Numerical Analysis HT24, Schur decomposition
Only mentioned in the notes as brief aside on the section on computing eigenvalues, covered in more depth in Notes - NLA MT25, Schur decompositionU.
Flashcards
Suppose that $A \in \mathbb C^{n \times n}$. What is the Schur decomposition of $A$, and when does it exist?
where $Q$ is a unitary matrix and $T$ is triangular.
It exists for every square matrix.
Suppose that $T$ is a normal matrix. What is special about how it can be diagonalised?
It can be diagonalised by a unitary similarity transformation.
Bite-sized
For a Schur decomposition $A = QTQ^*$, eigenvalues are read off as $\mathrm{diag}(T)$, since $T$ is upper triangular (eigenvalues = diagonal entries) and $T$ is similar to $A$.
Why is computing the Schur decomposition backward stable?
Because it’s a sequence of unitary similarity transformations, each of which is backward stable ($\mathrm{fl}(QX) = Q(X + \Delta X)$ with $\|\Delta X\| = O(\epsilon \|X\|)$). Composition of stable orthogonal operations is also stable — no $\kappa$-amplification, because $\|Q\| _ 2 = 1$.