Let \(A\) be an \(n\times n\) complex matrix all of whose eigenvalues are equal to \(1\). Suppose the set \[ \{A^n:n=1,2,\ldots\} \] is bounded. Prove that \(A=I\).
Solution
Let \[ N=A-I. \]
1. The matrix \(N\) is nilpotent.
Every eigenvalue of \(N=A-I\) is obtained by subtracting \(1\) from an eigenvalue of \(A\), so every eigenvalue of \(N\) is \(0\). Thus the characteristic polynomial of \(N\) is \(t^n\). By the Cayley–Hamilton theorem, \[ N^n=0. \]
2. If \(N\neq0\), there exist an integer \(r\ge1\), a vector \(v\),and a linear functional \(\ell\) such that\[ N^r\neq0, \qquad N^{r+1}=0, \qquad \ell(N^r v)\neq0. \]
By step 1, the positive powers of \(N\) eventually vanish. If \(N\neq0\), let \(r\ge1\) be maximal with \(N^r\neq0\). Choose \(v\) with \(N^r v\neq0\). Since linear functionals separate points in a finite-dimensional vector space, there is a linear functional \(\ell\) such that \(\ell(N^r v)\neq0\).
3. Under the assumption \(N\neq0\), the sequence\[ \ell(A^m v), \qquad m=1,2,\ldots, \]is unbounded.
2. Hence its absolute value is unboundedas \(m\to\infty\).
4. One must have \(N=0\).
If the matrices \(A^m\) form a bounded set, then for every fixed vector \(v\) the vectors \(A^m v\) form a bounded set, and applying a fixed linear functional \(\ell\) preserves boundedness. Step 3 would contradict this if \(N\neq0\). Therefore \(N=0\).
5. Consequently,\[ \boxed{A=I}. \]
By definition, \(N=A-I\). Step 4 gives \(A-I=0\).
6. Q.E.D.
Step 5 is the required conclusion.