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

Powers of Matrices

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The matrix powers $ A^n$, $ n=0,1,\ldots$, converge to the zero matrix if and only if the absolute value of each eigenvalue $ \lambda$ is smaller than $ 1$.

The sequence $ (A^n)$ is bounded if $ \vert\lambda\vert\le 1$ and algebraic and geometric multiplicity are equal for eigenvalues of modulus $ 1$.

If there exists an eigenvalue of modulus greater than $ 1$, then the sequence diverges.

(Authors: Burkhardt/Höllig/Hörner)
By means of the Jordan form $ J = Q^{-1} A Q$ of $ A$ matrix powers can be written in the form

$\displaystyle A^n = (Q J Q^{-1}) (Q J Q^{-1}) \cdots (Q J Q^{-1}) = Q J^n Q^{-1}\,. $

Hence, it suffices to analyse the convergence of the powers of $ J$. Because of the block form of $ J$ we can consider each block $ J_i$ separately. We write

$\displaystyle J_i = (\lambda_i E) + D \,, $

where $ D$ contains the secondary diagonal. It can easily be proved that $ D^m=0$ for a block of dimension $ m$. Thus, we have

$\displaystyle (J_i)^n = \lambda_i^n E + \binom{n}{1} \lambda_i^{n-1} D + \cdots + \binom{n}{m-1} \lambda_i^{n-m+1} D^{m-1} \,. $

From this identity we can read off the properties of convergence.

For $ \vert\lambda_i\vert<1$ we have

$\displaystyle \lim_{n\to\infty} \binom{n}{j} \lambda_i^{n-j} = 0\,, $

since $ \binom{n}{j}$ increases only polynomially, whereas $ \lambda_i^{n-j}$ decreases exponentially.

If $ \vert\lambda_i\vert=1$, then the sequence is bounded if and only if $ m=1$, that is, if there is no secondary diagonal with ones.

If $ \vert\lambda_i\vert>1$, then we have for the corresponding eigenvector $ v_i$

$\displaystyle \vert J^n v_i\vert = \vert\lambda_i\vert^n \vert v_i\vert \to \infty $

for $ n\to\infty$.

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