Let \(k\) be a field and \(n\geq 1\). Matrices \(A,B\in\Mat_{n\times n}(k)\) are similar if there exists \(P\in\GL_n(k)\) such that \(A = PBP^{-1}\).
Let \(k\) be an algebraically closed field and \(A,B\in\Mat_{n\times n}(k)\). Then \(A\) and \(B\) are similar if and only if they have the same Jordan canonical form, up to the order of the Jordan blocks.
Every matrix over \(k\) is similar to a Jordan canonical form, and similarity is an equivalence relation, so matrices with the same Jordan form are similar. Conversely, for \(\lambda\in k\) and \(j\geq 1\), the number of Jordan blocks of \(A\) with eigenvalue \(\lambda\) and size at least \(j\) is \(\dim\ker(A-\lambda I)^j-\dim\ker(A-\lambda I)^{j-1}\). If \(A=PBP^{-1}\), then \((A-\lambda I)^j=P(B-\lambda I)^jP^{-1}\), so these dimensions agree for \(A\) and \(B\), and hence so do their Jordan blocks.