## Linear Algebra and Its Applications, Exercise 2.6.17

Exercise 2.6.17. Find a matrix $A$ corresponding to the linear transformation of cyclically permuting vectors in $\mathbb{R}^4$ such that $A$ applied to $\begin{bmatrix} x_1&x_2&x_3&x_4 \end{bmatrix}^T$ produces $\begin{bmatrix} x_2&x_3&x_4&x_1 \end{bmatrix}^T$. Determine the effect of $A^2$ and $A^3$ and explain why $A^3 = A^{-1}$.

Answer: We can construct $A$ by considering its effect on the elementary vectors in $\mathbb{R}^4$ $e_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} \qquad e_2 = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} \qquad e_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix} \qquad e_4 = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}$

Applying the given cyclic permutation is equivalent to shifting each entry of a vector up or to the left (depending on how you look at it), so applying $A$ to each of $e_1$ through $e_4$ will produce the following vectors respectively: $\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} \qquad \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} \qquad \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} \qquad \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}$

In other words we must have $Ae_1 = e_4$, $Ae_2 = e_1$, $Ae_3 = e_2$, and $Ae_4 = e_3$.

As discussed in the previous exercise, we can do this by having each column of $A$ be the vector into which the corresponding elementary vector should be transformed. In other words, the first column of $A$ should be $e_4$ (since $A$ transforms $e_1$ into $e_4$), the second column should be $e_1$ (since $A$ transforms $e_2$ into $e_1$), and similarly the third and fourth columns should be $e_2$ and $e_3$ respectively.

We thus have $A = \begin{bmatrix} 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \\ 1&0&0&0 \end{bmatrix}$

Consider $A^2$. We have $A^2e_1 = A(Ae_1) = Ae_4 = e_3$ $A^2e_2 = A(Ae_2) = Ae_1 = e_4$ $A^2e_3 = A(Ae_3) = Ae_2 = e_1$ $A^2e_4 = A(Ae_4) = Ae_3 = e_2$

Constructing the columns of $A^2$ as we did for $A$ we see that $A^2 = \begin{bmatrix} 0&0&1&0 \\ 0&0&0&1 \\ 1&0&0&0 \\ 0&1&0&0 \end{bmatrix}$

and that $A^2$ has the effect of shifting the entries of a vector up (or to the left) two places: $A^2x = \begin{bmatrix} 0&0&1&0 \\ 0&0&0&1 \\ 1&0&0&0 \\ 0&1&0&0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} x_3 \\ x_4 \\ x_1 \\ x_2 \end{bmatrix}$

For $A^3$ we have $A^3e_1 = A^2(Ae_1) = A^2e_4 = e_2$ $A^3e_2 = A^2(Ae_2) = A^2e_1 = e_3$ $A^3e_3 = A^2(Ae_3) = A^2e_2 = e_4$ $A^3e_4 = A^2(Ae_4) = A^2e_3 = e_1$

Constructing the columns of $A^3$ as we did for $A$ and $A^2$ we see that $A^3 = \begin{bmatrix} 0&0&0&1 \\ 1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \end{bmatrix}$

and that $A^3$ has the effect of shifting the entries of a vector up (or to the left) three places: $A^3 = \begin{bmatrix} 0&0&0&1 \\ 1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} x_4 \\ x_1 \\ x_2 \\ x_3 \end{bmatrix}$

Finally for $A^4$ we have $A^4e_1 = A^3(Ae_1) = A^3e_4 = e_1$ $A^4e_2 = A^3(Ae_2) = A^3e_1 = e_2$ $A^4e_3 = A^3(Ae_3) = A^3e_2 = e_3$ $A^4e_4 = A^3(Ae_4) = A^3e_3 = e_4$

so that $A^4 = \begin{bmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix} = I$

In other words, $A^4$ has the effect of shifting the entries of vectors in $\mathbb{R}^4$ up (or to the left) four places, restoring the original vectors.

We then have $A^3A = A^4 = I$ and $AA^3 = A^4 = I$. Since $A^3$ is both a left and right inverse of $A$ we have $A^3 = A^{-1}$.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition , Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books .

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