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Generalized Eigenvectors and Jordan Form · Tutorial 845 of 1000

Common Errors in Jordan Chain Calculations

Use the chain equations and small matrix checks to diagnose common errors in Jordan form calculations.

Advanced 9 min read

What You'll Learn

  • Write the defining equations for a Jordan chain in the correct order
  • Check which chain vectors are eigenvectors and which are generalized eigenvectors
  • Relate an ordered chain basis to the superdiagonal Jordan block convention
  • Interpret geometric multiplicity as the number of Jordan blocks for an eigenvalue
  • Identify what geometric multiplicity does not determine about block sizes

Start with the Chain Equations

Jordan-chain calculations are especially vulnerable to errors that look plausible in a matrix: reversing a chain can reverse the off-diagonal pattern, a generalized eigenvector can be mistaken for an eigenvector, and a correct geometric multiplicity can be used to draw an unjustified conclusion about block sizes. The safest check is to return to the defining equations and test each vector directly.

Let \(T:V\to V\) be a linear map on a finite-dimensional vector space over \(\mathbb{C}\), and let \(\lambda\) be an eigenvalue. (The same discussion applies over \(\mathbb{R}\) when the eigenvalue and vectors lie in \(\mathbb{R}\).) A Jordan chain of length \(k\) for \(\lambda\), in the convention used here, is an ordered list \((\mathbf{v}_1,\dots,\mathbf{v}_k)\) satisfying

$$ (T-\lambda\operatorname{id}_V)\mathbf{v}_1=\mathbf{0}, \qquad (T-\lambda\operatorname{id}_V)\mathbf{v}_j=\mathbf{v}_{j-1} \quad (2\le j\le k), $$

with \(\mathbf{v}_1\ne\mathbf{0}\). The first vector is an eigenvector. Each later vector is a generalized eigenvector: repeated application of \(T-\lambda\operatorname{id}_V\) eventually sends it to \(\mathbf{0}\). In particular, the equations imply \((T-\lambda\operatorname{id}_V)^j\mathbf{v}_j=\mathbf{0}\), while \((T-\lambda\operatorname{id}_V)^{j-1}\mathbf{v}_j=\mathbf{v}_1\ne\mathbf{0}\).

Definition: A Jordan chain is ordered from the eigenvector at its head to the most generalized vector at its end: applying \(T-\lambda\operatorname{id}_V\) to each vector after the first gives the preceding vector. In this convention, the chain basis is \((\mathbf{v}_1,\dots,\mathbf{v}_k)\).

These equations are a useful diagnostic in their own right. If a proposed chain does not satisfy the recurrence for every index, it is not a chain in this order. If the recurrence holds, only the first vector is required to be an eigenvector; later vectors usually are not.

Why the Standard Jordan Block Has a Superdiagonal

The Jordan block convention places ones immediately above the diagonal. This pattern follows from the chain equations and the order of the basis; it is not a separate rule to memorize.

Theorem: Suppose \((\mathbf{v}_1,\dots,\mathbf{v}_k)\) is a linearly independent Jordan chain for \(T\) and \(\lambda\). The matrix of \(T\) restricted to the span of the chain, in the ordered basis \((\mathbf{v}_1,\dots,\mathbf{v}_k)\), is the Jordan block with \(\lambda\) on the diagonal and \(1\) on the superdiagonal.

Proof. Write the matrix of \(T\) in the stated basis by expressing the image of each basis vector as a column of coordinates. The chain equations give \(T\mathbf{v}_1=\lambda\mathbf{v}_1\). For each \(j\) with \(2\le j\le k\), they give \(T\mathbf{v}_j=\lambda\mathbf{v}_j+\mathbf{v}_{j-1}\). Therefore column \(1\) has \(\lambda\) in its first entry and zeros elsewhere. Column \(j\) has \(\lambda\) in row \(j\), \(1\) in row \(j-1\), and zeros elsewhere. Those entries are exactly \(\lambda\) on the diagonal and \(1\) immediately above it. \(\square\)

Worked Example: Checking the Chain and Its Block

Consider

$$ \mathbf{A}= \begin{bmatrix} 2&1&0\\ 0&2&1\\ 0&0&2 \end{bmatrix}, \qquad \mathbf{N}=\mathbf{A}-2\mathbf{I}_3= \begin{bmatrix} 0&1&0\\ 0&0&1\\ 0&0&0 \end{bmatrix}. $$

Let \(\mathbf{e}_1=(1,0,0)^{\top}\), \(\mathbf{e}_2=(0,1,0)^{\top}\), and \(\mathbf{e}_3=(0,0,1)^{\top}\). Multiplying by \(\mathbf{N}\) gives

$$ \mathbf{N}\mathbf{e}_1=\mathbf{0},\qquad \mathbf{N}\mathbf{e}_2=\mathbf{e}_1,\qquad \mathbf{N}\mathbf{e}_3=\mathbf{e}_2. $$

Thus \((\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3)\) is a chain for \(\lambda=2\). To check its matrix representation, compute the images under \(\mathbf{A}\):

$$ \mathbf{A}\mathbf{e}_1=2\mathbf{e}_1,\qquad \mathbf{A}\mathbf{e}_2=\mathbf{e}_1+2\mathbf{e}_2,\qquad \mathbf{A}\mathbf{e}_3=\mathbf{e}_2+2\mathbf{e}_3. $$

The coordinates of these images are the columns of the matrix in the chain basis, giving the displayed superdiagonal pattern. As a direct check, multiplying \(\mathbf{A}\) by the three standard basis columns returns \((2,0,0)^{\top}\), \((1,2,0)^{\top}\), and \((0,1,2)^{\top}\), respectively, as required.

Reversing the basis order changes the displayed pattern. In the same example, use \((\mathbf{e}_3,\mathbf{e}_2,\mathbf{e}_1)\). Then \(\mathbf{A}\mathbf{e}_3=2\mathbf{e}_3+\mathbf{e}_2\), \(\mathbf{A}\mathbf{e}_2=2\mathbf{e}_2+\mathbf{e}_1\), and \(\mathbf{A}\mathbf{e}_1=2\mathbf{e}_1\). Their coordinates in the reversed basis form

$$ \begin{bmatrix} 2&0&0\\ 1&2&0\\ 0&1&2 \end{bmatrix}. $$

This is a lower-triangular pattern, not the standard superdiagonal Jordan block under the stated convention. The underlying map has not changed; only the basis order has. The reversed list is not ordered as a chain under the defining recurrence above. A common error is to reverse the vectors but still label the resulting matrix the standard Jordan block without checking which convention is being used.

Generalized Eigenvectors Are Not Automatically Eigenvectors

The defining equation for a generalized eigenvector is weaker than the eigenvector equation. For \(\mathbf{v}_2\) in a chain, \((T-\lambda\operatorname{id}_V)\mathbf{v}_2=\mathbf{v}_1\), and the right-hand side is nonzero. Thus \(\mathbf{v}_2\) is not an eigenvector. More generally, the chain equations guarantee that the first vector is an eigenvector; they do not guarantee that later vectors are.

Worked Example: Testing a Generalized Eigenvector

Let

$$ \mathbf{A}= \begin{bmatrix} 3&1\\ 0&3 \end{bmatrix}, \qquad \mathbf{N}=\mathbf{A}-3\mathbf{I}_2= \begin{bmatrix} 0&1\\ 0&0 \end{bmatrix}, \qquad \mathbf{v}_1= \begin{bmatrix}1\\0\end{bmatrix}. $$

To find a vector \(\mathbf{x}=(a,b)^{\top}\) with \(\mathbf{N}\mathbf{x}=\mathbf{v}_1\), compute

$$ \mathbf{N}\mathbf{x} = \begin{bmatrix}0&1\\0&0\end{bmatrix} \begin{bmatrix}a\\b\end{bmatrix} = \begin{bmatrix}b\\0\end{bmatrix} = \begin{bmatrix}1\\0\end{bmatrix}. $$

Equating entries gives \(b=1\); \(a\) is free. Choose \(a=0\), so \(\mathbf{v}_2=(0,1)^{\top}\). Then \(\mathbf{N}\mathbf{v}_2=\mathbf{v}_1\), as required for the second chain vector. But \(\mathbf{N}\mathbf{v}_2\ne\mathbf{0}\), so \(\mathbf{v}_2\) is not an eigenvector for \(3\). Equivalently,

$$ \mathbf{A}\mathbf{v}_2= \begin{bmatrix}1\\3\end{bmatrix} \ne \begin{bmatrix}0\\3\end{bmatrix} =3\mathbf{v}_2. $$

For a further check, \(\mathbf{N}\mathbf{v}_1=\mathbf{0}\), so \((\mathbf{v}_1,\mathbf{v}_2)\) is indeed a chain. The distinction is exactly that \(\mathbf{v}_1\) is an eigenvector and \(\mathbf{v}_2\) is a generalized eigenvector.

Common wrong calculation: from \((\mathbf{A}-3\mathbf{I}_2)\mathbf{v}_2=\mathbf{v}_1\), one might incorrectly conclude that \(\mathbf{A}\mathbf{v}_2=3\mathbf{v}_2\). Adding \(3\mathbf{v}_2\) to both sides instead gives \(\mathbf{A}\mathbf{v}_2=3\mathbf{v}_2+\mathbf{v}_1\). Since \(\mathbf{v}_1\ne\mathbf{0}\), this is not the eigenvector equation.

What Geometric Multiplicity Does—and Does Not—Tell You

The geometric multiplicity of an eigenvalue \(\lambda\) is the dimension of its eigenspace \(E_{\lambda}=\ker(T-\lambda\operatorname{id}_V)\). For Jordan form, it has a precise interpretation: it counts the Jordan blocks for \(\lambda\), not the total number of vectors in those blocks and not the size of the largest block.

Theorem: If a linear map on a finite-dimensional complex vector space is written in Jordan form, then the geometric multiplicity of \(\lambda\) equals the number of Jordan blocks associated with \(\lambda\).

Proof. In a Jordan basis, the space is a direct sum of the subspaces belonging to the Jordan blocks. On a block of size \(k\) for \(\lambda\), subtracting \(\lambda\) times the identity leaves a matrix with ones on the superdiagonal and zeros elsewhere. If a vector in that block has coordinates \((x_1,\dots,x_k)^{\top}\), the equation for belonging to the kernel is \(x_2=0,\dots,x_k=0\); \(x_1\) is free. Thus that block contributes exactly one independent eigenvector, the first vector of its chain. Blocks for other eigenvalues contribute no vectors to this eigenspace: their block matrices after subtracting \(\lambda\) have nonzero diagonal \(\mu-\lambda\) and are invertible, since their triangular systems have a unique zero solution. The full eigenspace is the direct sum of the one-dimensional contributions from the \(\lambda\)-blocks, so its dimension is the number of those blocks. \(\square\)

Consequently, knowing the geometric multiplicity gives the number of chains, but not their lengths. The total dimension of the generalized eigenspace gives the sum of the lengths, and several different lists of lengths can have the same number of chains and the same sum.

Worked Example: Same Geometric Multiplicity, Different Block Sizes

Compare the following two matrices, each with eigenvalue \(4\):

$$ \mathbf{A}= \begin{bmatrix} 4&1&0&0\\ 0&4&1&0\\ 0&0&4&0\\ 0&0&0&4 \end{bmatrix}, \qquad \mathbf{B}= \begin{bmatrix} 4&1&0&0\\ 0&4&0&0\\ 0&0&4&1\\ 0&0&0&4 \end{bmatrix}. $$

For \(\mathbf{A}\), subtract \(4\mathbf{I}_4\). If \(\mathbf{x}=(x_1,x_2,x_3,x_4)^{\top}\), then \((\mathbf{A}-4\mathbf{I}_4)\mathbf{x}=\mathbf{0}\) gives \(x_2=0\) and \(x_3=0\), while \(x_1,x_4\) are free. Hence \(E_4\) has basis \(\mathbf{e}_1,\mathbf{e}_4\) and geometric multiplicity \(2\). Its blocks have sizes \(3\) and \(1\): the first three coordinates form one chain, and the fourth forms a chain of length one.

For \(\mathbf{B}\), the equation \((\mathbf{B}-4\mathbf{I}_4)\mathbf{x}=\mathbf{0}\) gives \(x_2=0\) and \(x_4=0\); \(x_1,x_3\) are free. Thus \(E_4\) has basis \(\mathbf{e}_1,\mathbf{e}_3\), again giving geometric multiplicity \(2\). Here the blocks have sizes \(2\) and \(2\). Directly, \((\mathbf{A}-4\mathbf{I}_4)\mathbf{e}_2=\mathbf{e}_1\), \((\mathbf{A}-4\mathbf{I}_4)\mathbf{e}_3=\mathbf{e}_2\), while \((\mathbf{B}-4\mathbf{I}_4)\mathbf{e}_2=\mathbf{e}_1\) and \((\mathbf{B}-4\mathbf{I}_4)\mathbf{e}_4=\mathbf{e}_3\). These computations check the stated chain lengths.

Both matrices have dimension \(4\) and geometric multiplicity \(2\), but their block sizes differ. Therefore, concluding that geometric multiplicity \(2\) forces two blocks of size \(2\) is incorrect. It forces two blocks, while additional information is needed to determine their lengths.

A Short Diagnostic Routine

When checking a proposed Jordan calculation, use the defining equations before relying on a remembered pattern.

1
Check the eigenvector.
Verify \((\mathbf{A}-\lambda\mathbf{I})\mathbf{v}_1=\mathbf{0}\) and \(\mathbf{v}_1\ne\mathbf{0}\).
2
Check every recurrence.
For each \(j\ge2\), multiply and verify \((\mathbf{A}-\lambda\mathbf{I})\mathbf{v}_j=\mathbf{v}_{j-1}\).
3
Match the basis order to the matrix.
With the chain ordered from eigenvector to generalized vectors, the standard block has ones on the superdiagonal.
4
Interpret the multiplicity carefully.
Geometric multiplicity counts blocks for the eigenvalue; it does not, by itself, specify their sizes.

A frequent practical mistake is to inspect only the final matrix and assume that a familiar-looking block proves the chain is correct. A matrix pattern can be checked independently: multiply it by each proposed basis vector and compare the result with the chain equations. This catches both an order reversal and a misplaced off-diagonal entry. Likewise, a count of eigenvectors can check the number of blocks, but it cannot replace checks on the chain lengths.

Check Your Understanding

Use the chain equations and verify each claim by a short computation.

  1. For \(\mathbf{N}=\begin{bmatrix}0&1\\0&0\end{bmatrix}\), determine whether \((\mathbf{e}_1,\mathbf{e}_2)\) is a chain, and state which vector is an eigenvector.
  2. If a chain is written in the order \((\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3)\), which equation must \((\mathbf{A}-\lambda\mathbf{I})\mathbf{v}_3\) satisfy?
  3. Explain why reversing a chain basis can change a superdiagonal pattern to a subdiagonal pattern.
  4. A Jordan form has geometric multiplicity \(3\) for \(\lambda\). What does this determine, and what does it not determine?
  5. Give a direct matrix multiplication check that would expose the mistaken claim that a second chain vector must be an eigenvector.

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