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Mathematics-Online course: Linear Algebra - Normal Forms - Jordan Normal Form

Triangular Form

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A square matrix $ A$ can be brought to upper triangle form by similarity transformations

$\displaystyle R = \left(\begin{array}{cccc} \lambda_1 & r_{1,2} & \cdots & r_{... ..._{n-1,n} \\ 0 & \cdots & 0 & \lambda_n \end{array}\right) = Q^{-1} A Q\, , $

where the diagonal entries are the eigenvalues $ \lambda_i$ of $ A$.
(Authors: Burkhardt/Höllig/Hörner)
Starting with the trivial case of a $ (1\times1)$-matrix, we prove the assumption by induction on the dimension $ n$ of the matrix.

An $ (n\times n)$-matrix $ A$ has at least one eigenvalue and, hence, $ A$ has an eigenvector $ v$. Now, complete this vector to a basis $ \{v, w_1,\ldots ,w_{n-1}\}$. With respect to this basis mapping $ A$ has the representation

$\displaystyle \tilde{A} = T^{-1}AT= \left(\begin{array}{cc} \lambda & x\\ 0 & B\\ \end{array}\right)\,. $

$ B$ is an $ \big((n-1)\times (n-1)\big)$-matrix. By induction hypothesis there exists a transformation $ \tilde{Q}$ with $ \tilde{Q}^{-1}B\tilde{Q}=R_{n-1}$. With

$\displaystyle Q=T\left(\begin{array}{cc} 1 & 0\\ 0 & \tilde{Q} \end{array}\right) $

it follows that
$\displaystyle Q^{-1}AQ$ $\displaystyle =$ $\displaystyle \left(\begin{array}{cc} 1 & 0\\ 0 & \tilde{Q}^{-1} \end{array}\right) T^{-1} A T \left(\begin{array}{cc} 1 & 0\\ 0 & \tilde{Q} \end{array}\right)$  
  $\displaystyle =$ $\displaystyle \left(\begin{array}{cc} 1 & 0\\ 0 & \tilde{Q}^{-1} \end{array}\r... ...array}\right) \left(\begin{array}{cc} 1 & 0\\ 0 & \tilde{Q} \end{array}\right)$  
  $\displaystyle =$ $\displaystyle \left(\begin{array}{cc} 1 & 0\\ 0 & \tilde{Q}^{-1} \end{array}\r... ...(\begin{array}{cc} \lambda & x\tilde{Q}\\ 0 & B\tilde{Q}\\ \end{array}\right)$  
  $\displaystyle =$ $\displaystyle \left(\begin{array}{cc} \lambda & x\tilde{Q}\\ 0 & \tilde{Q}^{-1... ...egin{array}{cc} \lambda & x\tilde{Q}\\ 0 & R_{n-1}\\ \end{array}\right)= R\,.$  

(Authors: Burkhardt/Höllig/Hörner)
  automatically generated 4/21/2005