## Jordan Elementary Form

Suppose we have a square matrix
$A$
with eigenvalues
$\lambda_1 , \lambda_2 ,..., \lambda_k$
which are the solutions to
$det(A- \lambda I)=0$
.
Each eigenvalue gives rise to an 'Elementary Jordan Matrix', with main diagonal consisting of one of the eigenvalues, and the diagonal below all 1s. All other entries are the zero.
Example:
$\left( \begin{array}{ccccc} 3 & 0 & 0 & 0 & 0 \\ 1 & 3 & 0 & 0 & 0 \\ 0 & 1 & 3 & 0 & 0 \\ 0 & 0 & 1 & 3 & 0 \\ 0 & 0 & 0 & 1 & 3 \end{array} \right)$
is a Jordan elementary matrix, but
$\left( \begin{array}{ccccc} 3 & 0 & 0 & 0 & 0 \\ 1 & 3 & 0 & 0 & 0 \\ 0 & 1 & 3 & 0 & 0 \\ 0 & 0 & 1 & 3 & 5 \\ 0 & 0 & 0 & 1 & 3 \end{array} \right), \: \left( \begin{array}{ccccc} 3 & 0 & 0 & 0 & 0 \\ 1 & 3 & 0 & 0 & 0 \\ 0 & 1 & 3 & 0 & 0 \\ 0 & 0 & 1 & 6 & 0 \\ 0 & 0 & 0 & 1 & 3 \end{array} \right)$
are not.