Let \(M_{n\times n}\) be the space of real \(n\times n\) matrices with metric \[ d(A,B)=\sum_{i,j=1}^n|a_{ij}-b_{ij}|. \] Prove that the set of nilpotent matrices in \(M_{n\times n}\) is closed.
Solution
1. An \(n\times n\) matrix \(A\) is nilpotent if and only if\[ A^n=0. \]
If \(A^n=0\), then \(A\) is nilpotent by definition.
Conversely, suppose \(A\) is nilpotent, so \(A^m=0\) for some positive integer \(m\). Over \(\CC\), every eigenvalue \(\lambda\) of \(A\) satisfies \[ 0=A^m v=\lambda^m v \] for a corresponding nonzero eigenvector \(v\), hence \(\lambda=0\). Therefore the characteristic polynomial of \(A\) is \(t^n\). By the Cayley–Hamilton theorem, \[ A^n=0. \]
2. The map\[ \Phi:M_{n\times n}\longrightarrow M_{n\times n}, \qquad \Phi(A)=A^n, \]is continuous for the stated metric.
Each entry of \(A^n\) is a polynomial in the \(n^2\) entries of \(A\). Polynomial functions on the finite-dimensional space \(\RR^{n^2}\) are continuous, and the stated metric induces its usual topology.
3. The set of nilpotent matrices is\[ \Phi^{-1}(\{0\}). \]
This is exactly the equivalence proved in step 1.
4. The set of nilpotent matrices is closed.
The singleton \(\{0\}\) is closed in the metric space \(M_{n\times n}\). By step 2, \(\Phi\) is continuous, so its inverse image \[ \Phi^{-1}(\{0\}) \] is closed. Step 3 identifies this inverse image with the nilpotent matrices.
5. Q.E.D.
Step 4 is the required conclusion.