gea3stwg

Answered

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.

where

Answer & Explanation

Kyler Jacobson

Expert

2022-01-31Added 8 answers

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${X}^{2}=M\left(a\right)$ , 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$2\times 2$ 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${\pm}^{\prime}s$ are all the same. That is,

$X=\pm M\left(b\right)$

for some b. Note that$(\pm M\left(b\right))}^{2}=M{\left(b\right)}^{2$ , 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:$b=\frac{a}{2}$ . This gives us precisely two solutions:

$X=\left[\begin{array}{cc}1& \frac{a}{2}\\ 0& 1\end{array}\right],-\left[\begin{array}{cc}1& \frac{a}{2}\\ 0& 1\end{array}\right].$

Let

Note M(a) has one eigenvalue: 1, and is not diagonalisable. This means that, if

We also note that any eigenvectors of X will be an eigenvector for

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

on the right reveals the first column of a

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:

where the three

for some b. Note that

which was solved by Angel in their answer:

becky4208fj

Expert

2022-02-01Added 10 answers

Step 1

Let$X=(I+N)$ 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,

${X}^{2}=I+N+N+{N}^{2}=I+2N$

the answer is trivially obtained from that.

Let

the answer is trivially obtained from that.

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