📂Matrix Algebra

Schur Decomposition of Square Matrices

Definition

For a given unitary matrix $Q$ and upper triangular matrix $T$, if $A = Q T Q^{\ast}$, then $A$ is said to have a Schur Factorization.

Theorem

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

Explanation

The downside of eigenvalue diagonalization is that when it is decomposed into $A = S \Lambda S^{-1}$, it still requires the effort to find $S^{-1}$. Although it is true that the time to calculate powers significantly decreases, considering that all problems of matrix algebra boil down to finding the inverse, powers may not be needed except for problems that require an enormous amount of them, where the cure may be worse than the disease.

On the other hand, the Schur Factorization, given that $T$ is an upper triangular matrix and might not be as useful for powers, makes it very easy to find $Q^{\ast}$, so handling $T$ properly can potentially solve problems much faster than eigenvalue diagonalization. Considering that an upper triangular matrix is not that difficult to manage, its versatility can be considered significant. Moreover, beyond everything else, the Schur Factorization has quite relaxed conditions since being a square matrix is sufficient.

Proof

We prove this using mathematical induction for a finite-dimensional $m$. Let’s say $A_{m}$ is some $m \times m$ matrix.

If $m = 1$, then $A_{1}$ is a scalar, which is trivial.

Assume for $m \ge 2$ that $A_{m-1}$ has a Schur Factorization.

Let $\mathbf{x}$ be a unit eigenvector corresponding to the eigenvalue $\lambda$, and define $U := \begin{bmatrix} \mathbf{x} & U_{1} \end{bmatrix}$ as a unitary matrix, then $\mathbf{x}^{\ast} A_{m} \mathbf{x} = \lambda$ and $U_{1}^{\ast} \mathbf{x} = 0$

$$\begin{align*} U^{\ast} A_{m} U =& \begin{bmatrix} \mathbf{x}^{\ast} \\ U_{1}^{\ast} \end{bmatrix} A_{m} \begin{bmatrix} \mathbf{x} & U_{1} \end{bmatrix} \\ =& \begin{bmatrix} \mathbf{x}^{\ast} A_{m} \mathbf{x} & \mathbf{x}^{\ast} A_{m} U_{1} \\ U_{1}^{\ast} A_{m} \mathbf{x} & U_{1}^{\ast} A_{m} U_{1} \end{bmatrix} \\ =& \begin{bmatrix} \lambda & \mathbf{x}^{\ast} A_{m} U_{1} \\ U_{1}^{\ast} \lambda \mathbf{x} & A_{m-1} \end{bmatrix} \\ =& \begin{bmatrix} \lambda & B \\ \mathbb{0} & A_{m-1} \end{bmatrix} \end{align*}$$

$B \in \mathbb{C}^{m-1}$ is any vector, and by assumption, $A_{m-1}$ has a Schur Factorization.

Therefore, $A_{m-1}$ is represented by some unitary matrix $V$ and upper triangular matrix $T$ as $A_{m-1} = V T V^{\ast}$.

Then, for a unitary matrix $Q := \begin{bmatrix} 1 & 0 \\ 0 & V \end{bmatrix}$ and upper triangular matrix $\begin{bmatrix} \lambda & B V \\ 0 & T \end{bmatrix}$,

$$\begin{align*} Q \begin{bmatrix} \lambda & B V \\ 0 & T \end{bmatrix} Q^{\ast} &= \begin{bmatrix} 1 & 0 \\ 0 & V \end{bmatrix} \begin{bmatrix} \lambda & B V \\ 0 & T \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & V^{\ast} \end{bmatrix} \\ =& \begin{bmatrix} 1 & 0 \\ 0 & V \end{bmatrix} \begin{bmatrix} \lambda & B \\ 0 & T V^{\ast} \end{bmatrix} \\ =& \begin{bmatrix} \lambda & B \\ 0 & V T V^{\ast} \end{bmatrix} \\ =& \begin{bmatrix} \lambda & B \\ 0 & A_{m-1} \end{bmatrix} \end{align*}$$

In other words,

$$\begin{align*} A_{m} =& U Q \begin{bmatrix} \lambda & B V \\ 0 & T \end{bmatrix} Q^{\ast} U^{\ast} \\ =& ( U Q ) \begin{bmatrix} \lambda & B V \\ 0 & T \end{bmatrix} ( U Q )^{\ast} \end{align*}$$

If $A_{m-1}$ has a Schur Factorization, then $A_{m}$ also has a Schur Factorization, therefore, by mathematical induction, every square matrix has a Schur Factorization.

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