Open main menu
Home
Random
Recent changes
Special pages
Community portal
Preferences
About Wikipedia
Disclaimers
Incubator escapee wiki
Search
User menu
Talk
Dark mode
Contributions
Create account
Log in
Editing
Schur decomposition
(section)
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
== Notes == Although every square matrix has a Schur decomposition, in general this decomposition is not unique. For example, the eigenspace ''V<sub>λ</sub>'' can have dimension > 1, in which case any orthonormal basis for ''V<sub>λ</sub>'' would lead to the desired result. Write the triangular matrix ''U'' as ''U'' = ''D'' + ''N'', where ''D'' is diagonal and ''N'' is strictly upper triangular (and thus a [[nilpotent matrix]]). The diagonal matrix ''D'' contains the eigenvalues of ''A'' in arbitrary order (hence its Frobenius norm, squared, is the sum of the squared moduli of the eigenvalues of ''A'', while the Frobenius norm of ''A'', squared, is the sum of the squared [[singular value]]s of ''A''). The nilpotent part ''N'' is generally not unique either, but its [[Matrix norm#Frobenius norm|Frobenius norm]] is uniquely determined by ''A'' (just because the Frobenius norm of A is equal to the Frobenius norm of ''U'' = ''D'' + ''N'').<ref>{{cite web |last1=Higham |first1=Nick |title=What Is a Schur Decomposition? |date=11 May 2022 |url=https://nhigham.com/2022/05/11/what-is-a-schur-decomposition/}}</ref> It is clear that if ''A'' is a [[normal matrix]], then ''U'' from its Schur decomposition must be a [[diagonal matrix]] and the column vectors of ''Q'' are the [[eigenvector]]s of ''A''. Therefore, the Schur decomposition extends the [[Eigendecomposition of a matrix|spectral decomposition]]. In particular, if ''A'' is [[Positive-definite matrix|positive definite]], the Schur decomposition of ''A'', its spectral decomposition, and its [[singular value decomposition]] coincide. A [[commutative operation|commuting]] family {''A<sub>i</sub>''} of matrices can be simultaneously triangularized, i.e. there exists a unitary matrix ''Q'' such that, for every ''A<sub>i</sub>'' in the given family, ''Q A<sub>i</sub> Q*'' is upper triangular. This can be readily deduced from the above proof. Take element ''A'' from {''A<sub>i</sub>''} and again consider an eigenspace ''V<sub>A</sub>''. Then ''V<sub>A</sub>'' is invariant under all matrices in {''A<sub>i</sub>''}. Therefore, all matrices in {''A<sub>i</sub>''} must share one common eigenvector in ''V<sub>A</sub>''. Induction then proves the claim. As a corollary, we have that every commuting family of normal matrices can be simultaneously [[Diagonalizable matrix|diagonalized]]. In the infinite dimensional setting, not every [[bounded operator]] on a [[Banach space]] has an invariant subspace. However, the upper-triangularization of an arbitrary square matrix does generalize to [[compact operator]]s. Every [[compact operator]] on a complex Banach space has a [[Flag (linear algebra)#Subspace nest|nest]] of closed invariant subspaces.
Edit summary
(Briefly describe your changes)
By publishing changes, you agree to the
Terms of Use
, and you irrevocably agree to release your contribution under the
CC BY-SA 4.0 License
and the
GFDL
. You agree that a hyperlink or URL is sufficient attribution under the Creative Commons license.
Cancel
Editing help
(opens in new window)