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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?

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Can i generalize my conlusion that i have got fot $T :\mathbb {R^2}\rightarrow \mathbb {R^2}$ to $\mathbb {R^7}$? Do i need to find out $7\times 7$ matrix? –  srijan May 11 '12 at 22:02
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Consider $n=3$ case, may be this will help –  Norbert May 11 '12 at 22:07
    
Oh for $n = 3$ conclusion is different . Statemen 4 is true while i am not sure about statement 2 –  srijan May 11 '12 at 22:15
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Just write out the matrix, it has an obvious structure. Then multiply it by itself once. The determinant is trivial to compute. It is symmetric and real. –  copper.hat May 11 '12 at 22:16
    
Alternatively, download octave and just try numerically! –  copper.hat May 11 '12 at 22:17
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3 Answers

up vote 2 down vote accepted

One sees by inspection that $1$ and $-1$ are eigenvalues of $T$ (the vector $(1,0,0,0,0,0, -1)$ is an eigenvector for the eigenvalue $-1$).

Since, $T(e_i)=e_{8-i}$ the matrix representation of $T$ can be easily constructed: $$ A=\begin{bmatrix}0&0&0&0&0&0&1\\ 0&0&0&0&0&1&0 \\0&0&0&0&1&0&0\\0&0&0&1&0&0&0 \\0&0&1&0&0&0&0\\0&1&0&0&0&0&0\\1&0&0&0&0&0&0 \end{bmatrix} $$

For the eigenvalue $1$, we have $$ A-I=\begin{bmatrix}-1&0&0&0&0&0&1\\ 0&-1&0&0&0&1&0 \\0&0&-1&0&1&0&0\\0&0&0&0&0&0&0 \\0&0&1&0&-1&0&0\\0&1&0&0&0&-1&0\\1&0&0&0&0&0&-1 \end{bmatrix} $$ which has echelon form $$ \begin{bmatrix}-1&0&0&0&0&0&1\\ 0&-1&0&0&0&1&0 \\0&0&-1&0&1&0&0\\0&0&0&0&0&0&0 \\0&0&0&0&0&0&0\\0&0&0&0&0&0&0\\0&0&0&0&0&0&0 \end{bmatrix} $$ It can be deduced from the above that the eigenspace for the eigenvalue $1$ has dimension 4.

An echelon form of the matrix $T-(-1)I$ is $$ \begin{bmatrix}1&0&0&0&0&0&1\\ 0&1&0&0&0&1&0 \\0&0&1&0&1&0&0\\0&0&0&2&0&0&0 \\0&0&0&0&0&0&0\\0&0&0&0&0&0&0\\0&0&0&0&0&0&0 \end{bmatrix} $$ from which it follows that the dimension of the eigenspace for the eigenvalue $-1$ is 3.

Thus $T$ has exactly two eigenvalues $1$ and $-1$. Since the dimensions of the eigenspaces sum to 7, it follows that $T$ is diagonalizable. Since the product of the eigenvalues is $1^4\cdot(-1)^3=-1$, it follows that the determinant of $T$ is $-1$ (the determinant can also easily be calculated directly).

It also follows that the characteristic polynomial of $T$ is $(\lambda^2-1)^3(\lambda-1)$; and since $T\ne I$, we must have $T^2=I$ (which can be more readily seen looking at the definition of $T$). So $T^7=T$.

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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.

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Maybe I'm off, but three row interchanges will give the identity matrix; so the determinant is $-1$. –  David Mitra May 11 '12 at 22:41
    
Yes, it is indeed -1. –  copper.hat May 11 '12 at 23:47
    
Absolutely right. I didn't pay attention that the sign changes go in pairs. –  Martin Argerami May 11 '12 at 23:56
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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).

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