## Linear Algebra and Its Applications, Review Exercise 2.29

Review exercise 2.29. The following matrices

$A_1 = \begin{bmatrix} 1&0 \\ 0&-1 \end{bmatrix} \qquad A_2 = \begin{bmatrix} 1&0 \\ 2&1 \end{bmatrix} \qquad A_3 = \begin{bmatrix} 0&1 \\ -1&0 \end{bmatrix}$

represent linear transformations in the $x$$y$ plane with $e_1 = (1, 0)$ and $e_2 = (0, 1)$ as a basis. Describe the effect of each transformation.

Answer: When the matrix $A_1$ is applied to the vector $e_1 = (1, 0)$ we obtain

$A_1v_1 = \begin{bmatrix} 1&0 \\ 0&-1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$

When the matrix $A_1$ is applied to the vector $e_2 = (0, 1)$ we obtain

$A_1v_1 = \begin{bmatrix} 1&0 \\ 0&-1 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ -1 \end{bmatrix}$

The effect of the transformation represented by $A_1$ is to reflect all vectors through the $x$-axis.

When the matrix $A_2$ is applied to the vector $e_1 = (1, 0)$ we obtain

$A_2e_1 = \begin{bmatrix} 1&0 \\ 2&1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$

When the matrix $A_2$ is applied to the vector $e_2 = (0, 1)$ we obtain

$A_2e_2 = \begin{bmatrix} 1&0 \\ 2&1 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$

When the matrix $A_2$ is applied to the vector $v_1 = (1, 2)$ we obtain

$A_2v_1 = \begin{bmatrix} 1&0 \\ 2&1 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 1 \\ 4 \end{bmatrix}$

The transformation represented by $A_2$ is a shearing transformation that leaves all vectors on the $y$-axis unchanged but changes the $y$ coordinates of all other vectors in proportion to their distance from the $y$-axis.

When the matrix $A_3$ is applied to the vector $e_1 = (1, 0)$ we obtain

$A_3e_1 = \begin{bmatrix} 0&1 \\ -1&0 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ -1 \end{bmatrix}$

When the matrix $A_2$ is applied to the vector $e_2 = (0, 1)$ we obtain

$A_2e_2 = \begin{bmatrix} 0&1 \\ -1&0 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$

The effect of the transformation represented by $A_3$ is to rotate all vectors through an angle of -90 degrees. (Or, in other words, clockwise through 90 degrees.)

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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