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

Jordan Form

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A square matrix $ A$ can be brought to block diagonal form by a similarity transformation

$\displaystyle J = \left(\begin{array}{ccc} J_1 & & 0 \\ & \ddots & \\ 0 & & J_k \end{array}\right) = Q^{-1} A Q\,. $

Here the Jordan blocks have the form

$\displaystyle J_i = \left(\begin{array}{ccccc} \lambda_i & 1 & & & 0 \\ 0 & \... ...ots & \\ & & & \lambda_i & 1 \\ 0 & & & & \lambda_i \end{array}\right) \,, $

where $ \lambda_i$ is an eigenvalue of $ A$.
(Authors: Burkhardt/Höllig/Hörner)
Since the generalised eigenspaces $ H_\lambda$ are invariant under mapping $ A$, we can reach the block form with respect to a basis of generalised eigenvectors. Using the special cyclic bases

$\displaystyle B^{k_i}v_i,\,\ldots,\,Bv_i,\,v_i,\quad B = A - \lambda_i E \,, $

we obtain blocks of the desired form.

(Authors: Burkhardt/Höllig/Hörner)
Let

$\displaystyle A=\left(\begin{array}{rrr}-1 & 0 & 4\\ 2 & -1 & 0\\ 3 & 2 & -1 \end{array}\right)\,. $

Similarity transformation by matrix

$\displaystyle Q=\left(\begin{array}{rrr}6 & -6 & 1\\ 3 & 6 & 2\\ 6 & 3 & -2 \end{array}\right) $

yields the Jordan form

$\displaystyle J=Q^{-1}AQ= \left(\begin{array}{rrr}3 & 0 & 0\\ 0 & -3 & 1\\ 0 & 0 & -3 \end{array}\right)\,. $

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