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Jordan normal form

{c}

If $\beta$ has eigenvalues $\lambda,\mu\ne0$ and is diagonalizable on a two-dimensional complex vector space, then
$$
\operatorname{JNF}(\phi_\beta)
=\operatorname{diag}\left(1,1,\frac\mu\lambda,\frac\lambda\mu\right).
$$
If $\beta$ has one size-two Jordan block, then
$$
\operatorname{JNF}(\phi_\beta)=J_3(1)\oplus J_1(1).
$$


Jordan normal form of conjugation on two-by-two matrices

Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 4 / 8F / Solution

Jordan normal form of conjugation on two-by-two matrices

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Jordan normal form of conjugation on two-by-two matrices

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