Franklin's Notes


Schur Factorization

If $A\in\mathbb C^{m\times m}$, then there exists a factorization of $A$ in the form $A=QUQ^\ast$, where $Q$ is unitary and $U$ is upper triangular. This kind of decomposition is called a Schur Factorization. Additionally, the matrix $U$ is called the Schur form of $A$.

Proposition 1. Every square matrix $A\in\mathbb C^{m\times m}$ has a Schur factorization.

Proof. Suppose $A\in\mathbb C^{m\times m}$ with $m > 1$, and let $v$ be a unit eigenvector of $A$ corresponding to some eigenvalue $\lambda$. Further, let $V$ be an arbitrary unitary matrix whose first column is equal to $v$, so that we have $V=[v|v_2|\cdots|v_n]$ with $v,v_2,\cdots,v_n$ orthonormal. We then have that and therefore, by orthonormality, the matrix $V^\ast AV$ takes the form Now, if the matrix $A'$ has a Schur factorization $A'=Q'U'Q'^\ast$, then we may form a Schur factorization of $A=QUQ^\ast$ as follows, by letting and This means that if a Schur factorization exists for all $(m-1)\times (m-1)$ matrices, then it must exist for all $m\times m$ matrices. Since the Schur factorization is trivial for $1\times 1$ matrices, we have by induction that it exists for all arbitrary matrices $A\in\mathbb C^{m\times m}$. $\blacksquare$

linear-algebra

matrices

matrix-factorization

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