Chapter 2.4, Problem 3CFU

To determine

To find: the elements of the reflection matrices where the $1s,-1s\text{ and }0s$ belong.

Answer to Problem 3CFU

  $\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]$ .

Explanation of Solution

Given:

  $1s,-1s\text{ and }0s$ .

Concept used:

The image of linear transformation or matrix is the span of the vector of the linear transformation.

It can be written as $Im\left(A\right)$ .

For example, consider the matrix $A$ which is equal to:

  $\left[\begin{array}{cc}1& 0\\ 0& 2\\ 0& 1\end{array}\right]$

Multiplying this by $2\times 1$ gives a $3\times 1$ matrix. However, regardless of what vector is chosen to multiply by, there are some vectors that can’t be the result.

Result is not image of $A$ . The vector that are possible belong to the span of $A$ . In this case, the span be represented by a parametrized matrix, where $t\text{ and }s$ can be any number:

  $\left[\begin{array}{cc}1& 0\\ 0& 2\\ 0& 1\end{array}\right]\times \left[\begin{array}{c}s\\ t\end{array}\right]=\left[\begin{array}{c}s\\ 2t\\ t\end{array}\right]=Im\left(A\right)$ .

Calculation:

From the geometry that a rotation of a figure ona coordinate plane can be achieved by a combination of reflection.

For example, a ${90}^0$ counter clockwise rotation can be found by first reflecting the image over the $x$ axis and then reflecting the reflected image over the line $y=x$ .

The rotation matrix, $Ro{t}_{90}$ , can be found by a composition of reflections. Since, reflection matrices are applied using multiplication, the composition of reflections. Since, reflection matrices are applied using multiplication. The composition of two reflection matrices is a product.

Remember that $f\left(g\left(x\right)\right)$ means that find $g\left(x\right)$ first and then evaluate the result for $f\left(x\right)$ .

So, the define $Ro{t}_{90}$ ,

  $$

Similarly, a rotation of ${180}^0$ would be rotation of ${90}^0$ twice or $Ro{t}_{{90}^0}$ . $Ro{t}_{{90}^0}$ .

  $\begin{array}{l}Ro{t}_{{180}^0}=Ro{t}_{{90}^0}Ro{t}_{{90}^0}.\\ \Rightarrow \left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right].\\ \Rightarrow \left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right].\end{array}$

Hence, $\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]$ .