Let \(V\) be a finite-dimensional vector space over the field \(F\), and let \(T\) be a linear operator on \(V\) such that \(\operatorname{rank}(T) = 1\). Prove that either \(T\) is diagonalizable or \(T\) is nilpotent, not both.
Solution
1. \(\dim \operatorname{im} T = 1\), so \(\operatorname{im} T = \operatorname{span}\{v\}\) for some nonzero \(v \in V\).
rank \(1\).
2. \(T\) has at most one nonzero eigenvalue.
2.1. If \(T\) has a nonzero eigenvalue \(\lambda\) with eigenvector \(u\), then \(Tu = \lambda u \in \operatorname{im} T\), so \(u \in \operatorname{im} T = \operatorname{span}\{v\}\).
\(\lambda \neq 0\) and \(Tu = \lambda u\) implies \(u = \lambda^{-1} Tu \in \operatorname{im} T\).
2.2. Hence the eigenspace for any nonzero eigenvalue is contained in the \(1\)-dimensional space \(\operatorname{span}\{v\}\).
1.
2.3. Therefore there is at most one nonzero eigenvalue.
two distinct nonzero eigenvalues would give two linearly independent eigenvectors, contradicting 2.
3. Case 1: \(T\) has a nonzero eigenvalue \(\lambda\).
3.1. Then \(T\) is diagonalizable.
\(T\) has eigenvalues \(\lambda\) (with multiplicity \(1\)) and \(0\) (with multiplicity \(n-1\)); the geometric multiplicity of \(0\) is \(\dim \ker T = n - 1\) (by rank–nullity), and of \(\lambda\) is \(1\), so the geometric multiplicities sum to \(n\), hence \(T\) is diagonalizable.
3.2. \(T\) is not nilpotent.
a nilpotent operator has all eigenvalues \(0\), but \(T\) has the nonzero eigenvalue \(\lambda\).
4. Case 2: \(T\) has no nonzero eigenvalue.
4.1. Then all eigenvalues of \(T\) are \(0\), so \(T\) is nilpotent.
the characteristic polynomial is \(t^n\), so by Cayley–Hamilton \(T^n = 0\).
4.2. \(T\) is not diagonalizable (unless \(T = 0\)).
if \(T\) were diagonalizable with all eigenvalues \(0\), then \(T = 0\), contradicting \(\operatorname{rank} T = 1\).
5. Hence exactly one of “diagonalizable” or “nilpotent” holds.
3 and 4 are mutually exclusive and exhaustive.
6. Q.E.D.
5.