Let \(N\) be an operator on an \(n\)-dimensional vector space, \(n>1\), such that \[ N^n=0, \qquad N^{n-1}\ne0. \] Prove that there is no operator \(X\) with \[ X^2=N. \]
Solution
1. Every nilpotent operator \(T\) on an \(n\)-dimensional vector spacesatisfies\[ T^n=0. \]
Let \(m\) be the nilpotency index of \(T\), so \[ T^m=0, \qquad T^{m-1}\neq0. \] Choose \(v\) with \(T^{m-1}v\neq0\). We claim that \[ v,Tv,T^2v,\ldots,T^{m-1}v \] are linearly independent. Indeed, suppose \[ c_0v+c_1Tv+\cdots+c_{m-1}T^{m-1}v=0. \] If \(j\) is the least index with \(c_j\neq0\), applying \(T^{m-1-j}\) gives \[ c_jT^{m-1}v=0, \] because every term with index larger than \(j\) acquires exponent at least \(m\). This contradicts \(c_j\neq0\) and \(T^{m-1}v\neq0\). Therefore \(m\) linearly independent vectors lie in an \(n\)-dimensional space, so \(m\leq n\). Hence \(T^n=0\).
2. If an operator \(X\) satisfied \(X^2=N\), then \(X\) would benilpotent.
The hypothesis \(N^n=0\) would give \[ X^{2n} =(X^2)^n =N^n =0. \] Thus some positive power of \(X\) is zero.
3. Under the assumption in step 2,\[ X^n=0. \]
Apply step 1 to the nilpotent operator \(X\).
4. Under the assumption \(X^2=N\), one would have\[ N^{n-1}=0. \]
Since \(n>1\), \[ 2n-2\geq n. \] Therefore step 3 gives \[ N^{n-1} =(X^2)^{n-1} =X^{2n-2} =0. \]
5. No operator \(X\) satisfies \(X^2=N\).
Step 4 contradicts the hypothesis \[ N^{n-1}\neq0. \] Hence no such \(X\) exists.
6. Q.E.D.
Step 5 is the required conclusion.