Jordan normal form

Perturb each diagonal position of a Jordan normal form matrix by a different arbitrarily small scalar, avoiding the finitely many collisions between diagonal entries. The resulting upper triangular matrix has distinct eigenvalues, hence is diagonalizable. Conjugate back to approximate any complex matrix. Continuous identities proved for diagonalizable matrices can therefore extend to all matrices.

A complex two-dimensional linear operator whose two eigenvalues both equal $a$ is similar either to $aI$ or to $\left(\begin{array}{cc}a& 1\\ 0& a\end{array}\right)$. Choose an eigenvector and complete it to a basis. The matrix becomes $\left(\begin{array}{cc}a& c\\ 0& a\end{array}\right)$ because its characteristic polynomial is $(t-a{)}^2$. A nonzero $c$ can be normalized to one by rescaling the first basis vector. The two types are distinguished by eigenspace dimension.

Over an algebraically closed field, a linear operator $X$ has a unique decomposition into a commuting diagonalisable endomorphism $X_s$ and nilpotent endomorphism $X_n$. On its generalized eigenspace for $\lambda$, set $X_s=\lambda I$ and $X_n=X-\lambda I$. The Chinese remainder theorem makes both parts polynomials in $X$, so they preserve every $X$-invariant subspace. Uniqueness follows by restricting any other commuting decomposition to those generalized eigenspaces. Over a perfect field the semisimple part need only become diagonalizable after scalar extension.

The commutator action of $X_s$ is diagonalizable on endomorphisms: on $\mathrm{Hom}(V_{\lambda },V_{\mu })$ it acts by $\mu -\lambda$. The action of $X_n$ is nilpotent by nilpotence of commutation by a nilpotent endomorphism. The two actions commute, so uniqueness of the Additive Jordan decomposition identifies them as the parts of $\mathrm{ad}X$.

For a complex semisimple Lie algebra $\mathfrak{g}\subseteq \mathrm{End}(V)$ and $X\in \mathfrak{g}$, both parts of its Additive Jordan decomposition lie in $\mathfrak{g}$. The polynomial semisimple part of $\mathrm{ad}X$ shows $X_s$ normalizes $\mathfrak{g}$. Split $\mathrm{End}(V)=\mathfrak{g}\oplus M$ by the Weyl complete reducibility theorem; the $M$-component of $X_s$ centralizes $\mathfrak{g}$. On each irreducible summand of $V$ it is scalar by the Schur lemma. Its trace is zero because $\mathfrak{g}$ is a perfect Lie algebra and the nilpotent part has zero trace, so those scalars vanish. Nonsemisimple subalgebras may fail this property: $\mathbb{C}(I+N)$ with $N\ne 0$ nilpotent contains neither Jordan part of $I+N$.

The algebraic multiplicity of an eigenvalue is its multiplicity as a root of the characteristic polynomial. In Jordan form it is the sum of the sizes of all blocks carrying that eigenvalue.

The geometric multiplicity of an eigenvalue is the dimension of its eigenspace. In Jordan form it equals the number of blocks carrying that eigenvalue.

A Jordan block has one eigenvalue on its diagonal and ones on its superdiagonal.

The model matrix $B=\left(\begin{array}{cc}0& 1\\ ϵ& d\end{array}\right)$ has a defective double eigenvalue when $d=ϵ=0$. Its characteristic polynomial is ${\lambda }^2-d\lambda -ϵ$, giving the displayed roots. For fixed nonzero $d$ the small root is $-ϵ/d+O({ϵ}^2)$ and the other is $d+ϵ/d+O({ϵ}^2)$. These expansions cease to be ordered at $d=O(\sqrt{ϵ})$, the distinguished limit in which both roots split on a square-root scale. This is the generic leading behavior when a perturbation closes a two-dimensional Jordan block through a nonzero lower-left entry; it need not occur for a perturbation that leaves that entry zero.

A nilpotent Jordan block has zeros on its diagonal and ones on its superdiagonal. Its $k$th power has ones on the $k$th superdiagonal, and a block of size $d$ satisfies $J_d(0{)}^d=0$.

If $\beta$ has eigenvalues $\lambda ,\mu \ne 0$ and is diagonalizable on a two-dimensional complex vector space, thenIf $\beta$ has one size-two Jordan block, then

Over an algebraically closed field, the characteristic polynomial gives the total size of the Jordan blocks for each eigenvalue, while the minimal polynomial gives the largest block size. For total dimension at most three, these two numbers determine the partition into block sizes: a multiplicity-three eigenvalue has partition $3$, $2+1$, or $1+1+1$ according as its largest block has size $3$, $2$, or $1$. Hence two matrices of size at most three with the same characteristic and minimal polynomials are similar.

A generalized eigenvector lies in the kernel of a positive power of A minus lambda I.

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.

The generalized eigenspace of a linear operator $A$ for an eigenvalue $\lambda$ is the union of the kernels $\ker (A-\lambda I{)}^N$. In finite dimension this union stabilizes for sufficiently large $N$.

For endomorphisms of a finite-dimensional complex vector space, suppose repeated commutation with $T$ annihilates $S$. Write the blocks of $S$ using the generalized eigenspaces of $T$. On $\mathrm{Hom}(V_{\lambda },V_{\mu })$, the operator $\mathrm{ad}T$ is $(\mu -\lambda )I$ plus a nilpotent endomorphism. If $\lambda \ne \mu$, it is invertible, so $(\mathrm{ad}T{)}^{N}S=0$ forces that off-diagonal block of $S$ to vanish. Thus $S$ preserves each generalized eigenspace. This allows a nilpotent Lie algebra representation to be split by generalized eigenvalues even when its acting operators do not commute.

For a linear operator $f$, the generalized eigenspacesbelonging to distinct eigenvalues are linearly independent. On the generalized $\alpha$-eigenspace, $f-\beta I=(\alpha -\beta )I+N$ is invertible for $\beta \ne \alpha$ because the nilpotent part $N$ has a finite geometric-series inverse.