Here is an example with two Jordan blocks that have the same eigenvalue. If we had only a single Jordan block, the dimension resulting would be exactly $4,$ but this comes out a little bigger.
$$ \left( \begin{array}{rrrr} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array} \right) \left( \begin{array}{rrrr} a & b & c & d \\ e & f & g & h \\ i & j & k & l \\ m & n & o & p \end{array} \right) = \left( \begin{array}{rrrr} e & f & g & h \\ 0 & 0 & 0 & 0 \\ m & n & o & p \\ 0 & 0 & 0 & 0 \end{array} \right) $$$$ \left( \begin{array}{rr|rr} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \hline 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array} \right) \left( \begin{array}{rrrr} a & b & c & d \\ e & f & g & h \\ i & j & k & l \\ m & n & o & p \end{array} \right) = \left( \begin{array}{rrrr} e & f & g & h \\ 0 & 0 & 0 & 0 \\ m & n & o & p \\ 0 & 0 & 0 & 0 \end{array} \right) $$
$$ \left( \begin{array}{rrrr} a & b & c & d \\ e & f & g & h \\ i & j & k & l \\ m & n & o & p \end{array} \right) \left( \begin{array}{rrrr} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array} \right) = \left( \begin{array}{rrrr} 0 & a & 0 & c \\ 0 & e & 0 & g \\ 0 & i & 0 & k \\ 0 & m & 0 & l \end{array} \right) $$$$ \left( \begin{array}{rrrr} a & b & c & d \\ e & f & g & h \\ i & j & k & l \\ m & n & o & p \end{array} \right) \left( \begin{array}{rr|rr} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \hline 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array} \right) = \left( \begin{array}{rrrr} 0 & a & 0 & c \\ 0 & e & 0 & g \\ 0 & i & 0 & k \\ 0 & m & 0 & o \end{array} \right) $$
At first glance, I get nineeight linear equations required for equality, $$e,g,l,m,o = 0, \; a=f, c=h, i=n,k=p. $$$$e,g,m,o = 0, \; a=f, c=h, i=n,k=p. $$
Out of $16,$ what is left is dimension $7.$$8.$
$$ \left( \begin{array}{rr|rr} a & b & c & d \\ 0 & a & 0 & c \\ \hline i & j & k & l \\ 0 & i & 0 & k \end{array} \right) $$ Dimension $8$ for four little 2 by 2 Jordan blocks scattered about.