2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $n \times n$.

$\endgroup$
  • $\begingroup$ I've thought about supposing one exists. Then looking at the Jordan normal form of N. Then N is similar to J. Since $N=A^2$, $A^2$ is also similar to J. But then I get stuck trying to find a contradiction. $\endgroup$ – Annalena Apr 12 '18 at 0:00
  • $\begingroup$ @Annalena a single Jordan block with nonzero diagonal value has a square root, here is me: math.stackexchange.com/questions/2693088/… I was not careful about eigenvalue zero, and someone brought that up in a comment to my answer. The final conclusion is that a single Jordan block has a square root if and only if it is nonsingular. I guess you are considering only the $\lambda = 0$ case. $\endgroup$ – Will Jagy Apr 12 '18 at 1:30
  • $\begingroup$ @WillJagy Yeah. The question has been bugging me. My book mentions that this is not the case for Nilpotent matrices, and then places the proof in the exercises. I believe your answer is one of the many I ran into, but I'm still not sure how to adapt it to nilpotent matrices. $\endgroup$ – Annalena Apr 12 '18 at 2:10
  • $\begingroup$ @Annalena it appears that the task is impossible for nilpotent matrices. I do not think I have seen a proof myself. I suggest you prove the 2 by 2 case first, which can probably be done by direct calculation; take a 2 by 2 matrix with entries a,b;c,d, square it, and show it cannot be the Jordan block 01;00 $\endgroup$ – Will Jagy Apr 12 '18 at 2:14
  • $\begingroup$ @WillJagy So this may be the perfect problem for induction. I do agree that this will probably be proof by contradiction. $\endgroup$ – Annalena Apr 12 '18 at 2:28
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ I don't know if this is as illustrative as the 2x2 example. $\endgroup$ – Annalena Apr 12 '18 at 2:52
  • $\begingroup$ @Annalena well, no, so far I do not see a way to go forward. If something clicks, say overnight, that may reveal an induction. $\endgroup$ – Will Jagy Apr 12 '18 at 2:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.