Solving quadratic matrix equation for X X2=[1a01] where a∈R0. Solve for matrix X.

Step 1
Let
$M(a)=\left[\begin{array}{cc}1& a\\ 0& 1\end{array}\right].$
Note M(a) has one eigenvalue: 1, and is not diagonalisable. This means that, if ${\displaystyle X^2=M\left(a\right)}$, then X has eigenvalue 1 or -1, but not both (if it had both, then X would be diagonalisable, and so would ${\displaystyle X^2=M\left(a\right)}$)
We also note that any eigenvectors of X will be an eigenvector for ${\displaystyle X^2}$, and ${\displaystyle X^2}$ has only the eigenvector
$\left[\begin{array}{c}1\\ 0\end{array}\right]$
Thus, X must have only this eigenvector as well. (Of course, when I say a matrix has only one eigenvector, I mean that there is one eigenspace, and it is one-dimensional, so we can only find one linearly independent eigenvector.)
Now, multiplication by
$\left[\begin{array}{c}1\\ 0\end{array}\right]$
on the right reveals the first column of a ${\displaystyle 2\times 2}$ matrix. Therefore, the first column of X is
$X\left[\begin{array}{c}1\\ 0\end{array}\right]=\pm 1\cdot \left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{c}\pm 1\\ 0\end{array}\right],$
depending on whether the unique eigenvalue is 1 or -1. This makes the matrix X upper-triangular (due to the 0 in the bottom left corner), so the eigenvalues of X lie on the diagonal. That is, X takes the form:
$\left[\begin{array}{cc}\pm 1& \ast \\ 0& \pm 1\end{array}\right]=\pm \left[\begin{array}{cc}1& \ast \\ 0& 1\end{array}\right],$
where the three ${\displaystyle {\pm }^{\prime }s}$ are all the same. That is,
${\displaystyle X=\pm M\left(b\right)}$
for some b. Note that ${\displaystyle {(\pm M\left(b\right))}^2=M{\left(b\right)}^2}$, so we are looking for b such that
${\left[\begin{array}{cc}1& b\\ 0& 1\end{array}\right]}^2=\left[\begin{array}{cc}1& a\\ 0& 1\end{array}\right],$
which was solved by Angel in their answer: ${\displaystyle b=\frac{a}{2}}$. This gives us precisely two solutions:
$X=\left[\begin{array}{cc}1& \frac{a}{2}\\ 0& 1\end{array}\right],-\left[\begin{array}{cc}1& \frac{a}{2}\\ 0& 1\end{array}\right].$

Step 1
Let ${\displaystyle X=(I+N)}$ where I is the identy matrix and N is the nilpotent matrix with all the diagonal entries and the lower left coner being zero. Take the second power of X,
${\displaystyle X^2=I+N+N+N^2=I+2N}$
the answer is trivially obtained from that.