Square root of a nilpotent matrix

Question

This question has been asked a few times. In the former case I noticed that there was some argument trending towards using the Jordan form. For the sake of completeness, I was hoping to see if I'm on the right track.

Let $N$ be a nilpotent matrix such that $N^n=0$ and $N^{n-1}\neq0$. Suppose there exists $A^2=N$. Then I can write $A^2=\lambda I+N'$ where $N'$ is a nilpotent matrix. Using the Taylor expansion, I can see that $A^2=(\lambda I+N')^{1/2}=\sqrt{\lambda}\left(I+\frac{N'}{2\lambda}-\frac{N'^2}{8\lambda^2}+\frac{N'^3}{16\lambda^3}+....\right)$. This series is convergent since it only has finitely many terms. But, $A^2=N$ and $N$ is nilpotent. So $\lambda^i=0$ for all $i\in \{0,...,n\}.$ Hence such an $A^2$ doesn't exist.

Answer 1

I'm not 100% sure if I follow, but it looks invalid. When you write $A^2 = \lambda I + N'$, where $N'$ and $A^2$ are both nilpotent, the only scalar $\lambda$ that can possibly fit the bill here is $\lambda = 0$. Why? Because adding $\lambda I$ to a matrix increases all eigenvalues by $\lambda$, and nilpotent matrices have only $0$ as their eigenvalue. Therefore, from the outset, writing the series with $\lambda$ in the denominator is meaningless.

You realise this, and you use it to conclude that no square root exists. This doesn't really work logically. All you've figured out is that the method you've got for computing a square root cannot help you. There's no guarantee that another method won't help you.

What you need is an answer like the one you've linked to: one that assumes one exists, but arrives at a contradiction.

EDIT: To draw an analogy, consider the differential equation $$\frac{\mathrm{d}y}{\mathrm{d}x} = y.$$ One common way to solve this is by separation of variables, that is, $$\mathrm{d}{y} = y \, \mathrm{d}x \implies \frac{\mathrm{d}{y}}{y} = \mathrm{d}x \implies \int \frac{\mathrm{d}{y}}{y} = \int \mathrm{d}x \implies \ln|y| = x + C \implies y = \pm e^C e^x.$$ Notice that we divide by $y$, which means this method cannot arrive at the solution $y = 0$. However, $y = 0$ is definitely a solution; the fact that we divided by $y$ does not prove $y = 0$ is not a solution, just that if it is a solution, we've missed it using this method.

EDIT2: In order to prove it, what I'd do is suppose $A$ be a square root of $N$. Note that $A^{2n} = N^n = 0$, hence $A$ is nilpotent. Moreover, $A^{2n - 2} = N^{n - 1} \neq 0$. However, this contradicts the fact that $A^n = 0$, which is true since $A$ is nilpotent and $n \times n$.

Answer 2

just calculations over some field. If $$ \left( \begin{array}{cc} a & b \\ c & d \end{array} \right)^2 = \left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right) \; , $$ then $$ \left( \begin{array}{cc} a^2 + bc & b(a+d) \\ c(a+d) & d^2 + bc \end{array} \right) = \left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right) \; , $$ Since $b(a+d) = 1,$ we know $a+d \neq 0.$ Since $c(a+d) = 0$ but $a+d \neq 0,$ we know $c=0.$ But then $a^2 = 0$ and $a=0,$ also $d^2 = 0$ so $d=0.$ We have see that $a+d = 0,$ which gives a contradiction for the existence of this matrix square root. We used only field axioms, characteristic did not matter.

Answer 3

trying 3 by 3

$$ \left( \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array} \right)^2 = \left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array} \right) \; , $$ then $$ \left( \begin{array}{ccc} a^2 + bd+cg & ab+be+ch & ac+bf+ci \\ ad+de+fg & e^2+bd+fh & cd +ef +fi \\ ag+dh+gi & bg+eh+hi & i^2+cg +fh \end{array} \right) = \left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array} \right) \; , $$