$T :\mathbb {R^7}\rightarrow \mathbb {R^7} $ is defined by $T(x_1,x_2,\ldots x_6,x_7) = (x_7,x_6,\ldots x_2,x_1)$ pick out the true statements. Consider the linear  transformations $T :\mathbb {R^7}\rightarrow \mathbb {R^7} $ defined by $T(x_1,x_2,\ldots x_6,x_7) = (x_7,x_6,\ldots x_2,x_1)$. Which of the following statements are true. 
1-  $\det T = 1$
2 - There is a basis of $\mathbb {R^7}$ with respect to which $T$ is a diagonal matrix, 
3- $T^7=I$ 
4- The smallest $n$ such that $T^n = I$ is even.
What i have done so for is I have tried for $T :\mathbb {R^2}\rightarrow \mathbb {R^2} $
 and found that all the statments are true. Can i generalize my conclusion to $\mathbb {R^7} $. Do i need to find $7\times 7$  matrix? Is there any other approach?
 A: It is not hard to see that, in the canonical basis,
$$
T=\begin{bmatrix}0&0&0&0&0&0&1\\ 0&0&0&0&0&1&0 \\0&0&0&0&1&0&0\\0&0&0&1&0&0&0
\\0&0&1&0&0&0&0\\0&1&0&0&0&0&0\\1&0&0&0&0&0&0
\end{bmatrix}
$$
From this it is not hard to see that $\det T=-1$. Also, $T$ is symmetric and real, so it diagonalizable. For the last two questions, it is enough to notice that $T^2=I$ (from the definition), so 3 is false, and $n=2$ in 4. 
A: We can start guessing the eigenvectors: with eigenvalue $1$, we have eigenvectors $e_1 + e_7$, $e_2 + e_6$, $e_3 + e_5$, and $e_4$; with eigenvalue $-1$, we have eigenvectors $e_1 - e_7$, $e_2 - e_6$, $e_3 - e_5$. These seven eigenvectors form a basis of $\mathbb{R}^7$, so with respect to this basis $T$ will be diagonal. Also, since the determinant is the product of the eigenvalues, $\det T = 1^4 \cdot (-1)^3 = -1$. We can easily see that $T$ switches three pairs of coordinates, so in order to come back to $x \in \mathbb{R}^7$ after applying $T$ repeatedly $n$ times on $x$, $n$ has to be even and in particular it cannot be $7$ (or alternatively: if $T^n = I$, $\det T^n = (\det T)^n = (-1)^n = det I = 1$, so $n$ is even).
