The matrix
J=
\begin{pmatrix}
A11 & \ast & A13 \\
A21 & A22 & \otimes \\
A31 & A32 & A33
\end{pmatrix}
is the Jordan normal form of a matrix A which is not diagonalizable.
Determine the entries \ast und \otimes.
\ast
=
0
1
\otimes
=
1
0
Since the matrix A is not diagonalizable, we cannot have
\ast= 0 = \otimes.
The diagonal entries L1 and (double) L2 are the eigenvalues
of A.
The diagonal entries L2 and (double) L1 are the eigenvalues
of A.
Due to the non-diagonalizability, the double eigenvalue L2 has a Jordan block of length 2.
Due to the non-diagonalizability, the double eigenvalue L1 has a Jordan block of length 2.
Therefore, it follows that \ast = 0 and \otimes = 1.
Therefore, \ast = 1 and \otimes = 0.