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An invertible complex Jordan block $\lambda I+N$, with $N^s=0$, has logarithm $\ell I+{\sum }_{k=1}^{s-1}(-1{)}^{k+1}(N/\lambda {)}^k/k$, where $e^{\ell }=\lambda$. Applying this construction blockwise and conjugating back proves surjectivity of the matrix exponential on the complex general linear group. There is no globally single-valued continuous choice of logarithm.

For a continuous periodic coefficient matrix $A(t+L)=A(t)$, a fundamental matrix $X$ for $X^{\prime }=A(t)X$, normalized by $X(0)=I$, obeys $X(t+L)=X(t)X(L)$. The invertible matrix $X(L)$ has a complex matrix logarithm; set $B=L^{-1}\log X(L)$ and $P(t)=X(t)e^{-tB}$. Since $X(L)=e^{LB}$ commutes with $B$, direct substitution gives $P(t+L)=P(t)$ and proves the factorization. A real logarithm is not required for this complex representation.

The eigenvalues of $X(L)$ are Floquet multipliers. All solutions are bounded for forward time when every multiplier has modulus at most one and those on the unit circle have no nontrivial Jordan blocks. In a real undamped second-order scalar equation with no first-derivative term, the monodromy determinant is one. Distinct conjugate unit-modulus multipliers give bounded solutions; a nontrivial Jordan block at multiplier $1$ or $-1$ gives an unbounded second solution even when one periodic or antiperiodic solution exists.

A Jordan chain of a linear operator $T$ for an eigenvalue $\lambda$ is a list of nonzero vectors satisfying $(T-\lambda I)v_1=0$ and $(T-\lambda I)v_j=v_{j-1}$ for $2\le j\le r$. These vectors are linearly independent: in a vanishing linear combination, applying the largest relevant power of $T-\lambda I$ isolates the last coefficient times $v_1$, then descending induction removes all coefficients. On their span the operator has one Jordan block. This is different from a generalized eigenfunction that solves $(T-\lambda I)u=0$ outside the original function space.

If $(M-\lambda I)e=0$ and $(M-\lambda I)v=e$, the vector $v$ is a generalized eigenvector and $e^{\lambda t}(v+te)$ solves $\dot{x}=Mx$. The polynomial factor distinguishes a nontrivial Jordan block from an ordinary repeated eigenvalue with a full eigenvector basis.

On polynomials of degree at most one, let the generator of $\mathbb{Z}$ act by $p(x)\mapsto p(x+1)$. In the basis $(1,x)$ its matrix isThis nontrivial Jordan block is not diagonalizable, so the representation is not a direct sum of one-dimensional subrepresentations.