# $X^n= \begin{pmatrix}3&6\\ 2&4\end{pmatrix}$, How many solutions are there if n is odd?

$X^n= \begin{pmatrix}3&6\\ 2&4\end{pmatrix}$, $n \in N^*$

How many solutions are there if n is odd?

From the powers of $\begin{pmatrix}3&6\\ 2&4\end{pmatrix}$ I got that $X=\begin{pmatrix}\frac{3}{\sqrt[n]{7^{n-1}}}&\frac{6}{\sqrt[n]{7^{n-1}}}\\ \frac{2}{\sqrt[n]{7^{n-1}}}&\frac{4}{\sqrt[n]{7^{n-1}}}\end{pmatrix}$. But I'm not sure whether this is the only solution... Could I get some hints on how to get this done? Thank you

$\det(X^n)=(\det X)^n=\det \begin{pmatrix}3&6\\ 2&4\end{pmatrix} = 0$, so $\det X=0$, and $X$ is singular.

$X$ cannot be the zero matrix, so it has two real eigenvalues: $0$ and $a\neq 0$. $X$ is diagonalizable: there is some non-singular $P$ such that $X=P\begin{pmatrix}0&0\\ 0&a\end{pmatrix}P^{-1}$, yielding $X^n=P\begin{pmatrix}0&0\\ 0&a^n\end{pmatrix}P^{-1}$.

Then $trace(X^n)=7=a^n$. Since $n$ is odd, we have $a=7^{1/n}$, hence $trace(X)=7^{1/n}$.

By Cayley-Hamilton, $X^2-7^{1/n}X=0$, that is $X^2=7^{1/n}X$. It's easy to prove by induction that for all $m\geq 1$, $X^m=7^{(m-1)/n}X$.

With $m=n$, $\begin{pmatrix}3&6\\ 2&4\end{pmatrix}=X^n=7^{(n-1)/n}X$, so that $$X=\frac{1}{7^{(n-1)/n}}\begin{pmatrix}3&6\\ 2&4\end{pmatrix}$$

The only solution is the one found by the OP.

• Thank you. I really am not familiar with the eigenvalues though, could this be done in another way? Or could you let me know sections should I look up on en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors#Eigenvalues so that I can understand your answer? – Alexander Sep 5 '17 at 15:57
• @Alexander do you know Cayley-Hamilton theorem ? I don't think you can solve your problem without it. Btw, the solution you found is the only one. – Gabriel Romon Sep 5 '17 at 16:46
• I got the second method, thank you. – Alexander Sep 5 '17 at 20:16