Jordan normal form

Statement

Let $F$ be an algebraically closed field, for example $\C$, and let $A$ be an $n \times n$ matrix over $F$. Then $A$ is similar to a block diagonal matrix $$J = \begin{pmatrix} J_{k_1}(\lambda_1) & & \\ & \ddots & \\ & & J_{k_r}(\lambda_r) \end{pmatrix},$$ whose diagonal blocks are Jordan blocks. The matrix $J$ is unique up to the order of the blocks. It is the Jordan normal form of $A$.

Remarks

The $\lambda_i$ are the eigenvalues of $A$; they need not be distinct. The number of blocks for $\lambda$ is its geometric multiplicity, and the sum of their sizes is its algebraic multiplicity. The largest block for $\lambda$ has size equal to the multiplicity of $\lambda$ in the minimal polynomial. The theorem holds over any field $F$ for the matrices whose characteristic polynomial splits over $F$.

Explanations

L1 human

Every complex square matrix becomes almost diagonal in a good basis. The only entries off the diagonal are some $1$s just above it, inside the Jordan blocks. Two complex matrices are similar exactly when they have the same Jordan form, up to the order of the blocks.