Failure Example: A Nonzero Nilpotent Matrix

Nilpotent Does Not Mean Zero

A matrix can be nonzero while one of its powers is the zero matrix. Such matrices provide a useful failure example for diagonalization: their eigenvalues are restricted to zero, but their eigenspaces do not contain enough vectors to form a basis. The crucial distinction is between a matrix being zero and a power of that matrix being zero.

Throughout, matrices and vectors are over \(\mathbb{R}\), and \(n\ge 1\). The zero matrix is denoted by \(\mathbf{O}\), and \(\mathbf{I}_n\) is the \(n\times n\) identity matrix. For a square matrix \(\mathbf{N}\), its powers are defined by repeated multiplication: \(\mathbf{N}^2=\mathbf{N}\mathbf{N}\), \(\mathbf{N}^3=\mathbf{N}\mathbf{N}\mathbf{N}\), and so on.

A nonzero nilpotent matrix has nilpotency index at least \(2\): if its index were \(1\), then \(\mathbf{N}=\mathbf{O}\). The powers before the first zero power need not be zero. In particular, a nilpotent matrix can act nontrivially on vectors even though repeated applications eventually send every vector to \(\mathbf{0}\).

Recall that \(\lambda\in\mathbb{R}\) is an eigenvalue of \(\mathbf{N}\) if \(\mathbf{N}\mathbf{v}=\lambda\mathbf{v}\) for some \(\mathbf{v}\ne\mathbf{0}\). The eigenspace for \(\lambda\) is \(E_{\lambda}=\{\mathbf{v}:\mathbf{N}\mathbf{v}=\lambda\mathbf{v}\}\); it includes \(\mathbf{0}\). We now establish what nilpotence forces about eigenvalues and eigenspaces.

Zero Is the Only Eigenvalue

Proof. Choose a positive integer \(k\) such that \(\mathbf{N}^k=\mathbf{O}\). Suppose \(\lambda\) is an eigenvalue, with eigenvector \(\mathbf{v}\ne\mathbf{0}\). From \(\mathbf{N}\mathbf{v}=\lambda\mathbf{v}\), repeated application of \(\mathbf{N}\) gives \(\mathbf{N}^k\mathbf{v}=\lambda^k\mathbf{v}\). But \(\mathbf{N}^k\mathbf{v}=\mathbf{0}\), so \(\lambda^k\mathbf{v}=\mathbf{0}\). Since \(\mathbf{v}\ne\mathbf{0}\) and the scalars are real, this implies \(\lambda^k=0\), hence \(\lambda=0\). Thus there are no eigenvalues other than zero. To show that zero is in fact an eigenvalue, suppose instead that \(\ker \mathbf{N}=\{\mathbf{0}\}\). Then \(\mathbf{N}\) is one-to-one, so its composition with itself \(k\) times, the map given by \(\mathbf{N}^k\), is one-to-one. However, \(\mathbf{N}^k=\mathbf{O}\) sends both \(\mathbf{0}\) and the nonzero vector \(\mathbf{e}_1\) to \(\mathbf{0}\), so it is not one-to-one. This contradiction shows that \(\ker\mathbf{N}\) contains a nonzero vector. Since \(E_0=\ker\mathbf{N}\), zero is an eigenvalue. \(\square\)

The proof also gives a practical way to understand the eigenspace. For a nilpotent matrix, the eigenvector equation for its only possible eigenvalue, zero, is \(\mathbf{N}\mathbf{v}=\mathbf{0}\). Thus \(E_0\) is exactly the solution space of the homogeneous system with coefficient matrix \(\mathbf{N}\). A nonzero vector in this null space is an eigenvector; vectors outside it are not.

Finding the Eigenspace in a Longer Chain

These calculations illustrate why the eigenspace can be small. For the second matrix, many vectors are moved by \(\mathbf{N}\), but only vectors on the first coordinate axis are sent directly to zero. The matrix powers describe repeated action; the eigenspace describes vectors that are already sent to zero after one application. Those are different conditions.

Why a Nonzero Nilpotent Matrix Is Not Diagonalizable

A square matrix \(\mathbf{A}\in\mathbb{R}^{n\times n}\) is diagonalizable over \(\mathbb{R}\) if there is an invertible matrix \(\mathbf{P}\) and a diagonal matrix \(\mathbf{D}\) such that \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\). Equivalently, its columns in \(\mathbf{P}\) are a basis of eigenvectors. The next result shows directly why a nonzero nilpotent matrix cannot have such a representation.

Proof. Let \(\mathbf{N}\ne\mathbf{O}\) be nilpotent. The preceding theorem shows that its only eigenvalue is zero and that \(E_0\) contains a nonzero vector. Since \(\mathbf{N}\ne\mathbf{O}\), at least one column of \(\mathbf{N}\) is nonzero; consequently there is a standard basis vector \(\mathbf{e}_j\) with \(\mathbf{N}\mathbf{e}_j\ne\mathbf{0}\). This vector does not belong to \(E_0=\ker\mathbf{N}\), so \(E_0\ne\mathbb{R}^n\). Now suppose, for contradiction, that \(\mathbf{N}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\) for an invertible \(\mathbf{P}\) and diagonal \(\mathbf{D}=\operatorname{diag}(d_1,\dots,d_n)\). For each \(i\), the vector \(\mathbf{P}\mathbf{e}_i\) is nonzero because \(\mathbf{P}\) is invertible, and \(\mathbf{N}(\mathbf{P}\mathbf{e}_i)=\mathbf{P}\mathbf{D}\mathbf{e}_i=d_i\mathbf{P}\mathbf{e}_i\). Thus each \(d_i\) is an eigenvalue of \(\mathbf{N}\). The only eigenvalue is zero, so every \(d_i=0\), which gives \(\mathbf{D}=\mathbf{O}\). It follows that \(\mathbf{N}=\mathbf{P}\mathbf{O}\mathbf{P}^{-1}=\mathbf{O}\), contradicting \(\mathbf{N}\ne\mathbf{O}\). Therefore \(\mathbf{N}\) is not diagonalizable. \(\square\)

This argument makes the eigenspace obstruction explicit: all eigenvectors must belong to \(E_0\), and \(E_0\) is a proper subspace. It cannot supply a basis for \(\mathbb{R}^n\). The same conclusion follows from the diagonal form: if a diagonalization existed, every diagonal entry would have to be the only eigenvalue, zero, making the diagonal matrix itself zero.

Common Pitfall: A Zero Power Is Not a Zero Matrix

A tempting but incorrect calculation is to infer \(\mathbf{N}=\mathbf{O}\) from \(\mathbf{N}^2=\mathbf{O}\). For \(\mathbf{N}=\begin{bmatrix}0&1\\0&0\end{bmatrix}\), the calculation above gives \(\mathbf{N}^2=\mathbf{O}\), but \(\mathbf{N}\begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}1\\0\end{bmatrix}\ne\mathbf{0}\). So \(\mathbf{N}\ne\mathbf{O}\). The correction is to keep track of how many times the matrix acts: \(\mathbf{N}^2=\mathbf{O}\) says \(\mathbf{N}(\mathbf{N}\mathbf{x})=\mathbf{0}\) for every \(\mathbf{x}\), not that \(\mathbf{N}\mathbf{x}=\mathbf{0}\) for every \(\mathbf{x}\).

A related error is to conclude that a matrix is diagonalizable merely because its only eigenvalue is zero. The eigenvalue alone is not enough: diagonalizability requires enough linearly independent eigenvectors to form a basis. For a nonzero nilpotent matrix, the zero eigenspace is proper, so it cannot provide that basis.