## 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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