nilpotent matrix

The square matrix $A$ is said to be nilpotent if $A^n=\underset{\text{n times}}{\underbrace{AA\mathrm{\cdots }A}}=\mathrm{𝟎}$ for some positive integer $n$ (here $\mathrm{𝟎}$denotes the matrix where every entry is 0).

Theorem (Characterization of nilpotent matrices).

A matrix is nilpotent iff its eigenvalues are all 0.

Proof.

Let $A$ be a nilpotent matrix. Assume $A^n=\mathrm{𝟎}$. Let $\lambda$ be an eigenvalue of $A$.Then $A𝐱=\lambda 𝐱$ for some nonzero vector $𝐱$.By induction ${\lambda }^n𝐱=A^n𝐱=0$, so $\lambda =0$.

Conversely, suppose that all eigenvalues of $A$ are zero. Then the chararacteristicpolynomial of $A$: $det(\lambda I-A)={\lambda }^n$. It now follows from theCayley-Hamilton theorem that $A^n=\mathrm{𝟎}$.∎

Since the determinant is the product of the eigenvalues it follows that a nilpotent matrix has determinant 0. Similarly, since the trace of a square matrix is the sum of the eigenvalues, it follows that it has trace 0.

One class of nilpotent matrices are the http://planetmath.org/node/4381strictly triangular matrices (lower or upper), this follows from the fact that the eigenvalues of a triangular matrix are the diagonal elements, and thus are all zero in the case of strictly triangular matrices.

Note for $2\times 2$ matrices $A$ the theorem implies that $A$ is nilpotent iff $A=\mathrm{𝟎}$ or $A^2=\mathrm{𝟎}$.

Also it is worth noticing that any matrix that is similar to a nilpotent matrix is nilpotent.