gea3stwg

2022-01-30

Solving quadratic matrix equation for X
${X}^{2}=\left[\begin{array}{cc}1& a\\ 0& 1\end{array}\right]$
where $a\in \frac{\mathbb{R}}{0}$. Solve for matrix X.

Kyler Jacobson

Expert

Step 1
Let
$M\left(a\right)=\left[\begin{array}{cc}1& a\\ 0& 1\end{array}\right].$
Note M(a) has one eigenvalue: 1, and is not diagonalisable. This means that, if ${X}^{2}=M\left(a\right)$, then X has eigenvalue 1 or -1, but not both (if it had both, then X would be diagonalisable, and so would ${X}^{2}=M\left(a\right)$)
We also note that any eigenvectors of X will be an eigenvector for ${X}^{2}$, and ${X}^{2}$ 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 $2×2$ matrix. Therefore, the first column of X is
$X\left[\begin{array}{c}1\\ 0\end{array}\right]=±1\cdot \left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{c}±1\\ 0\end{array}\right],$
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:
$\left[\begin{array}{cc}±1& \ast \\ 0& ±1\end{array}\right]=±\left[\begin{array}{cc}1& \ast \\ 0& 1\end{array}\right],$
where the three ${±}^{\prime }s$ are all the same. That is,
$X=±M\left(b\right)$
for some b. Note that ${\left(±M\left(b\right)\right)}^{2}=M{\left(b\right)}^{2}$, 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: $b=\frac{a}{2}$. This gives us precisely two solutions:
$X=\left[\begin{array}{cc}1& \frac{a}{2}\\ 0& 1\end{array}\right],-\left[\begin{array}{cc}1& \frac{a}{2}\\ 0& 1\end{array}\right].$

becky4208fj

Expert

Step 1
Let $X=\left(I+N\right)$ 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,
${X}^{2}=I+N+N+{N}^{2}=I+2N$
the answer is trivially obtained from that.

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