Shifting Nilpotent Blocks to Eigenvalue Blocks

The Basic Shift

A nilpotent Jordan block records a chain of vectors whose successive images eventually become zero. Adding \(\lambda\mathbf{I}\) changes the operator’s action on each vector in the chain by adding \(\lambda\) times that vector. Subtracting \(\lambda\mathbf{I}\) then removes exactly this added part, so the chain structure is unchanged. This simple identity is the link between nilpotent blocks and Jordan blocks for a specified eigenvalue.

Work over \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{C}\), with \(\lambda\in\mathbb{F}\). A nilpotent Jordan block of size \(m\) is the \(m\times m\) matrix with ones immediately above the diagonal and zeros elsewhere. We write it as \(\mathbf{J}_m(0)\). The Jordan block with eigenvalue \(\lambda\) is obtained by adding \(\lambda\mathbf{I}_m\):

Thus the diagonal entries of \(\mathbf{J}_m(\lambda)\) are \(\lambda\), and its entries immediately above the diagonal remain \(1\). The block’s chain is visible by applying \(\mathbf{J}_m(0)\) to the standard basis: \(\mathbf{e}_1\) maps to \(\mathbf{0}\), and \(\mathbf{e}_j\) maps to \(\mathbf{e}_{j-1}\) for \(2\le j\le m\). The shifted block instead maps \(\mathbf{e}_1\) to \(\lambda\mathbf{e}_1\), and \(\mathbf{e}_j\) to \(\lambda\mathbf{e}_j+\mathbf{e}_{j-1}\).

The order of the chain matters: each vector after the first maps to the vector immediately before it. If the chain vectors form a basis of an \(m\)-dimensional space, the matrix of \(T\) in that ordered basis has ones above the diagonal, not below it.

Why the Shift Produces a Jordan Block

Proof. Since \(N\mathbf{v}_1=\mathbf{0}\), we have \(T\mathbf{v}_1=N\mathbf{v}_1+\lambda\mathbf{v}_1=\lambda\mathbf{v}_1\). For every \(j\) with \(2\le j\le m\), \(T\mathbf{v}_j=N\mathbf{v}_j+\lambda\mathbf{v}_j=\mathbf{v}_{j-1}+\lambda\mathbf{v}_j\). In the ordered basis \(\mathcal{B}\), the first equation gives the first column \((\lambda,0,\dots,0)^{\top}\); each later equation gives a \(\lambda\) in position \(j\), a \(1\) in position \(j-1\), and zeros elsewhere. These are exactly the columns of \(\mathbf{J}_m(\lambda)\). Also, \(T-\lambda\operatorname{id}_V=N\), so the stated equations for \(N\) are precisely the generalized eigenvector chain equations for \(T\). \(\square\)

The converse is the same identity used in reverse. If an ordered basis gives the matrix \(\mathbf{J}_m(\lambda)\) for \(T\), then subtracting \(\lambda\mathbf{I}_m\) gives \(\mathbf{J}_m(0)\). In that basis, \(T-\lambda\operatorname{id}_V\) therefore acts as a nilpotent Jordan block. This is a statement about a block already represented in a suitable basis; it does not assert that every operator has such a basis.

Proof. If the list is a generalized eigenvector chain, its defining equations are \((T-\lambda\operatorname{id}_V)\mathbf{v}_1=\mathbf{0}\) and \((T-\lambda\operatorname{id}_V)\mathbf{v}_j=\mathbf{v}_{j-1}\) for \(2\le j\le m\). Setting \(N=T-\lambda\operatorname{id}_V\) gives the nilpotent chain equations. Conversely, substituting \(N=T-\lambda\operatorname{id}_V\) into the nilpotent chain equations gives the generalized eigenvector chain equations. The condition \(\mathbf{v}_1\ne\mathbf{0}\) is retained in both definitions. \(\square\)

These results explain how to translate a calculation rather than redo it: find a chain for the nilpotent map \(T-\lambda\operatorname{id}_V\), and the same ordered vectors form a generalized eigenvector chain for \(T\). The next examples show this in standard coordinates, in a nonstandard basis, and for a direct sum of blocks.

Reading the Chain in Coordinates

The block size records the chain length. The scalar shift changes the eigenvalue but not which vector maps to which under \(T-\lambda\operatorname{id}_V\). The same conclusion holds when the chain is expressed in a basis that is not the standard basis.

Several Blocks and a Shared Eigenvalue

A block diagonal nilpotent matrix may contain more than one nilpotent block. Adding \(\lambda\mathbf{I}\) shifts every diagonal block by the same scalar; it does not join their chains into a single longer chain. Each block retains its own chain length.

Common Pitfall: Changing More Than the Eigenvalue

A frequent error is to multiply a nilpotent block by \(\lambda\) instead of adding \(\lambda\mathbf{I}\). For example, with \(\mathbf{N}=\begin{bmatrix}0&1\\0&0\end{bmatrix}\) and \(\lambda=2\), the incorrect calculation \(2\mathbf{N}=\begin{bmatrix}0&2\\0&0\end{bmatrix}\) still has zero diagonal. It has not shifted the eigenvalue to \(2\). The correct calculation is

Another mistake is to assume that the chain vectors themselves are unchanged by the action of \(T\). They are unchanged as vectors in the chain, but \(T\mathbf{v}_j\) includes the added term \(\lambda\mathbf{v}_j\). The chain equations apply to \(T-\lambda\operatorname{id}_V\), not to \(T\) alone.

The practical procedure is therefore precise: identify the scalar \(\lambda\), subtract \(\lambda\mathbf{I}\) from the matrix (or \(\lambda\operatorname{id}_V\) from the map), and perform the nilpotent-chain calculation on that difference. If the nilpotent part has a Jordan block in a basis, adding \(\lambda\mathbf{I}\) gives the corresponding block for eigenvalue \(\lambda\) in the same basis. This translation is useful because it lets a chain calculation be reused without changing its vectors or its length.