If \(N\) is a nilpotent \(3 \times 3\) matrix over \(C\), prove that \(A = I + \frac{1}{2}N - \frac{1}{8}N^2\) satisfies \(A^2 = I + N\), i.e., \(A\) is a square root of \(I + N\). Use the binomial series for \((1 + t)^{1/2}\) to obtain a similar formula for a square root of \(I + N\), where \(N\) is any nilpotent \(n \times n\) matrix over \(C\).
Solution
1. Verify the displayed formula in size three.
For a nilpotent \(3\times3\) matrix, \(N^3=0\) [@HK71]. Put \[ A=I+\frac12N-\frac18N^2. \] Because all powers of \(N\) commute, \[ \begin{aligned} A^2 &=I+N+\left(\frac14-\frac14\right)N^2 -\frac18N^3+\frac1{64}N^4\\ &=I+N. \end{aligned} \] Thus the stated \(A\) is a square root of \(I+N\).
2. Truncate the binomial series at the nilpotency index.
Suppose generally that \(N^r=0\). The formal binomial series satisfies \[ (1+t)^{1/2}=\sum_{j=0}^{\infty}\binom{1/2}{j}t^j \] and its square is \(1+t\). Define the finite polynomial \[ q(t)=\sum_{j=0}^{r-1}\binom{1/2}{j}t^j. \] The power-series identity implies \[ q(t)^2-(1+t) \] has no terms of degree below \(r\), so it is divisible by \(t^r\) in \(\mathbb C[t]\) modulo terms irrelevant after substitution. Equivalently, \[ q(t)^2\equiv1+t\pmod{t^r}. \] Substituting \(t=N\) and using \(N^r=0\) gives \[ q(N)^2=I+N. \] Hence \[ \boxed{ (I+N)^{1/2} =\sum_{j=0}^{r-1}\binom{1/2}{j}N^j} \] is a polynomial square root of \(I+N\).