an example for Schur decomposition

Let

$A=\begin{pmatrix}5&7\\ -2&-4\end{pmatrix}.$

We will find an orthogonal matrix $P$ and an upper triangular matrix $T$ such that $P^{t}AP=T$ applying the proof of Schur’s decomposition. We ’re following the steps below

•

We find the eigenvalues of $A$
The eigenvalues of a matrix are precisely the solutions to the equation

$\det(\lambda I-A)=0\leftrightarrow\lambda^{2}-\lambda-6=0$

Hence the roots of the quadratic equation (http://planetmath.org/QuadraticFormula) are the eigenvalues $\lambda_{1}=-2,\lambda_{2}=3$

•

We find the eigenvectors
For each eigenvalue $\lambda_{i}$ , solving the system

$(A-\lambda_{i}I)X_{i}=0$

So we have that for $\lambda_{1}=-2$

$(A+2I)=0\leftrightarrow\begin{pmatrix}7&7\\ -2&-2\end{pmatrix}\begin{pmatrix}x_{1}\\ x_{2}\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix}\rightarrow X_{1}=(1,-1)$

Analogously for $\lambda_{2}=3$ the eigenvector $X_{2}=(7,-2)$

•

We get an orthonormal set of eigenvectors using Gram-Schmidt orthogonalization
Consider the above two eigenvectors which are linearly independent but are not orthogonal

	$\displaystyle X_{1}$	$\displaystyle=(1,-1)$
	$\displaystyle X_{2}$	$\displaystyle=(7,-2)$

First we take $w_{1}=X_{1}=(1,-1)$ . Therefore

$w_{2}=X_{2}-\frac{w_{1}\cdot X_{2}}{\|w_{1}\|^{2}}w_{1}$

that is,

$w_{2}=(\frac{5}{2},\frac{5}{2})$

and finally the orthonormal set is $\{w_{1}/\|w_{1}\|,w_{2}/\|w_{2}\|\}=\{(\frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}})% ,(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})\}$
So

$P=\frac{1}{\sqrt{2}}\begin{pmatrix}1&1\\ -1&1\end{pmatrix}.$

Then

$T=P^{t}AP=\begin{pmatrix}-2&9\\ 0&3\end{pmatrix}.$

Title	an example for Schur decomposition
Canonical name	AnExampleForSchurDecomposition
Date of creation	2013-03-22 15:27:02
Last modified on	2013-03-22 15:27:02
Owner	georgiosl (7242)
Last modified by	georgiosl (7242)
Numerical id	8
Author	georgiosl (7242)
Entry type	Application
Classification	msc 15-00
Related topic	SchurDecomposition
Related topic	GramSchmidtOrthogonalization