 Solution

Two square matrices are similar when $B=P^{-1}AP$ for some invertible matrix $P$. A Jordan normal form is a block-diagonal matrix whose blocks have one eigenvalue on the diagonal, ones on the superdiagonal, and zeros elsewhere; over an algebraically closed field every matrix is similar to such a form, unique up to reordering its Jordan blocks.

Both displayed matrices have characteristic polynomialHowever,so their eigenspace dimensions are respectively two and one. Thus $A$ has Jordan form $J_2(1)\oplus J_1(1)$, whereas $B$ has Jordan form $J_3(1)$. By uniqueness of Jordan normal form,