The Jordan Form Existence Theorem

The Existence Question

A Jordan form describes a linear operator using blocks that record its eigenvalues and the way generalized eigenvectors are linked into chains. The existence theorem says that over the complex numbers, every square matrix has such a description in a suitable basis. Its proof has two parts: primary decomposition separates the space according to eigenvalues, and the chain-basis theorem provides a basis for the nilpotent part on each resulting subspace.

All vector spaces and matrices in this tutorial are over \(\mathbb{C}\). Let \(\mathbf{A}\in\mathbb{C}^{n\times n}\), with \(n\geq 1\), and regard it as the matrix of a linear operator \(T:\mathbb{C}^n\to\mathbb{C}^n\) in the standard basis. Two matrices \(\mathbf{A}\) and \(\mathbf{J}\) are similar if \(\mathbf{J}=\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\) for some invertible matrix \(\mathbf{P}\). This means that they represent the same operator in different bases.

For example, the convention for a block of size three is

The entries above the diagonal reflect the chain convention used here: a chain \(\mathbf{v}_1,\dots,\mathbf{v}_k\) for \(T\) at \(\lambda\) satisfies \(T\mathbf{v}_1=\lambda\mathbf{v}_1\) and \(T\mathbf{v}_j=\lambda\mathbf{v}_j+\mathbf{v}_{j-1}\) for \(2\leq j\leq k\). The chain-basis theorem for nilpotent operators, established earlier in the material on Jordan chains, says that a nilpotent operator on a finite-dimensional vector space has a basis made of chains with this shift relation for the nilpotent operator. We use that theorem below.

From a Chain to a Block

Proof. Each \(T\mathbf{v}_j\) is a linear combination of vectors from the list, so \(T\) maps their span into itself. Since the list is linearly independent, it is a basis of its span, and each displayed chain equation gives the coordinates of \(T\mathbf{v}_j\) in this basis. The first coordinate column is \((\lambda,0,\dots,0)^{\top}\). For \(j\geq2\), column \(j\) has \(\lambda\) in row \(j\), \(1\) in row \(j-1\), and \(0\) elsewhere. These are exactly the columns of \(\mathbf{J}_k(\lambda)\). \(\square\)

The independence hypothesis is essential: chain equations alone do not establish that a list can serve as a basis. In the existence proof, the chain-basis theorem supplies chains that together form a basis, so the lemma applies to each chain separately.

Why Primary Decomposition Produces the Right Pieces

Over \(\mathbb{C}\), the characteristic polynomial splits into linear factors. Write its distinct roots as \(\lambda_1,\dots,\lambda_s\), with algebraic multiplicities \(a_1,\dots,a_s\). The primary decomposition theorem gives the direct sum of generalized eigenspaces

Here a direct sum means that every vector has a unique expression as a sum of vectors from the listed subspaces. This is the primary decomposition result established earlier. On the \(i\)-th subspace, the operator \(N_i=T-\lambda_i\operatorname{id}\) is nilpotent: by definition, every vector in that subspace is killed by \((T-\lambda_i\operatorname{id})^{a_i}\). Each subspace is \(T\)-invariant because \(T\) commutes with \(T-\lambda_i\operatorname{id}\), and therefore also commutes with its powers.

We can now state the main result. A direct sum of matrices means the block-diagonal matrix with those matrices as its diagonal blocks.

Proof. Let \(T\) be the operator represented by \(\mathbf{A}\), and use the primary decomposition above. Fix one of its summands \(V_i=\ker(T-\lambda_i\operatorname{id})^{a_i}\). On \(V_i\), the operator \(N_i=T-\lambda_i\operatorname{id}\) is nilpotent. The chain-basis theorem for nilpotent operators applies because \(V_i\) is finite-dimensional; it gives a basis of \(V_i\) consisting of chains \(\mathbf{v}_1,\dots,\mathbf{v}_k\) satisfying \(N_i\mathbf{v}_1=\mathbf{0}\) and \(N_i\mathbf{v}_j=\mathbf{v}_{j-1}\) for \(2\leq j\leq k\). Since \(T=N_i+\lambda_i\operatorname{id}\) on this subspace, these equations become \(T\mathbf{v}_1=\lambda_i\mathbf{v}_1\) and \(T\mathbf{v}_j=\lambda_i\mathbf{v}_j+\mathbf{v}_{j-1}\). Each chain is linearly independent because it is part of a basis. By the lemma, the matrix on the span of each chain is a Jordan block with eigenvalue \(\lambda_i\). Concatenate the chain bases for all the summands \(V_i\). Primary decomposition ensures that the resulting list is a basis of \(\mathbb{C}^n\). In that basis, each summand is invariant and each chain span has the block matrix just described, so the full matrix is block diagonal with Jordan blocks on its diagonal. Let \(\mathbf{P}\) have this ordered basis as its columns in standard coordinates. It is invertible, and its change-of-basis matrix is \(\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\). Every block eigenvalue is one of the \(\lambda_i\), hence a root of the characteristic polynomial. \(\square\)

This is an existence result: it guarantees at least one Jordan form and a basis producing it. It does not yet establish that the sizes of the blocks are uniquely determined. That is a separate question.

Worked Examples of the Basis Construction

Common Pitfall: A Chain Is Not Automatically a Basis

A tempting but invalid argument is to check the chain equations and then announce that the vectors give a Jordan block. For a concrete failure, take the \(1\times1\) zero matrix \(\mathbf{A}=[0]\), choose \(\lambda=0\), and set \(\mathbf{v}_1=\mathbf{0}\). The equation \(\mathbf{A}\mathbf{v}_1=\lambda\mathbf{v}_1\) is true, but \((\mathbf{v}_1)\) is not a basis: it is linearly dependent and does not span \(\mathbb{C}\). Therefore it cannot be used to form a change-of-basis matrix, and the chain-to-block lemma does not apply. The correction is to verify independence when using an individual chain, or, in the existence proof, to rely on the chain-basis theorem, which supplies a basis made of chains.

There is a second useful check on the argument: the direct sum in primary decomposition is what lets the separate chain bases combine into one basis of the whole space. Bases chosen independently inside generalized eigenspaces would not by themselves imply this. The direct-sum property guarantees both that the combined list spans \(\mathbb{C}^n\) and that it is independent.