Let \(n\) be a positive integer, \(n \geq 2\), and let \(N\) be an \(n \times n\) matrix over the field \(F\) such that \(N^n = 0\) but \(N^{n-1} \neq 0\). Prove that \(N\) has no square root, i.e., that there is no \(n \times n\) matrix \(A\) such that \(A^2 = N\).
Solution
1. \(N\) is nilpotent of index \(n\) (i.e. \(N^n = 0\) but \(N^{n-1} \neq 0\)), so its Jordan form is a single Jordan block \(J_n(0)\) of size \(n\).
a nilpotent matrix of index \(n\) on an \(n\)-dimensional space has a single Jordan block of size \(n\).
2. Suppose for contradiction that \(A^2 = N\).
assume a square root exists.
3. Then \(A^{2n} = N^n = 0\), so \(A\) is nilpotent.
\(A^{2n} = (A^2)^n = N^n = 0\).
4. A nilpotent \(n \times n\) matrix has nilpotency index at most \(n\), so \(A^n = 0\).
the nilpotency index of an \(n \times n\) nilpotent matrix is at most \(n\).
5. Hence \(N^{n-1} = (A^2)^{n-1} = A^{2n-2} = 0\) (since \(2n - 2 \ge n\) for \(n \ge 2\)).
\(A^{2n-2} = A^n A^{n-2} = 0 \cdot A^{n-2} = 0\) (using \(A^n = 0\) from <1>4).
6. This contradicts \(N^{n-1} \neq 0\).
5 and the hypothesis.
7. Hence no such \(A\) exists.
<1>2–<1>6.
8. Q.E.D.
7.