## Linear Algebra and Its Applications, Exercise 2.6.18

Exercise 2.6.18. Given a vector $\begin{bmatrix} x_1&x_2&x_3 \end{bmatrix}^T$ in $\mathbb{R}^3$ find a matrix $A$ that produces a corresponding vector $\begin{bmatrix} 0&x_1&x_2&x_3 \end{bmatrix}^T$ in $\mathbb{R}^4$ in which all entries are shifted right one place. Find a second matrix $B$ that takes a  vector $\begin{bmatrix} x_1&x_2&x_3&x_4 \end{bmatrix}^T$ in $\mathbb{R}^4$ and produces the vector $\begin{bmatrix} x_2&x_3&x_4 \end{bmatrix}^T$ in $\mathbb{R}^3$ in which all entries are shifted left one place. What are the product matrices $AB$ and $BA$ and what effects do they have?

Answer: We can construct $A$ by considering its effect on the elementary vectors in $\mathbb{R}^3$ $e_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \qquad e_2 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \qquad e_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$

Applying $A$ to each of $e_1$ through $e_3$ respectively will shift each of their entries to the right (or down, depending on your point of view) and produce the following vectors in $\mathbb{R}^4$: $\begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} \qquad \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix} \qquad \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}$

As discussed in the answer to exercise 2.6.16, we can do this by having each column of $A$ be the vector into which the corresponding elementary vector should be transformed. We thus have $A = \begin{bmatrix} 0&0&0 \\ 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix}$

so that $Ax = \begin{bmatrix} 0&0&0 \\ 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ x_1 \\ x_2 \\ x_3 \end{bmatrix}$

Now consider $B$. We can construct $B$ 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 $B$ to each of $e_1$ through $e_4$ respectively will shift each of their entries to the left (or up) and produce the following vectors in $\mathbb{R}^3$: $\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \qquad \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \qquad \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \qquad \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$

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

so that $Bx = \begin{bmatrix} 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} x_2 \\ x_3 \\ x_4 \end{bmatrix}$

Constructing the product matrices we have $AB = \begin{bmatrix} 0&0&0 \\ 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix} \begin{bmatrix} 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix} = \begin{bmatrix} 0&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix}$

and $BA = \begin{bmatrix} 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} 0&0&0 \\ 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix} = \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix}= I$

The product $matrix BA$ corresponds to taking a vector $\begin{bmatrix} x_1&x_2&x_3 \end{bmatrix}^T$ in $\mathbb{R}^3$ and shifting the entries right to produce a corresponding vector $\begin{bmatrix} 0&x_1&x_2&x_3 \end{bmatrix}^T$ in $\mathbb{R}^4$ and then shifting the entries left to produce the original vector $\begin{bmatrix} x_1&x_2&x_3 \end{bmatrix}^T$ in $\mathbb{R}^3$. So $BA$ preserves all vectors and is thus equal to the identity matrix on $\mathbb{R}^3$.

On the other hand the product $matrix AB$ corresponds to taking a vector $\begin{bmatrix} x_1&x_2&x_3&x_4 \end{bmatrix}^T$ in $\mathbb{R}^4$ and shifting the entries left to produce a corresponding vector $\begin{bmatrix} x_2&x_3&x_4 \end{bmatrix}^T$ in $\mathbb{R}^3$ and then shifting the entries right to produce the vector $\begin{bmatrix} 0&x_2&x_3&x_4 \end{bmatrix}^T$ in $\mathbb{R}^4$. So the net effect of $AB$ is to change the first entry of a vector in $\mathbb{R}^4$ to zero and preserve the second, third, and fourth entries.

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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