POTW Solution to a Matrix Quadratic Equation

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Let ##A## be a complex nilpotent ##n\times n##-matrix. Show that there is a unique nilpotent solution to the quadratic equation ##X^2 + 2X = A## in ##M_n(\mathbb{C})##, and write the solution explicitly (that is, in terms of ##A##).
 
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(X+E)^2=A+E
X=(A+E)^{1/2}-E
But I am not certain that any matrix has its "square root".
 
To be clearer, the solution should be expressed as a polynomial in ##A##.
 
We are given that A^k = 0 for some minimal 0 &lt; k \leq n. So we need a polynomial <br /> p(A) = \sum_{r=0}^{k-1} a_rA^r such that <br /> p(A)^2 = \sum_{r=0}^{k-1} A^r \left(\sum_{s=0}^{r} a_sa_{r-s}\right) = A + I. Hence <br /> X = (a_0 - 1)I + \frac{1}{2a_0}A + \dots where a_0 = \pm 1. That gives two solutions, so we need to show that only one is nilpotent; I suspect the one with a_0 = 1 since then <br /> X^k = (\tfrac12 A + \dots)^k = \tfrac{1}{2^k} A^k + \dots = 0 whereas with a_0 = -1, <br /> X^r = (-2I + \dots)^r = (-2)^rI + \dots which does not appear to be zero for any r.
 
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anuttarasammyak said:
(X+E)^2=A+E
X=(A+E)^{1/2}-E
But I am not certain that any matrix has its "square root".
I think you can just take

\begin{align*}
X = \pm (I + A)^{1/2} - I
\end{align*}

and expand the ##(I + A)^{1/2}##. It convergences because ##A## is nilpotent.

Then you can use that the eigenvalues of ##-(I + A)^{1/2} - I## are non-zero implying it is not nilpotent. The nilpotent solution, for ##A^k=0##, is:

\begin{align*}
X & = \sum_{r=1}^{k-1} (-1)^{r-1} \dfrac{(2r)!}{(2r-1) (2^r r!)^2} A^r
\nonumber \\
& = \frac{A}{2} - \frac{A^2}{8} + \frac{A^3}{16} - \frac{5A^4}{128} + \cdots + (-1)^k \dfrac{(2k-2)!}{(2k-3) (2^{k-1} (k-1)!)^2} A^{k-1}
\end{align*}

You can check this. Say ##A^2=0##, so that ##X=a_0 I + a_1 A##, then:

\begin{align*}
(a_0 I + a_1 A)^2 + 2 (a_0 I + a_1 A) = A
\end{align*}
or
\begin{align*}
(a_0^2+2a_0) I + 2 a_1 (a_0 + 1) A = A
\end{align*}
so that
\begin{align*}
(a_0^2+2a_0) =0 , \quad 2 a_1 (a_0 + 1) = 1 .
\end{align*}
Implying ##a_0=-2 , a_1= -\frac{1}{2}## or ##a_0=0 , a_1= \frac{1}{2}##.

Say ##A^3=0##, so that ##X=a_0 I + a_1 A + a_2 A^2##, then:

\begin{align*}
(a_0 I + a_1 A + a_2 A^2)^2 + 2 (a_0 I + a_1 A + a_2 A^3) = A
\end{align*}
or
\begin{align*}
(a_0^2+2a_0) I + 2 a_1 (a_0 + 1) A + (a_1^2 + 2a_0 a_2 + 2 a_2) A^2 = A
\end{align*}
so that
\begin{align*}
(a_0^2+2a_0) =0 , \quad 2 a_1 (a_0 + 1) = 1 , \quad a_1^2 + 2a_0 a_2 + 2 a_2= 0 .
\end{align*}
Implying ##a_0=-2 , a_1= -\frac{1}{2} , a_2 = \frac{1}{8}## or ##a_0=0 , a_1= \frac{1}{2} , a_2 = -\frac{1}{8}##.

The conditions you get on ##a_0,a_1, \dots, a_{k-1}## are the same conditions you would get by plugging ##\alpha = \sum_{r=0}^\infty a_r x^r## into

\begin{align*}
\alpha^2 + 2 \alpha = x
\end{align*}

and considering terms of powers up to ##x^{k-1}## on the LHS. This is why the approach of plugging ##X=\sum_{r=0}^{k-1} a_r A^r## into ##X^2 + 2X = A## will give the same answer as expanding ##\pm (I + A)^{1/2} - I##.
 
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