mathbf{y} \rangle = \mathbf{x}^{*}\mathbf{y}$. This inner product is conjugate-linear in the first argument, $\langle c\mathbf{x}, \mathbf{y} \rangle = \overline{c}\langle \mathbf{x}, \mathbf{y} \rangle$. The plain product $\mathbf{v}^{\top}\mathbf{v} = v_1^2 + \cdots + v_n^2$ can be zero, negative or nonreal for a nonzero $\mathbf{v}$, so it cannot serve as a squared length.