Schur Decomposition

Definition A square matrix $Q \in \mathbb{C}^{n \times n}$ is unitary if

where $Q^{*} = \overline{Q}^{\mathsf{T}}$ is the conjugate transpose. With the inner product $\langle x, y \rangle = x^{*}y$ on $\mathbb{C}^n$, the entry in position $(i, j)$ of $Q^{*}Q$ is $\langle q_i, q_j \rangle$, so $Q$ is unitary exactly when its columns $q_1, \dots, q_n$ form an orthonormal basis of $\mathbb{C}^n$. A unitary matrix is invertible with $Q^{-1} = Q^{*}$, so $QQ^{*} = I_n$ as well, and it preserves inner products,

Products of unitary matrices are unitary, because $(Q_1Q_2)^{*}Q_1Q_2 = Q_2^{*}Q_1^{*}Q_1Q_2 = I_n$, and the inverse $Q^{*}$ of a unitary matrix is unitary, because $(Q^{*})^{*}Q^{*} = QQ^{*} = I_n$. Every unitary matrix has $\lvert \det Q \rvert = 1$, since $\lvert \det Q \rvert^2 = \det(Q^{*})\det(Q) = \det(Q^{*}Q) = 1$. A real unitary matrix is an orthogonal matrix, $Q^{\mathsf{T}}Q = I_n$. Squareness matters. A $3 \times 2$ matrix with orthonormal columns satisfies $Q^{*}Q = I_2$, but $QQ^{*}$ is a $3 \times 3$ matrix of rank $2$, not $I_3$. Two matrices $A$ and $Q^{*}AQ$ with $Q$ unitary are called unitarily similar. They are similar, since $Q^{*} = Q^{-1}$.

Definition Schur’s theorem says the following. Let $A \in \mathbb{C}^{n \times n}$ with $n \ge 1$. There exist a unitary matrix $Q \in \mathbb{C}^{n \times n}$ and an upper triangular matrix $R \in \mathbb{C}^{n \times n}$ such that

or equivalently $Q^{*}AQ = R$. Such a pair is a Schur decomposition of $A$. The diagonal entries of $R$ are the eigenvalues of $A$, each listed as often as its algebraic multiplicity, and the proof can place them in any chosen order. The decomposition is not unique. Since $AQ = QR$ and column $j$ of $R$ has zeros below row $j$, each $Aq_j$ is a combination of $q_1, \dots, q_j$. So for every $k$ the first $k$ columns of $Q$ span a $k$-dimensional subspace that $A$ maps into itself. Only the first column of $Q$ is guaranteed to be an eigenvector. The theorem applies to real matrices too, with $Q$ and $R$ complex in general. If $A$ is real and all its eigenvalues are real, the same construction works with real vectors and gives a real orthogonal $Q$ and a real $R$. The proof uses the fundamental theorem of algebra, taken here without proof, which says that every nonconstant polynomial with complex coefficients has a root in $\mathbb{C}$. Applied to $\chi_A$, it supplies an eigenvalue.

Definition Let $V$ be a finite-dimensional inner product space over $\mathbb{F}$, where $\mathbb{F}$ is $\mathbb{R}$ or $\mathbb{C}$, let $T \in \operatorname{Hom}(V, V)$, let $W$ be a subspace of $V$, and let $P_W$ be the orthogonal projection onto $W$. The compression of $T$ to $W$ is the operator on $W$ given by

It is defined for every subspace $W$, invariant or not. If $(w_1, \dots, w_k)$ is an orthonormal basis of $W$, its matrix in that basis has entries $\langle w_i, Tw_j \rangle$. The compression equals the restriction $T|_W$ exactly when $W$ is $T$-invariant, since then $P_W(Tw) = Tw$.

The compression appears in the first step of Schur’s theorem. Let $A \in \mathbb{C}^{n \times n}$ with $n \ge 2$ and $Aq_1 = \lambda q_1$ for a unit vector $q_1$, and complete $q_1$ to an orthonormal basis $(q_1, \dots, q_n)$, the columns of a unitary $Q_0$. Then

with $a \in \mathbb{C}^{n-1}$ and $B \in \mathbb{C}^{(n-1) \times (n-1)}$. The first column is $(\lambda, 0, \dots, 0)$ because $Aq_1 = \lambda q_1$, and $B$, with entries $\langle q_i, Aq_j \rangle$ for $i, j \ge 2$, is the matrix of the compression of $A$ to $q_1^{\perp}$ in the basis $(q_2, \dots, q_n)$. Similar matrices share their characteristic polynomial, and the block form gives

so the eigenvalues of $B$ are those of $A$ with one copy of $\lambda$ removed.

Definition Induction on dimension proves a statement $S(n)$ for every $n \ge 1$ in two parts. The base case proves $S(1)$, and the induction step proves, for every $n \ge 2$, that $S(n - 1)$ implies $S(n)$. The hypothesis $S(n - 1)$ may be applied to any object of the smaller size, so the smaller object need not be a restriction of the original one. For Schur’s theorem, $S(n)$ says that every matrix in $\mathbb{C}^{n \times n}$ has a Schur decomposition.

1. The base case. A $1 \times 1$ matrix is upper triangular, so $Q = [1]$ and $R = A$ work.
2. The first eigenvector. For $n \ge 2$ the polynomial $\chi_A$ has degree $n$, so by the fundamental theorem of algebra it has a root $\lambda$. Then $A - \lambda I$ is singular, and a nonzero vector in its kernel, divided by its length, is a unit eigenvector $q_1$.
3. The block form. Completing $q_1$ to an orthonormal basis gives a unitary $Q_0$ for which $Q_0^{*}AQ_0$ has first column $(\lambda, 0, \dots, 0)$, first row $(\lambda, a^{*})$ and lower right block $B$, the compression of $A$ to $q_1^{\perp}$.
4. The induction. The hypothesis $S(n - 1)$ gives a unitary $Z$ with $Z^{*}BZ = R_B$ upper triangular. Extending $Z$ by a $1$ in the corner gives a unitary $Q_1$, and $Q = Q_0Q_1$ is unitary as a product of unitary matrices.

With these choices,

which is upper triangular. The step is legitimate because $S(n - 1)$ covers every $(n - 1) \times (n - 1)$ matrix and $B$ is one, even though $q_1^{\perp}$ need not be invariant. Since $\chi_B = \chi_A / (t - \lambda)$, choosing $\lambda$ freely at each step puts the eigenvalues on the diagonal in any order.

Definition Let $A \in \mathbb{C}^{n \times n}$ with $n \ge 1$. By the fundamental theorem of algebra the characteristic polynomial factors as

with distinct $\lambda_1, \dots, \lambda_k \in \mathbb{C}$ and $m_1 + \cdots + m_k = n$. The exponent $m_j$ is the algebraic multiplicity of the eigenvalue $\lambda_j$. The geometric multiplicity is $g_j = \dim \ker(A - \lambda_j I)$, and

For a real matrix the same definitions use the complex roots of $\chi_A$. Similar matrices have the same characteristic polynomial and eigenspaces of equal dimension, so both multiplicities are unchanged by similarity. The characteristic polynomial of a triangular matrix is the product of the factors $t - r_{ii}$ over its diagonal entries, so in every Schur form, and in every triangular matrix similar to $A$, the eigenvalue $\lambda_j$ appears exactly $m_j$ times on the diagonal. To see $g_j \le m_j$, extend a basis of the eigenspace of $\lambda_j$ to a basis of $\mathbb{C}^n$. The eigenspace is invariant, so the matrix in that basis is block upper triangular with upper left block $\lambda_j I_{g_j}$, and $(t - \lambda_j)^{g_j}$ divides $\chi_A$.

Definition Let $A \in \mathbb{C}^{n \times n}$ with $n \ge 1$, and let $\lambda_1, \dots, \lambda_n$ be its eigenvalues listed with algebraic multiplicity. The trace is the sum and the determinant is the product of the eigenvalues,

No diagonalizability is needed, and for a real matrix the list includes its nonreal eigenvalues. For the proof, take a Schur decomposition $Q^{*}AQ = R$. The determinant is multiplicative and $\det(Q^{*})\det(Q) = \det(Q^{*}Q) = 1$, so $\det A = \det R$. For the trace, whenever $X$ is $m \times n$ and $Y$ is $n \times m$,

so $\operatorname{tr}(Q^{*}AQ) = \operatorname{tr}(AQQ^{*}) = \operatorname{tr} A$. The determinant of the triangular matrix $R$ is the product and its trace is the sum of its diagonal entries, which are $\lambda_1, \dots, \lambda_n$.