## 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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