A Worked Nilpotent Jordan Form

From a Nilpotent Matrix to Its Chains

A nilpotent matrix has a power equal to the zero matrix, but its action can still link vectors together in chains before they reach zero. Finding those chains is a concrete way to determine the matrix’s nilpotent Jordan form. The main tasks are to calculate successive kernels, choose vectors whose images form chains, and check directly that the resulting vectors really are a basis.

We work over \(\mathbb{R}\) with a finite-dimensional space. A matrix \(\mathbf{N}\) is nilpotent if \(\mathbf{N}^r=\mathbf{O}\) for some positive integer \(r\). Its nilpotency index is the smallest such \(r\). A nilpotent Jordan block of size \(s\), in the convention used here, has ones immediately above the main diagonal and zeros elsewhere. Thus, in its ordered chain basis \((\mathbf{v}_1,\dots,\mathbf{v}_s)\), the map sends \(\mathbf{v}_1\) to \(\mathbf{0}\) and \(\mathbf{v}_j\) to \(\mathbf{v}_{j-1}\) for \(2\le j\le s\).

Consider the following \(5\times5\) matrix, acting on column vectors in \(\mathbb{R}^5\):

For \(\mathbf{x}=(x_1,x_2,x_3,x_4,x_5)^{\top}\), the matrix product is

This formula makes it possible to compute powers and kernels without guessing a Jordan basis first. The kernel of a matrix is the set of vectors it sends to \(\mathbf{0}\); the kernel of \(\mathbf{N}^k\) records which vectors reach zero after at most \(k\) applications.

Successive Kernels Reveal the Chain Lengths

Here the kernel dimensions are \(2,4,5\), so their successive increases are \(2,2,1\). These increases indicate that two chains are needed, that both chains have length at least \(2\), and that just one has length at least \(3\). The next result explains the dimension calculation when a basis has already been arranged into chains.

Proof. Consider first one chain \((\mathbf{v}_1,\dots,\mathbf{v}_s)\). Repeatedly applying the chain relations gives \(\mathbf{N}^k\mathbf{v}_j=\mathbf{v}_{j-k}\) if \(j>k\), and \(\mathbf{N}^k\mathbf{v}_j=\mathbf{0}\) if \(j\le k\). Thus the first \(\min(k,s)\) chain vectors lie in the kernel. Every remaining chain vector, when there is one, maps to a distinct basis vector among \(\mathbf{v}_1,\dots,\mathbf{v}_{s-k}\). Their images are linearly independent, so no nonzero combination of those remaining vectors can be in the kernel. The kernel within this chain’s span therefore has dimension \(\min(k,s)\). Since all the chains together form a basis, their spans have independent coordinates: a sum of vectors from the separate chain spans is zero only if every component is zero. The kernel dimensions from the chains consequently add, giving the formula. \(\square\)

For the computed dimensions, the increase from \(\dim\ker\mathbf{N}^{k-1}\) to \(\dim\ker\mathbf{N}^k\) counts the chains of length at least \(k\): each chain contributes one new dimension at step \(k\) exactly when its length is at least \(k\). The increases \(2,2,1\) therefore suggest one chain of length \(3\) and one of length \(2\). This is a guide to construction, not a substitute for checking the chain vectors and their basis property.

Construct Chains and Check They Form a Basis

Checking both independence and spanning can be done by coordinate calculations, as above. Once the chain vectors are known to be a basis, their order determines the matrix of the map. The following result justifies reading off the Jordan blocks directly from the chain relations.

Proof. In the matrix of a map, column \(j\) contains the coordinates of the image of the \(j\)th basis vector. In a chain \((\mathbf{v}_1,\dots,\mathbf{v}_s)\), the first image is zero, so the first column within that chain’s block is zero. For \(j\ge2\), the image is \(\mathbf{v}_{j-1}\), whose coordinate column has a \(1\) in position \(j-1\) and zeros elsewhere. These are exactly the columns of the nilpotent Jordan block with ones immediately above the diagonal. Each chain vector maps to a vector in the same chain, so its image has zero coordinates in every other chain. Therefore all entries between distinct chain blocks are zero, proving the stated block-diagonal form. \(\square\)

Verify the Jordan Form by Matrix Multiplication

Common Pitfall: Applying the Matrix Twice Incorrectly

A common error is to read the first application formula as if the second application still depends on \(x_2\). For example, the incorrect calculation \(\mathbf{N}^2\mathbf{x}=x_2\mathbf{e}_1\) would say that \(\mathbf{e}_2\notin\ker\mathbf{N}^2\). But the actual calculation is \(\mathbf{N}\mathbf{e}_2=\mathbf{e}_1\), followed by \(\mathbf{N}\mathbf{e}_1=\mathbf{0}\), so \(\mathbf{e}_2\in\ker\mathbf{N}^2\). More generally, applying \(\mathbf{N}\) to \((x_2,x_3,0,x_3+x_5,0)^{\top}\) gives \((x_3,0,0,0,0)^{\top}\), not \((x_2,0,0,0,0)^{\top}\).

Another check that should not be skipped is basis verification. A collection of valid chains does not by itself establish a Jordan form: their vectors might be dependent or fail to span the space. In the example, the coordinate equations establish independence, the explicit expression for any \(\mathbf{x}\) establishes spanning, and \(\mathbf{N}\mathbf{P}=\mathbf{P}\mathbf{J}\) then confirms the resulting matrix.