Generalized Eigenvectors and Jordan Chains

From Eigenvectors to Generalized Eigenvectors

An eigenvector records a direction that a matrix acts on by simple scaling: if \(\mathbf{v}\ne\mathbf{0}\) and \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\), then \(\mathbf{v}\) is an eigenvector for \(\lambda\). But a matrix need not have enough ordinary eigenvectors to describe all of its action. A related condition allows repeated application of \(\mathbf{A}-\lambda\mathbf{I}\): a nonzero vector may be sent to zero after several applications, even when it is not sent to zero after just one. Such vectors are generalized eigenvectors.

Unless stated otherwise, let \(\mathbf{A}\in\mathbb{F}^{n\times n}\), where \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\), and let \(\lambda\in\mathbb{F}\) be an eigenvalue of \(\mathbf{A}\). Write \(\mathbf{N}=\mathbf{A}-\lambda\mathbf{I}_n\). This shift subtracts the eigenvalue from the action of the matrix: ordinary eigenvectors for \(\lambda\) are precisely the nonzero vectors sent to \(\mathbf{0}\) by \(\mathbf{N}\).

The order is a smallest exponent, not just any exponent that works. For example, once a power sends a vector to zero, all higher powers do too. The exact order tells us how many applications are needed before the vector disappears, and this number determines the length of the chain that can be linked to it.

Reading the Powers in a Small Matrix

The example also shows why the eigenvalue must be specified. The same vector can be generalized for one eigenvalue but not another. The condition is not merely that some power of a matrix sends the vector to zero; it is that a power of the particular shifted matrix \(\mathbf{A}-\lambda\mathbf{I}_n\) does so.

Jordan Chains and Their Link Equations

A Jordan chain is an ordered list in which each vector is linked to the preceding one by the shifted matrix. The order matters: the first vector is an ordinary eigenvector, and each later vector is mapped to the one immediately before it. This convention makes the chain equations explicit and avoids confusing a chain with an unordered set of generalized eigenvectors.

In terms of \(\mathbf{N}=\mathbf{A}-\lambda\mathbf{I}_n\), the chain equations are \(\mathbf{N}\mathbf{v}_1=\mathbf{0}\), \(\mathbf{N}\mathbf{v}_2=\mathbf{v}_1\), continuing through \(\mathbf{N}\mathbf{v}_p=\mathbf{v}_{p-1}\). Repeated application gives \(\mathbf{N}^j\mathbf{v}_j=\mathbf{0}\), while \(\mathbf{N}^{j-1}\mathbf{v}_j=\mathbf{v}_1\ne\mathbf{0}\). Consequently, the last vector has exact order \(p\), and each vector \(\mathbf{v}_j\) has exact order \(j\).

Proof of the repeated-application claim. For \(j=1\), the equation \(\mathbf{N}\mathbf{v}_1=\mathbf{0}\) is the first chain condition. For \(j\ge2\), applying \(\mathbf{N}\) to \(\mathbf{v}_j\) gives \(\mathbf{v}_{j-1}\); applying it again gives \(\mathbf{v}_{j-2}\) when \(j\ge3\), and continuing gives \(\mathbf{N}^{j-1}\mathbf{v}_j=\mathbf{v}_1\). One further application gives \(\mathbf{N}^j\mathbf{v}_j=\mathbf{N}\mathbf{v}_1=\mathbf{0}\). Since \(\mathbf{v}_1\ne\mathbf{0}\), the previous power is nonzero, so the order is exactly \(j\). \(\square\)

This argument also runs in reverse: a generalized eigenvector of exact order \(p\) supplies a chain of length \(p\). It is useful because it gives a direct way to identify all the linked vectors, rather than checking only the final power condition.

Proof. Write \(\mathbf{N}=\mathbf{A}-\lambda\mathbf{I}_n\). Exact order \(p\) means \(\mathbf{N}^p\mathbf{v}=\mathbf{0}\) and \(\mathbf{N}^{p-1}\mathbf{v}\ne\mathbf{0}\). Thus \(\mathbf{v}_1=\mathbf{N}^{p-1}\mathbf{v}\ne\mathbf{0}\), and \(\mathbf{N}\mathbf{v}_1=\mathbf{N}^p\mathbf{v}=\mathbf{0}\). For each \(j=2,\dots,p\), the definition gives \(\mathbf{v}_j=\mathbf{N}^{p-j}\mathbf{v}\), so \(\mathbf{N}\mathbf{v}_j=\mathbf{N}^{p-j+1}\mathbf{v}=\mathbf{v}_{j-1}\). Finally, \(\mathbf{v}_p=\mathbf{N}^0\mathbf{v}=\mathbf{v}\). These are exactly the chain conditions. \(\square\)

Why the Vectors in a Chain Are Independent

A chain contains more than repeated applications of the same vector: its linked vectors provide distinct directions. In fact, every Jordan chain is linearly independent. This claim is important when chains are used to organize vectors into a basis, but its proof needs only the chain equations and the definition of linear independence.

Proof. Let \(\mathbf{N}=\mathbf{A}-\lambda\mathbf{I}_n\), and suppose \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p=\mathbf{0}\). If some coefficient is nonzero, let \(r\) be the largest index with \(c_r\ne0\). Apply \(\mathbf{N}^{r-1}\) to the equation. For every \(j<r\), the chain equations give \(\mathbf{N}^j\mathbf{v}_j=\mathbf{0}\), so \(\mathbf{N}^{r-1}\mathbf{v}_j=\mathbf{0}\). The \(r\)-th term becomes \(c_r\mathbf{N}^{r-1}\mathbf{v}_r=c_r\mathbf{v}_1\), by the chain equations. There are no terms with index greater than \(r\), because their coefficients are zero by the choice of \(r\). Therefore the resulting equation is \(c_r\mathbf{v}_1=\mathbf{0}\). Since \(\mathbf{v}_1\ne\mathbf{0}\), this forces \(c_r=0\), contradicting the choice of \(r\). No nonzero coefficient can exist, so all the coefficients are zero and the chain is independent. \(\square\)

Common Mistake: Requiring the First Power to Vanish

A common error is to treat the generalized-eigenvector condition as though it were the ordinary eigenvector condition. For instance, with \(\mathbf{A}=\begin{bmatrix}2&1\\0&2\end{bmatrix}\), \(\lambda=2\), and \(\mathbf{v}=\begin{bmatrix}0\\1\end{bmatrix}\), a mistaken calculation stops after finding \((\mathbf{A}-2\mathbf{I}_2)\mathbf{v}=\begin{bmatrix}1\\0\end{bmatrix}\ne\mathbf{0}\) and concludes that \(\mathbf{v}\) is not generalized. That conclusion uses the wrong test: it checks only the first power.

The correction is to continue applying the shifted matrix:

So \(\mathbf{v}\) is a generalized eigenvector of order \(2\), not an ordinary eigenvector. The linked chain is \(\left(\begin{bmatrix}1\\0\end{bmatrix},\begin{bmatrix}0\\1\end{bmatrix}\right)\): the shifted matrix sends the second vector to the first and the first to zero. The order of the vectors is part of the definition; reversing them would not satisfy these equations.

Generalized eigenvectors and their chains give a precise record of matrix action that ordinary eigenvectors alone may not capture. Their defining calculations are local: shift by the chosen eigenvalue, apply the shifted matrix, and verify each link. Later results use these chains to build canonical forms, but the chain equations themselves already explain the essential structure: an eigenvector at the start, followed by vectors that map one step back along the chain.