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Generalized Eigenvectors and Jordan Form · Tutorial 827 of 1000

Kernel Filtrations of an Eigenvalue

Build the kernel filtration for a fixed eigenvalue, compute its spaces in examples, and interpret what enters at each power.

Advanced 10 min read

What You'll Learn

  • Define the kernel filtration associated with a fixed eigenvalue.
  • Prove that successive kernels are nested and their dimensions cannot decrease.
  • Compute kernels of powers using exact matrix calculations.
  • Distinguish eigenvectors from vectors that first enter a higher kernel.
  • Use newly appearing vectors to identify generalized eigenvector chains.

From an Eigenvalue to a Sequence of Kernels

For an eigenvalue \(\lambda\), the equation \((\mathbf{A}-\lambda\mathbf{I})\mathbf{x}=\mathbf{0}\) identifies its eigenvectors together with the zero vector. Taking higher powers of the same shifted matrix asks a broader question: which vectors become zero after applying the shift repeatedly? These successive kernels reveal how vectors beyond the ordinary eigenvectors enter the picture.

Throughout, let \(\mathbf{A}\in\mathbb{R}^{n\times n}\), and fix a real eigenvalue \(\lambda\) of \(\mathbf{A}\). Set \(\mathbf{N}=\mathbf{A}-\lambda\mathbf{I}_n\). A kernel here is the set of vectors sent to zero by a matrix acting on \(\mathbb{R}^n\). The zeroth power \(\mathbf{N}^0\) means \(\mathbf{I}_n\).

Definition: The kernel filtration for \(\lambda\) is the sequence of subspaces \(K_0,K_1,K_2,\ldots\) defined by \(K_0=\{\mathbf{0}\}\) and \(K_k=\ker(\mathbf{N}^k)=\{\mathbf{x}\in\mathbb{R}^n:\mathbf{N}^k\mathbf{x}=\mathbf{0}\}\) for each integer \(k\geq 1\). A nonzero vector in \(K_k\) is a generalized eigenvector for \(\lambda\) of order at most \(k\); it is of exact order \(k\) if it belongs to \(K_k\) but not \(K_{k-1}\).

The word “filtration” refers to an increasing sequence of subspaces. It does not mean that every inclusion must be strict: an example may have \(K_k=K_{k+1}\). The main information recorded at stage \(k\) is which vectors have entered by that point and the dimension \(\dim K_k\). Here \(\dim K_k\) is the number of vectors in any basis of \(K_k\), so it measures the number of independent directions in that subspace.

Two Basic Facts About the Filtration

Theorem: For every \(k\geq 0\), \(K_k\subseteq K_{k+1}\). Consequently, \(\dim K_k\leq\dim K_{k+1}\).

Proof. For \(k=0\), \(K_0=\{\mathbf{0}\}\), which is contained in every subspace. Now take \(k\geq1\) and \(\mathbf{x}\in K_k\). By definition, \(\mathbf{N}^k\mathbf{x}=\mathbf{0}\). Therefore \(\mathbf{N}^{k+1}\mathbf{x}=\mathbf{N}(\mathbf{N}^k\mathbf{x})=\mathbf{N}\mathbf{0}=\mathbf{0}\), so \(\mathbf{x}\in K_{k+1}\). For the dimension statement, a basis of \(K_k\) is a linearly independent list in \(K_{k+1}\). In a finite-dimensional space, a linearly independent list can be extended to a basis by adding vectors from the larger space until it is spanned. Thus \(K_{k+1}\) has at least as many basis vectors as \(K_k\). \(\square\)

The first kernel has a familiar interpretation: its nonzero vectors are precisely the eigenvectors for \(\lambda\). Higher kernels can contain vectors that are not eigenvectors themselves, but that the shifted matrix sends into a lower kernel.

Proposition: \(K_1=\ker(\mathbf{A}-\lambda\mathbf{I}_n)\) consists of the zero vector and all eigenvectors of \(\mathbf{A}\) with eigenvalue \(\lambda\).

Proof. A vector \(\mathbf{x}\) belongs to \(K_1\) exactly when \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{x}=\mathbf{0}\), which rearranges to \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\). If \(\mathbf{x}\ne\mathbf{0}\), this is the definition of an eigenvector with eigenvalue \(\lambda\). Conversely, every such eigenvector satisfies the equation and therefore belongs to \(K_1\); the zero vector also satisfies it. \(\square\)

There is a useful way to read a strict increase. If \(\mathbf{x}\in K_{k+1}\) but \(\mathbf{x}\notin K_k\), then the successive vectors obtained by applying \(\mathbf{N}\) to \(\mathbf{x}\) form a chain down to a nonzero eigenvector.

Proposition: If \(\mathbf{x}\in K_{k+1}\setminus K_k\), where \(k\geq1\), then \(\mathbf{N}^k\mathbf{x}\ne\mathbf{0}\), \(\mathbf{N}^{k+1}\mathbf{x}=\mathbf{0}\), and \(\mathbf{x},\mathbf{N}\mathbf{x},\ldots,\mathbf{N}^k\mathbf{x}\) are all nonzero. Moreover, applying \(\mathbf{N}\) to each vector gives the next one, ending at zero.

Proof. Membership in \(K_{k+1}\) gives \(\mathbf{N}^{k+1}\mathbf{x}=\mathbf{0}\), while nonmembership in \(K_k\) gives \(\mathbf{N}^k\mathbf{x}\ne\mathbf{0}\). If \(\mathbf{N}^j\mathbf{x}=\mathbf{0}\) for some \(0\leq j\leq k\), applying \(\mathbf{N}^{k-j}\) would give \(\mathbf{N}^k\mathbf{x}=\mathbf{0}\), a contradiction. So all the listed vectors are nonzero. Finally, \(\mathbf{N}(\mathbf{N}^j\mathbf{x})=\mathbf{N}^{j+1}\mathbf{x}\) for \(0\leq j<k\), and \(\mathbf{N}(\mathbf{N}^k\mathbf{x})=\mathbf{0}\). \(\square\)

Computing the Kernels One Power at a Time

Worked Example: A Three-Stage Filtration

Let \(\mathbf{A}=\begin{bmatrix}2&1&0\\0&2&1\\0&0&2\end{bmatrix}\) and \(\lambda=2\). Compute \(K_1,K_2,K_3\).

The shifted matrix and its first two powers are

$$ \mathbf{N}=\mathbf{A}-2\mathbf{I}_3= \begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}, \qquad \mathbf{N}^2= \begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix}, \qquad \mathbf{N}^3=\mathbf{O}. $$

For \(\mathbf{x}=(x_1,x_2,x_3)^\top\), the equation \(\mathbf{N}\mathbf{x}=\mathbf{0}\) reads \(x_2=0\), \(x_3=0\), with \(x_1\) free. Hence \(K_1=\operatorname{span}\{(1,0,0)^\top\}\) and \(\dim K_1=1\). For the second power, \(\mathbf{N}^2\mathbf{x}=(x_3,0,0)^\top\), so \(x_3=0\) while \(x_1,x_2\) are free. Thus \(K_2=\operatorname{span}\{(1,0,0)^\top,(0,1,0)^\top\}\) and \(\dim K_2=2\). Since \(\mathbf{N}^3=\mathbf{O}\), every vector is in \(K_3=\mathbb{R}^3\), which has dimension \(3\).

Check: \(\mathbf{N}(1,0,0)^\top=\mathbf{0}\), \(\mathbf{N}(0,1,0)^\top=(1,0,0)^\top\), and \(\mathbf{N}(0,0,1)^\top=(0,1,0)^\top\). These calculations verify the stated kernel memberships and show how the third coordinate direction enters at the third power.

To compute \(K_k\), form the power \(\mathbf{N}^k\), then solve the homogeneous system \(\mathbf{N}^k\mathbf{x}=\mathbf{0}\). Row reduction is valid because elementary row operations preserve the solution set of a homogeneous system. Keep the original matrix columns as vectors; the unknowns in this system are the coordinates of \(\mathbf{x}\), not coefficients in a list of unrelated row vectors.

Worked Example: A New Direction at the Second Power

Let \(\mathbf{A}=\begin{bmatrix}4&1&0\\0&4&0\\0&0&2\end{bmatrix}\) and \(\lambda=4\). Find the first two kernels and exhibit a vector in \(K_2\setminus K_1\).

Subtracting \(4\mathbf{I}_3\) gives \(\mathbf{N}=\begin{bmatrix}0&1&0\\0&0&0\\0&0&-2\end{bmatrix}\), and direct multiplication gives \(\mathbf{N}^2=\begin{bmatrix}0&0&0\\0&0&0\\0&0&4\end{bmatrix}\). For \(\mathbf{x}=(x_1,x_2,x_3)^\top\), the first equation is \(\mathbf{N}\mathbf{x}=(x_2,0,-2x_3)^\top=\mathbf{0}\). Thus \(x_2=x_3=0\), \(x_1\) is free, and \(K_1=\operatorname{span}\{(1,0,0)^\top\}\), of dimension \(1\). At the second power, \(\mathbf{N}^2\mathbf{x}=(0,0,4x_3)^\top=\mathbf{0}\) requires only \(x_3=0\). Therefore \(K_2=\operatorname{span}\{(1,0,0)^\top,(0,1,0)^\top\}\), of dimension \(2\).

The vector \(\mathbf{e}_2=(0,1,0)^\top\) is in \(K_2\) but not \(K_1\): \(\mathbf{N}\mathbf{e}_2=(1,0,0)^\top\ne\mathbf{0}\), whereas \(\mathbf{N}^2\mathbf{e}_2=\mathbf{0}\). Check: multiplying by \(\mathbf{N}\) once gives the eigenvector \(\mathbf{e}_1\), and multiplying by \(\mathbf{N}\) again gives zero. This confirms both the claimed order and the kernel calculations.

When the Kernels Are Not Coordinate Axes

The kernels need not be spanned by standard basis vectors. The calculation is still the same: compute the powers, solve the resulting equations, and express the solutions as spans. A vector appearing in a higher kernel should be checked against the lower one; satisfying the higher-power equation alone does not establish its exact order.

Worked Example: A Filtration in Nonstandard Directions

Take \(\mathbf{A}=\mathbf{N}=\begin{bmatrix}1&-1&0\\1&-1&1\\0&0&0\end{bmatrix}\) and \(\lambda=0\). Find \(K_1,K_2,K_3\).

Multiplication gives

$$ \mathbf{N}^2= \begin{bmatrix}0&0&-1\\0&0&-1\\0&0&0\end{bmatrix}, \qquad \mathbf{N}^3=\mathbf{O}. $$

For the first kernel, \(\mathbf{N}\mathbf{x}=(x_1-x_2,\;x_1-x_2+x_3,\;0)^\top\). Setting this equal to zero gives \(x_1=x_2\) and \(x_3=0\). Thus \(K_1=\operatorname{span}\{(1,1,0)^\top\}\), with dimension \(1\). For the second, \(\mathbf{N}^2\mathbf{x}=(-x_3,-x_3,0)^\top\), so \(x_3=0\) and \(x_1,x_2\) are free. Hence \(K_2=\operatorname{span}\{(1,0,0)^\top,(0,1,0)^\top\}\), with dimension \(2\). The third power is zero, so \(K_3=\mathbb{R}^3\), with dimension \(3\).

For a direct check of a vector entering at the second stage, let \(\mathbf{v}=(1,0,0)^\top\). Then \(\mathbf{N}\mathbf{v}=(1,1,0)^\top\ne\mathbf{0}\), while \(\mathbf{N}^2\mathbf{v}=\mathbf{0}\). A vector entering at the third stage is \(\mathbf{w}=(1,0,-1)^\top\): multiplication gives \(\mathbf{N}\mathbf{w}=(1,0,0)^\top\), \(\mathbf{N}^2\mathbf{w}=(1,1,0)^\top\), and \(\mathbf{N}^3\mathbf{w}=\mathbf{0}\). These checks verify that the calculated kernels contain the claimed vectors, with \(\mathbf{v}\notin K_1\) and \(\mathbf{w}\notin K_2\).

Reading the Dimension Data Carefully

For a fixed eigenvalue, the list \(\dim K_1,\dim K_2,\ldots\) records how many independent directions have appeared by each power. In the first and third examples, the dimensions at the displayed stages are \(1,2,3\); in the second, the first two dimensions are \(1,2\). The spaces themselves matter as well: the dimensions alone do not identify which vectors lie in a kernel. Computing a basis for each kernel makes the new directions explicit.

These calculations are useful when finding generalized eigenvectors and organising them into chains. The proposition above explains why: a vector that first appears in \(K_{k+1}\) has successive nonzero images under \(\mathbf{N}\), ending at an eigenvector. The full theory relates patterns of dimension growth to the structure of Jordan form, but that interpretation requires additional results. At this stage, use the filtration to compute and compare the kernels rather than inferring more structure than the calculations establish.

Common Pitfall: Reusing the First-Kernel Equations

In the third example, the correct equation for \(K_2\) is \(\mathbf{N}^2\mathbf{x}=(-x_3,-x_3,0)^\top=\mathbf{0}\), which requires \(x_3=0\). A common incorrect calculation is to use the equations for \(\mathbf{N}\mathbf{x}=\mathbf{0}\) again and conclude that \(K_2\) still requires \(x_1=x_2\). That would give only the line spanned by \((1,1,0)^\top\). The correction is to multiply by \(\mathbf{N}\) a second time before solving: for example, \((1,0,0)^\top\) satisfies \(\mathbf{N}^2\mathbf{x}=\mathbf{0}\) but not \(\mathbf{N}\mathbf{x}=\mathbf{0}\). Thus it lies in \(K_2\setminus K_1\), and the second kernel is a plane, not the first-kernel line.

For each new power, check the matrix multiplication and solve that power’s equation from scratch. Then verify the resulting basis vectors by multiplying them by the stated power, and check at least one new vector against the preceding kernel. This separates a correct kernel computation from an unsupported claim about exact order.

Check Your Understanding

For each calculation, state the kernel dimension and verify any vector you claim has exact order.

  1. For \(\mathbf{A}=\begin{bmatrix}3&1\\0&3\end{bmatrix}\) and \(\lambda=3\), compute \(K_1\) and \(K_2\).
  2. Explain why \(K_k\subseteq K_{k+1}\) for every \(k\geq1\).
  3. For the matrix in the third worked example, give a vector in \(K_2\setminus K_1\) and verify both conditions.
  4. If \(\mathbf{x}\in K_4\setminus K_3\), which of \(\mathbf{x},\mathbf{N}\mathbf{x},\mathbf{N}^2\mathbf{x},\mathbf{N}^3\mathbf{x},\mathbf{N}^4\mathbf{x}\) must be nonzero?
  5. Why is solving \(\mathbf{N}\mathbf{x}=\mathbf{0}\) not enough to determine \(K_2\)?

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