Undergrad Solution to a Matrix Quadratic Equation

[Euge]

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 $A$).

 

[anuttarasammyak]

(X+E)^2=A+E
X=(A+E)^{1/2}-E
But I am not certain that any matrix has its "square root".

 

[Euge]

To be clearer, the solution should be expressed as a polynomial in $A$.

 

[pasmith]

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.

 

[julian]

(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 $\pm (I + A)^{1/2} - I$.