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Let $T:\mathbb R^7 \longrightarrow \mathbb R^7$ be the linear transformation given by $$ T(x_1, x_2, x_3, x_4 ,x_5, x_6, x_7) = (x_7, x_6, x_5 ,x_4 ,x_3 ,x_2 ,x_1) $$ Which of the following statements are true?

a. Determinant of $T$ is $1$

b. Basis of $\mathbb R^7$ w.r.t. $T$ is a diagonal matrix

c. $T^7 =I$

d. Smallest $n$ s.t. $T^n =I$ is even.

I am stuck on this problem and don't know where to begin. Can anyone help me please?

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It seems like you dont want to do anything and just grasp the answer here. Show some effort, try as much as you can and you will get hints from the community – Applied mathematician Dec 17 '12 at 9:49
At least you should try (c), which is pretty straightforward. Just compute $T(T(T(T(T(T(T(x_1,x_2,x_3,x_4,x_5,x_6,x_7)))))))$ and see if it is identical to $\left(x_1,x_2,x_3,x_4,x_5,x_6,x_7\right)$. – user1551 Dec 17 '12 at 9:53

Think about the problem visually. One of the ways of understanding a structured linear transformation is studying its action carefully on a generic vector. The following comments are aimed at solving this problem with minimum computation.

The linear transformation $T$ flips the vector vertically along the horizontal axis passing through the 4th co-ordinate. Clearly if you flip a vector twice, it becomes that vector again.

So $T^2 = I$ and option D is true. And thus option C must be false (Why?).

For option B, observe that for the matrix to be diagonal w.r.t to the standard basis, the vectors of the standard basis must be eigenvectors. Geometrically speaking, a flip cannot scale the length of a vector. Since, $T^2 = I$, eigenvectors of a flip can either fix a vector or scale the vector by $-1$. Therefore a flip cannot maintain the same direction, unless the vectors are symmetric/anti-symmetric about the horizontal axis passing through 4th co-ordinate. Hence the standard vectors cannot be eigenvectors.

However, the above reasoning tells us that there are four linearly independent eigenvectors associated with eigenvalue $1$: $$(1,0,0,0,0,0,1)^T,(1,1,0,0,0,1,1)^T, (1,1,1,0,1,1,1)^T,(1,1,1,1,1,1,1)^T$$

and 3 linearly independent eigenvectors associated eigenvalue $-1$:

$$(1,0,0,0,0,0,-1)^T,(1,1,0,0,0,-1,-1)^T, (1,1,1,0,-1,-1,-1)^T$$

Therefore $\det(T) = (-1)^3(1)^4 = -1$ and option A is wrong.

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With respect to standard bases, $$T=\begin{pmatrix}0&0&0&0&0&0&1\\0&0&0&0&0&1&0\\0&0&0&0&1&0&0\\0&0&0&1&0&0&0\\0&0&1&0&0&0&0\\0&1&0&0&0&0&0\\1&0&0&0&0&0&0 \end{pmatrix}$$

As mentioned by @Learner, $T$ is a permutation matrix. Also note that $T^2x=T(Tx)=x$. Can you take from here?

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I am confuse determinant of T is 1 or not.......By starting first row first column its 0 and By starting last row last column its 1??? – prakash Dec 17 '12 at 10:28

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