Let \(M\) be an \(n\times n\) complex matrix and define \[ G_M=\{\lambda\in\CC:\lambda M\text{ is similar to }M\}. \]
- Determine \(G_M\) when \[ M=\begin{pmatrix} 0&0&4\\ 0&0&0\\ 0&0&0 \end{pmatrix}. \]
- Assume \(M\) is not nilpotent. Prove that \(G_M\) is finite.
Solution
1. For the matrix in part (1), every nonzero scalar belongs to \(G_M\).
Let \(\lambda\in\CC^\times\) and set \[ P_\lambda \coloneqq \begin{pmatrix} 1&0&0\\ 0&1&0\\ 0&0&\lambda \end{pmatrix}. \] Then \(P_\lambda\) is invertible, and direct multiplication gives \[ P_\lambda^{-1}MP_\lambda = \begin{pmatrix} 0&0&4\lambda\\ 0&0&0\\ 0&0&0 \end{pmatrix} = \lambda M. \] Thus \(\lambda M\) is similar to \(M\).
2. For the matrix in part (1), \(0\notin G_M\).
The matrix \(M\) has rank \(1\), whereas \(0M\) has rank \(0\). Similar matrices have the same rank, so \(0M\) is not similar to \(M\).
3. Therefore, in part (1),\[ \boxed{G_M=\CC^\times}. \]
Step 1 proves \(\CC^\times\subseteq G_M\), and step 2 excludes the only remaining scalar.
4. If a complex matrix \(M\) is not nilpotent, then \(M\) has a nonzeroeigenvalue.
If every eigenvalue of \(M\) were \(0\), then its characteristic polynomial would be \[ \chi_M(t)=t^n. \] By the Cayley–Hamilton theorem, \[ M^n=0, \] which would make \(M\) nilpotent. The contrapositive gives the claim.
5. Fix a nonzero eigenvalue \(\alpha\) of the nonnilpotent matrix \(M\).If \(\lambda\in G_M\), then\[ \lambda\alpha\in\Spec(M). \]
Choose a nonzero vector \(v\) with \[ Mv=\alpha v. \] Then \[ (\lambda M)v=\lambda\alpha v, \] so \(\lambda\alpha\) is an eigenvalue of \(\lambda M\). Since \(\lambda\in G_M\), the matrices \(\lambda M\) and \(M\) are similar and hence have the same eigenvalues. Therefore \(\lambda\alpha\) is an eigenvalue of \(M\).
6. If \(M\) is not nilpotent, then \(G_M\) is finite.
By step 4, choose a nonzero eigenvalue \(\alpha\) of \(M\). Step 5 implies \[ G_M \subseteq \left\{ \frac{\beta}{\alpha}: \beta\in\Spec(M) \right\}. \] The spectrum of an \(n\times n\) complex matrix is a finite set, so the set on the right is finite. Hence \(G_M\) is finite.
7. Q.E.D.
Step 3 answers part (1), and step 6 proves part (2).