Solve the matrix equation for $X$

Question

Solve $X^{7}=\begin{pmatrix} 1 & 0 & 1 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}$.
We dont know the field where the entries of $X$ are. The matrix $A=X^{7}$ is not diagonalizable. I tried taking the Jordan Decompotion. Is there a way to find all solutions for $X$ taking every case for the field?

Answer 1

Suppose that $\alpha$ is a solution of $x^7=-1$ and $\beta$ is a solution of $x^7=1$ in a field $F$. If $F$ has characteristic not divisible by $7$, then the matrix $$M= \begin{pmatrix} \beta & 0 &\frac{\beta}{7} \\ 0 & \alpha & 0 \\ 0 & 0 & \beta \end{pmatrix} $$ is a solution to $X^7=A$ where $$ A= \begin{pmatrix} 1 & 0 & 1 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix} $$ in $M_3(F)$.

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It is easier (I think) to apply a change of basis

$$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}$$

Then $A$ takes the form

$$ \left(\begin{array}{cc|c} 1 & 1 & 0\\ 0 & 1 & 0\\ \hline 0 & 0 & -1 \end{array}\right) $$

which is block diagonal. My first guess was to try block diagonal solutions to match this, and you can show that solutions have to in fact be block diagonal.

The $2\times 2$ block is also easily seen to be upper-triangular, and it has to have a repeated eigenvalue that is a $7$th root of unity.

Answer 2

Let $F$ be the ambient field.

As we work in the subalgebra of upper triangular matrices, it is not difficult to see that all solutions $M$ have to be upper triangular matrices, moreover with $M_{1,1}=M_{3,3}=1$. In this way, the matrix equation becomes, in a more general form :

$$M^n=\pmatrix{1&a&b\\0&c&d\\0&0&1}^n=\pmatrix{1&0&1\\0&-1&0\\0&0&1}\tag{1}$$

(we will set $n=7$ later on).

Let us consider the two cases :

• either $c=1$ :

$$\pmatrix{1&a&b\\0&1&d\\0&0&1}^n=\pmatrix{1&0&1\\0&-1&0\\0&0&1}\tag{2}$$

which becomes, by an immediate induction :

$$\pmatrix{1&na&\tfrac12n(n-1)ad+nb\\0&1&nd\\0&0&1}=\pmatrix{1&0&1\\0&-1&0\\0&0&1}\tag{3}$$

With $n=7$, we get :

$$\begin{cases}1&=&-1\\7a&=&0\\7(3ad+b)&=&-1\\7d&=&0\end{cases}\tag{4}$$

The first equation means that we must be in a field with characteristic 2.In this case, $a=d=0$ and $b=-1/7$ is the unique solution to equation (2).

• or $c \ne 1$

In this case, (1) can be written :

$$\pmatrix{1&A&B\\0&C&D\\0&0&1}^n=\pmatrix{1&0&1\\0&-1&0\\0&0&1}\tag{5}$$

with $$\begin{cases}A&=&a(1-c^n)/(1-c)\\B&=&ad(c^n-nc+n-1)/(1-c)^2+nb\\C&=&c^n\\D&=&d(1-c^n)/(1-c)\end{cases}.\tag{6}$$

Now, it remains to solve in all generality (with $n=7$) the equations

$$\begin{cases}A&=&0\\B&=&1\\C&=&-1\\D&=&0\end{cases}\tag{7}$$

Edit 1 :

The third equation expresses that $c$ is a 7th root of $-1$ (for example $e^{i \pi k/7}$ if the ambient field is $F=\mathbb{C}$ for some $k$). Two cases now :

• if $c^n=7$, which means that $c$ is also a $7$th root of $+1$, we have $c^7=-1=+1$ which is (again !) only possible if the characteristic of the field is $2$. As a consequence, as $7=1$ in such a field, the second equation (the last one to be verified) is reduced to :

$$\frac{ad}{1-c}+b=1\tag{8}$$

(8) has solutions $a,b,c,d$ for example in a finite field $F_{2^q}$ (see Edit 2).

• If instead $c^n \ne 1$, which means that it isn't a 7th root of $-1$, we must have $a=d=0$ and the second equation is reduced to $7b=-1$ which means that $b=-1/7$ (unless the characteristic of the ambient field $F$ is $7$ in which case there is no solution :

$$\pmatrix{1&0&-1/7\\0&c&0\\0&0&1}$$

as in the other answer with $c$ a $7$th root of $-1$, if it exists in ambient field $F$.

Edit 2 :

Here is an example in the case of a field with characteristic $2$ :

$$F=GF(2^3)=\{0,x,x^2,x+1,x^2+x,x^2+x+1,x^2+1,1\}$$

$$M^7=\pmatrix{1&x&0\\0&x+1&1\\0&0&1}^7=\pmatrix{1&0&1\\0&\color{red}{1}&0\\0&0&1}$$

Remark : please note that in characteristic 2, $\color{red}{1=-1}$, and that all elements, except $0$ have their $n$th power equal to $1$ whatever $n$).

One can check that (8) is fulfilled.

You can check this solution using the following SAGE program :

F.<x>=GF(2^3)
L=list(F);show(L)
M=matrix([[1,x,0],[0,x+1,1],[0,0,1]]);show(M);
N=M^7;show(N)