There exists a $2 \times 2$ real matrix $A$ such that $A^5 = I$.

a) $A$ must be identity.

b) $A$ must be similar to an element of $SO(2)$ .

c) $A$ must be diagonalisable.

I have checked that minimal polynomial must divide $(x^5-1)=(x-1)(x^4+\dots+1)$ so it must be $x-1$ so $A=I$, and of course diagonalizable so a and c are correct options, but I am not sure about b.

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    $\begingroup$ If $\zeta = a + bi$ is a fifth root of unity in $\mathbb C$, try $A = \left(\begin{matrix}a & b \\ -b & a \end{matrix}\right)$. $\endgroup$ – Charles Staats Jun 22 '13 at 5:13
  • $\begingroup$ @Bicycle Theif: Please see here for how to typeset common math expressions with MathJax, and see here for how to use Markdown formatting. $\endgroup$ – Zev Chonoles Jun 22 '13 at 5:15
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    $\begingroup$ @Bicycle Theif: Note that although $A=I$ is one solution, you did not show that it is the only solution (in fact there are others, as Charles points out). In order for statement a to be true, the identity matrix would have to be the only $2\times 2$ real matrix satisfying $A^5=I$. $\endgroup$ – Zev Chonoles Jun 22 '13 at 5:16
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    $\begingroup$ @Bicycle Theif: Lastly, please make the titles of your posts actual questions, not just the first introductory line of your post, so that people can understand what you are asking from looking at the front page. $\endgroup$ – Zev Chonoles Jun 22 '13 at 5:19
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    $\begingroup$ @BicycleTheif: Just because no real numbers satisfy the polynomial $x^4 + \dotsb + 1$, it does not follow that no real matrices satisfy this polynomial. $\endgroup$ – Charles Staats Jun 22 '13 at 5:54

Consider a matrix that rotates by 72 degrees. I.e. the matrix A so that Ax has rotated the vector x by 72 degrees about the origin. We then $A^5 = I.$

a) and c) are false for this matrix.

b) is true however. So the question is then is b) always true?


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